Course 421: Academic Year 1998-9 Part I: Topological Spaces

D. R. Wilkins

Contents

1 Topological Spaces 2 1.1 The Concept of Continuity . . . . . . . . . . . . . . . . . . . . 2 1.2 Topological Spaces . . . . . . . . . . . . . . . . . . . . . . . . 2 1.3 Subsets of Euclidean Space . . . . . . . . . . . . . . . . . . . . 3 1.4 Open Sets in Metric Spaces . . . . . . . . . . . . . . . . . . . 4 1.5 Further Examples of Topological Spaces . . . . . . . . . . . . 6 1.6 Closed Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.7 Hausdorff Spaces . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.8 Subspace Topologies . . . . . . . . . . . . . . . . . . . . . . . 7 1.9 Continuous Functions between Topological Spaces . . . . . . . 9 1.10 Continuous Functions between Metric Spaces . . . . . . . . . . 9 1.11 A Criterion for Continuity . . . . . . . . . . . . . . . . . . . . 10 1.12 Homeomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . 12 1.13 Sequences and Convergence . . . . . . . . . . . . . . . . . . . 12 1.14 Neighbourhoods, Closures and Interiors . . . . . . . . . . . . . 14

2 Product Topologies 16 2.1 The Cartesian Product of Subsets of Euclidean Space . . . . . 16 2.2 Product Topologies . . . . . . . . . . . . . . . . . . . . . . . . 18

3 Identification Maps and Quotient Topologies 24 3.1 Cut and Paste Constructions . . . . . . . . . . . . . . . . . . . 24 3.2 Identification Maps and Quotient Topologies . . . . . . . . . . 26

4 Compactness 29 4.1 Compact Topological Spaces . . . . . . . . . . . . . . . . . . . 29 4.2 The Lebesgue Lemma and Uniform Continuity . . . . . . . . . 35

5 Connectedness 37 5.1 Connected Topological Spaces . . . . . . . . . . . . . . . . . . 37

1 Topological Spaces

1.1 The Concept of Continuity

The concept of continuity plays an important role in mathematics. There is a precise definition of continuity for functions of a real variable. A function f: D → ℝ, defined on a subset D of the real line ℝ, is said to be continuous at a point p of D if, given any real number ε satisfying ε > 0, there exists a real number δ satisfying δ > 0 such that |f(x) − f(p)| < ε for all points x of D satisfying |x − p| < δ. This definition of continuity can easily be adapted so as to apply to functions of a complex variable. It can also be generalized to as to apply to functions of several real or complex variables. We thus obtain a definition of continuity for functions between subsets of Euclidean spaces.

This definition of continuity generalizes directly to functions between metric spaces. A metric space is a set provided with a distance function, measuring the distance between any two points of the set. This distance function is required to satisfy certain axioms: the distance between any two points of a metric space is always non-negative, and is zero if and only if those points coincide; the distance from a point x to a point y is the same as the distance from y to x; given any three points x, y and z of a metric space, the distance from x to z is required to be less than or equal to the sum of the distance from x to y and the distance from y to z. A function from a metric space X to a metric space Y is continuous at a point p of X if and only if, given any real number ε satisfying ε > 0, there exists a real number δ satisfying δ > 0 such that the distance from f(x) to f(p) is less than ε for all points x of X whose distance from p is less than δ.

We shall introduce the concept of a topological space, and give a definition of continuity for functions from one topological space to another which generalizes the definitions of continuity discussed above for functions of a real variable, for functions of a complex variable, for functions between subsets of Euclidean spaces, and for functions from one metric space to another.

The theory of topological spaces has proved itself to be very useful in many areas of mathematics.

1.2 Topological Spaces

Definition A topological space X consists of a set X together with a collection of subsets, referred to as open sets, such that the following conditions are satisfied:—

(i) the empty set ∅ and the whole set X are open sets, (ii) the union of any collection of open sets is itself an open set,

(iii) the intersection of any finite collection of open sets is itself an open set.

The collection consisting of all the open sets in a topological space X is referred to as a topology on the set X.

Remark If it is necessary to specify explicitly the topology on a topological space then one denotes by (X, τ) the topological space whose underlying set is X and whose topology is τ. However if no confusion will arise then it is customary to denote this topological space simply by X.

1.3 Subsets of Euclidean Space

Let X be a subset of n-dimensional Euclidean space ℝ^{n}. The Euclidean distance |x − y| between two points x and y of X is defined as follows:

|x − y| = \sqrt{∑_{i=1}^{n}(x_{i} − y_{i})^{2}},

where x = (x_{1}, x_{2}, …, x_{n}) and y = (y_{1}, y_{2}, …, y_{n}). The Euclidean distances between any three points x, y and z of X satisfy the Triangle Inequality:

|x − z| ≤ |x − y| + |y − z|.

A subset V of X is said to be open in X if, given any point v of V, there exists some δ > 0 such that

{x ∈ X : |x − v| < δ} ⊂ V.

The empty set is also considered to be open in X.

Both ∅ and X are open sets in X. Also it is not difficult to show that any union of open sets in X is open in X, and that any finite intersection of open sets in X is open in X. (This will be proved in more generality for open sets in metric spaces.) Thus the collection of open sets in a subset X of a Euclidean space ℝ^{n} satisfies the topological space axioms. Thus every subset X of ℝ^{n} is a topological space with these open sets. This topology on a subset X of ℝ^{n} is referred to as the usual topology on X, generated by the Euclidean distance function.

In particular ℝ^{n} is itself a topological space.

1.4 Open Sets in Metric Spaces

Definition A metric space (X, d) consists of a set X together with a distance function d: X × X → [0, +∞) on X satisfying the following axioms:

(i) d(x, y) ≥ 0 for all x, y ∈ X,

(ii) d(x, y) = d(y, x) for all x, y ∈ X,

(iii) d(x, z) ≤ d(x, y) + d(y, z) for all x, y, z ∈ X,

(iv) d(x, y) = 0 if and only if x = y.

The quantity d(x, y) should be thought of as measuring the distance between the points x and y. The inequality d(x, z) ≤ d(x, y)+d(y, z) is referred to as the Triangle Inequality. The elements of a metric space are usually referred to as points of that metric space.

An n-dimensional Euclidean space ℝ^{n} is a metric space with with respect to the Euclidean distance function d, defined by

d(x, y) = |x − y| = \sqrt{∑_{i=1}^{n}(x_{i} − y_{i})^{2}}

for all x, y ∈ ℝ^{n}. Any subset X of ℝ^{n} may be regarded as a metric space whose distance function is the restriction to X of the Euclidean distance function on ℝ^{n} defined above.

Definition Let (X, d) be a metric space. Given a point x of X and r ≥ 0, the open ball B_{X}(x, r) of radius r about x in X is defined by

B_{X}(x, r) ≡ {x′ ∈ X : d(x′, x) < r}.

Definition Let (X, d) be a metric space. A subset V of X is said to be an open set if and only if the following condition is satisfied:

• given any point v of V there exists some δ > 0 such that B_{X}(v, δ) ⊂ V.

By convention, we regard the empty set ∅ as being an open subset of X. (The criterion given above is satisfied vacuously in this case.)

Lemma 1.1 Let X be a metric space with distance function d, and let x_{0} be a point of X. Then, for any r > 0, the open ball B_{X}(x_{0}, r) of radius r about x_{0} is an open set in X.

Proof Let x ∈ B_{X}(x_{0}, r). We must show that there exists some δ > 0 such that B_{X}(x, δ) ⊂ B_{X}(x_{0}, r). Now d(x, x_{0}) < r, and hence δ > 0, where δ = r − d(x, x_{0}). Moreover if x′ ∈ B_{X}(x, δ) then

d(x′, x_{0}) ≤ d(x′, x) + d(x, x_{0}) < δ + d(x, x_{0}) = r,

by the Triangle Inequality, hence x′ ∈ B_{X}(x_{0}, r). Thus B_{X}(x, δ) ⊂ B_{X}(x_{0}, r), showing that B_{X}(x_{0}, r) is an open set, as required.

Proposition 1.2 Let X be a metric space. The collection of open sets in X has the following properties:—

(i) the empty set ∅ and the whole set X are both open sets;

(ii) the union of any collection of open sets is itself an open set;

(iii) the intersection of any finite collection of open sets is itself an open set.

Proof The empty set ∅ is an open set by convention. Moreover the definition of an open set is satisfied trivially by the whole set X. Thus (i) is satisfied.

Let A be any collection of open sets in X, and let U denote the union of all the open sets belonging to A. We must show that U is itself an open set. Let x ∈ U. Then x ∈ V for some open set V belonging to the collection A. Therefore there exists some δ > 0 such that B_{X}(x, δ) ⊂ V. But V ⊂ U, and thus B_{X}(x, δ) ⊂ U. This shows that U is open. Thus (ii) is satisfied.

Finally let V_{1}, V_{2}, V_{3}, …, V_{k} be a finite collection of open sets in X, and let V = V_{1} ∩ V_{2} ∩ ⋯ ∩ V_{k}. Let x ∈ V. Now x ∈ V_{j} for all j, and therefore there exist strictly positive real numbers δ_{1}, δ_{2}, …, δ_{k} such that B_{X}(x, δ_{j}) ⊂ V_{j} for j = 1, 2, …, k. Let δ be the minimum of δ_{1}, δ_{2}, …, δ_{k}. Then δ > 0. (This is where we need the fact that we are dealing with a finite collection of open sets.) Moreover B_{X}(x, δ) ⊂ B_{X}(x, δ_{j}) ⊂ V_{j} for j = 1, 2, …, k, and thus B_{X}(x, δ) ⊂ V. This shows that the intersection V of the open sets V_{1}, V_{2}, …, V_{k} is itself open. Thus (iii) is satisfied.

Any metric space may be regarded as a topological space. Indeed let X be a metric space with distance function d. We recall that a subset V of X is an open set if and only if, given any point v of V, there exists some δ > 0 such that {x ∈ X : d(x, v) < δ} ⊂ V. Proposition 1.2 shows that the topological space axioms are satisfied by the collection of open sets in any metric space. We refer to this collection of open sets as the topology generated by the distance function d on X.

1.5 Further Examples of Topological Spaces

Example Given any set X, one can define a topology on X where every subset of X is an open set. This topology is referred to as the discrete topology on X.

Example Given any set X, one can define a topology on X in which the only open sets are the empty set ∅ and the whole set X.

1.6 Closed Sets

Definition Let X be a topological space. A subset F of X is said to be a closed set if and only if its complement X \ F is an open set.

We recall that the complement of the union of some collection of subsets of some set X is the intersection of the complements of those sets, and the complement of the intersection of some collection of subsets of X is the union of the complements of those sets. The following result therefore follows directly from the definition of a topological space.

Proposition 1.3 Let X be a topological space. Then the collection of closed sets of X has the following properties:—

(i) the empty set ∅ and the whole set X are closed sets,

(ii) the intersection of any collection of closed sets is itself a closed set,

(iii) the union of any finite collection of closed sets is itself a closed set.

1.7 Hausdorff Spaces

Definition A topological space X is said to be a Hausdorff space if and only if it satisfies the following Hausdorff Axiom:

• if x and y are distinct points of X then there exist open sets U and V such that x ∈ U, y ∈ V and U ∩ V = ∅.

Lemma 1.4 All metric spaces are Hausdorff spaces.

Proof Let X be a metric space with distance function d, and let x and y be points of X, where x ≠ y. Let ε = \frac{1}{2}d(x, y). Then the open balls B_{X}(x, ε) and B_{X}(y, ε) of radius ε centred on the points x and y are open sets (see Lemma 1.1). If B_{X}(x, ε) ∩ B_{X}(y, ε) were non-empty then there would exist z ∈ X satisfying d(x, z) < ε and d(z, y) < ε. But this is impossible, since it would then follow from the Triangle Inequality that d(x, y) < 2ε, contrary to the choice of ε. Thus x ∈ B_{X}(x, ε), y ∈ B_{X}(y, ε), B_{X}(x, ε) ∩ B_{X}(y, ε) = ∅. This shows that the metric space X is a Hausdorff space.

We now give an example of a topological space which is not a Hausdorff space.

Example The Zariski topology on the set ℝ of real numbers is defined as follows: a subset U of ℝ is open (with respect to the Zariski topology) if and only if either U = ∅ or else ℝ \ U is finite. It is a straightforward exercise to verify that the topological space axioms are satisfied, so that the set ℝ of real numbers is a topological space with respect to this Zariski topology. Now the intersection of any two non-empty open sets in this topology is always non-empty. (Indeed if U and V are non-empty open sets then U = ℝ \ F_{1} and V = ℝ \ F_{2}, where F_{1} and F_{2} are finite sets of real numbers. But then U ∩ V = ℝ \ (F_{1} ∪ F_{2}), which is non-empty, since F_{1} ∪ F_{2} is finite and ℝ is infinite.) It follows immediately from this that ℝ, with the Zariski topology, is not a Hausdorff space.

1.8 Subspace Topologies

Let X be a topological space with topology τ, and let A be a subset of X. Let τ_{A} be the collection of all subsets of A that are of the form V ∩ A for V ∈ τ. Then τ_{A} is a topology on the set A. (It is a straightforward exercise to verify that the topological space axioms are satisfied.) The topology τ_{A} on A is referred to as the subspace topology on A.

Any subset of a Hausdorff space is itself a Hausdorff space (with respect to the subspace topology).

Lemma 1.5 Let X be a metric space with distance function d, and let A be a subset of X. A subset W of A is open with respect to the subspace topology on A if and only if, given any point w of W, there exists some δ > 0 such that

{a ∈ A : d(a, w) < δ} ⊂ W.

Thus the subspace topology on A coincides with the topology on A obtained on regarding A as a metric space (with respect to the distance function d).

Proof Suppose that W is open with respect to the subspace topology on A. Then there exists some open set U in X such that W = U ∩ A. Let w be a point of W. Then there exists some δ > 0 such that

{x ∈ X : d(x, w) < δ} ⊂ U.

But then

{a ∈ A : d(a, w) < δ} ⊂ U ∩ A = W.

Conversely, suppose that W is a subset of A with the property that, for any w ∈ W, there exists some δ_{w} > 0 such that

{a ∈ A : d(a, w) < δ_{w}} ⊂ W.

Define U to be the union of the open balls B_{X}(w, δ_{w}) as w ranges over all points of W, where

B_{X}(w, δ_{w}) = {x ∈ X : d(x, w) < δ_{w}}.

The set U is an open set in X, since each open ball B_{X}(w, δ_{w}) is an open set in X (Lemma 1.1), and any union of open sets is itself an open set. Moreover

B_{X}(w, δ_{w}) ∩ A = {a ∈ A : d(a, w) < δ_{w}} ⊂ W

for any w ∈ W. Therefore U ∩ A ⊂ W. However W ⊂ U ∩ A, since, W ⊂ A and {w} ⊂ B_{X}(w, δ_{w}) ⊂ U for any w ∈ W. Thus W = U ∩ A, where U is an open set in X. We deduce that W is open with respect to the subspace topology on A.

Example Let X be any subset of n-dimensional Euclidean space ℝ^{n}. Then the subspace topology on X coincides with the topology on X generated by the Euclidean distance function on X. We refer to this topology as the usual topology on X.

Let X be a topological space, and let A be a subset of X. One can readily verify the following:—

• a subset B of A is closed in A (relative to the subspace topology on A) if and only if B = A ∩ F for some closed subset F of X;

• if A is itself open in X then a subset B of A is open in A if and only if it is open in X;

• if A is itself closed in X then a subset B of A is closed in A if and only if it is closed in X.

1.9 Continuous Functions between Topological Spaces

Definition A function f: X → Y from a topological space X to a topological space Y is said to be continuous if f^{−1}(V) is an open set in X for every open set V in Y, where

f^{−1}(V) ≡ {x ∈ X : f(x) ∈ V }.

A continuous function from X to Y is often referred to as a map from X to Y.

Lemma 1.6 Let X, Y and Z be topological spaces, and let f: X → Y and g: Y → Z be continuous functions. Then the composition g ∘ f: X → Z of the functions f and g is continuous.

Proof Let V be an open set in Z. Then g^{−1}(V) is open in Y (since g is continuous), and hence f^{−1}(g^{−1}(V)) is open in X (since f is continuous). But f^{−1}(g^{−1}(V)) = (g ∘ f)^{−1}(V). Thus the composition function g ∘ f is continuous.

Lemma 1.7 Let X and Y be topological spaces, and let f: X → Y be a function from X to Y. The function f is continuous if and only if f^{−1}(G) is closed in X for every closed subset G of Y.

Proof If G is any subset of Y then X \ f^{−1}(G) = f^{−1}(Y \ G) (i.e., the complement of the preimage of G is the preimage of the complement of G). The result therefore follows immediately from the definitions of continuity and closed sets.

1.10 Continuous Functions between Metric Spaces

The following definition of continuity for functions between metric spaces generalizes that for functions of a real or complex variable.

Definition Let X and Y be metric spaces with distance functions d_{X} and d_{Y} respectively. A function f: X → Y from X to Y is said to be continuous at a point x of X if and only if the following criterion is satisfied:—

• given any real number ε satisfying ε > 0 there exists some δ > 0 such that d_{Y} (f(x), f(x′)) < ε for all points x′ of X satisfying d_{X}(x, x′) < δ.

The function f: X → Y is said to be continuous on X if and only if it is continuous at x for every point x of X.

This definition can be rephrased in terms of open balls: a function f: X → Y from a metric space X to a metric space Y is continuous at a point x of X if and only if, given any ε > 0, there exists some δ > 0 such that f maps B_{X}(x, δ) into B_{Y} (f(x), ε) (where B_{X}(x, δ) and B_{Y} (f(x), ε) denote the open balls of radius δ and ε about x and f(x) respectively).

Let f: X → Y be a function from a set X to a set Y. Given any subset V of Y, we denote by f^{−1}(V) the preimage of V under the map f, defined by

f^{−1}(V) = {x ∈ X : f(x) ∈ V }.

The following result shows that the definition of continuity given above for functions between metric spaces is consistent with the more general definition of continuity for functions between topological spaces.

Proposition 1.8 Let X and Y be metric spaces, and let f: X → Y be a function from X to Y. The function f is continuous if and only if f^{−1}(V) is an open set in X for every open set V of Y.

Proof Suppose that f: X → Y is continuous. Let V be an open set in Y. We must show that f^{−1}(V) is open in X. Let x be a point belonging to f^{−1}(V). We must show that there exists some δ > 0 with the property that B_{X}(x, δ) ⊂ f^{−1}(V). Now f(x) belongs to V. But V is open, hence there exists some ε > 0 with the property that B_{Y} (f(x), ε) ⊂ V. But f is continuous at x. Therefore there exists some δ > 0 such that f maps the open ball B_{X}(x, δ) into B_{Y} (f(x), ε) (see the remarks above). Thus f(x′) ∈ V for all x′ ∈ B_{X}(x, δ), showing that B_{X}(x, δ) ⊂ f^{−1}(V). We have thus shown that if f: X → Y is continuous then f^{−1}(V) is open in X for every open set V in Y.

Conversely suppose that f: X → Y has the property that f^{−1}(V) is open in X for every open set V in Y. Let x be any point of X. We must show that f is continuous at x. Let ε > 0 be given. The open ball B_{Y} (f(x), ε) is an open set in Y, by Lemma 1.1, hence f^{−1} (B_{Y} (f(x), ε)) is an open set in X which contains x. It follows that there exists some δ > 0 such that B_{X}(x, δ) ⊂ f^{−1} (B_{Y} (f(x), ε)). We have thus shown that, given any ε > 0, there exists some δ > 0 such that f maps the open ball B_{X}(x, δ) into B_{Y} (f(x), ε). We conclude that f is continuous at x, as required.

1.11 A Criterion for Continuity

We now show that, if a topological space X is the union of a finite collection of closed sets, and if a function from X to some topological space is continuous on each of these closed sets, then that function is continuous on X.

Lemma 1.9 Let X and Y be topological spaces, let f: X → Y be a function from X to Y, and let X = A_{1} ∪A_{2} ∪⋯∪A_{k}, where A_{1}, A_{2}, …, A_{k} are closed sets in X. Suppose that the restriction of f to the closed set A_{i} is continuous for i = 1, 2, …, k. Then f: X → Y is continuous.

Proof Let V be an open set in Y. We must show that f^{−1}(V) is open in X. Now the preimage of the open set V under the restriction f|A_{i} of f to A_{i} is f^{−1}(V) ∩ A_{i}. It follows from the continuity of f|A_{i} that f^{−1}(V) ∩ A_{i} is relatively open in A_{i} for each i, and hence there exist open sets U_{1}, U_{2}, …, U_{k} in X such that f^{−1}(V)∩A_{i} = U_{i}∩A_{i} for i = 1, 2, …, k. Let W_{i} = U_{i}∪(X\A_{i}) for i = 1, 2, …, k. Then W_{i} is an open set in X (as it is the union of the open sets U_{i} and X \ A_{i}), and W_{i} ∩ A_{i} = U_{i} ∩ A_{i} = f^{−1}(V) ∩ A_{i} for each i. We claim that f^{−1}(V) = W_{1} ∩ W_{2} ∩ ⋯ ∩ W_{k}.

Let W = W_{1} ∩ W_{2} ∩ ⋯ ∩ W_{k}. Then f^{−1}(V) ⊂ W, since f^{−1}(V) ⊂ W_{i} for each i. Also

W = ⋃_{i=1}^{k}(W ∩ A_{i}) ⊂ ⋃_{i=1}^{k}(W_{i} ∩ A_{i}) = ⋃_{i=1}^{k}(f^{−1}(V) ∩ A_{i}) ⊂ f^{−1}(V),

since X = A_{1} ∪A_{2} ∪⋯∪A_{k} and W_{i} ∩A_{i} = f^{−1}(V)∩A_{i} for each i. Therefore f^{−1}(V) = W. But W is open in X, since it is the intersection of a finite collection of open sets. We have thus shown that f^{−1}(V) is open in X for any open set V in Y. Thus f: X → Y is continuous, as required.

Alternative Proof A function f: X → Y is continuous if and only if f^{−1}(G) is closed in X for every closed set G in Y (Lemma 1.7). Let G be an closed set in Y. Then f^{−1}(G) ∩ A_{i} is relatively closed in A_{i} for i = 1, 2, …, k, since the restriction of f to A_{i} is continuous for each i. But A_{i} is closed in X, and therefore a subset of A_{i} is relatively closed in A_{i} if and only if it is closed in X. Therefore f^{−1}(G) ∩ A_{i} is closed in X for i = 1, 2, …, k. Now f^{−1}(G) is the union of the sets f^{−1}(G) ∩ A_{i} for i = 1, 2, …, k. It follows that f^{−1}(G), being a finite union of closed sets, is itself closed in X. It now follows from Lemma 1.7 that f: X → Y is continuous.

Example Let Y be a topological space, and let α: [0, 1] → Y and β: [0, 1] → Y be continuous functions defined on the interval [0, 1], where α(1) = β(0). Let γ: [0, 1] → Y be defined by

γ(t) = \{ \begin{matrix}α(2t) & if 0 ≤ t ≤ \frac{1}{2}; \\ β(2t − 1) & if \frac{1}{2} ≤ t ≤ 1. \end{matrix}

Now γ|[0, \frac{1}{2}] = α ∘ ρ where ρ: [0, \frac{1}{2}] → [0, 1] is the continuous function defined by ρ(t) = 2t for all t ∈ [0, \frac{1}{2}]. Thus γ|[0, \frac{1}{2}] is continuous, being a composition of two continuous functions. Similarly γ|[\frac{1}{2}, 1] is continuous. The subintervals [0, \frac{1}{2}] and [\frac{1}{2}, 1] are closed in [0, 1], and [0, 1] is the union of these two subintervals. It follows from Lemma 1.9 that γ: [0, 1] → Y is continuous.

1.12 Homeomorphisms

Definition Let X and Y be topological spaces. A function h: X → Y is said to be a homeomorphism if and only if the following conditions are satisfied:

• the function h: X → Y is both injective and surjective (so that the function h: X → Y has a well-defined inverse h^{−1}: Y → X),

• the function h: X → Y and its inverse h^{−1}: Y → X are both continuous.

Two topological spaces X and Y are said to be homeomorphic if there exists a homeomorphism h: X → Y from X to Y.

If h: X → Y is a homeomorphism between topological spaces X and Y then h induces a one-to-one correspondence between the open sets of X and the open sets of Y. Thus the topological spaces X and Y can be regarded as being identical as topological spaces.

1.13 Sequences and Convergence

Definition Let X be a topological space. A sequence x_{1}, x_{2}, x_{3}, … of points in a topological space X is said to converge to a point p of X if, given any open set U containing the point p, there exists some natural number N such that x_{j} ∈ U for all j ≥ N. If the sequence (x_{j}) converges to p then we refer to p as a limit of the sequence.

We now show that this definition of convergence for a sequence of points in a topological space is consistent with the standard definition of convergence for a sequence of points in a metric space.

Lemma 1.10 Let X be a metric space with distance function d. A sequence x_{1}, x_{2}, x_{3}, … of points in a metric space X converges to a point p of X if and only if, given any real number ε satisfying ε > 0, there exists some natural number N such that d(x_{n}, p) < ε whenever n ≥ N.

Proof Let x_{1}, x_{2}, x_{3}, … be a sequence of points in X that converges to the point p of X. Let ε > 0 be given. The open ball B_{X}(p, ε) of radius ε about p is an open set (see Lemma 1.1). Therefore there exists some natural number N such that, if j ≥ N, then x_{j} ∈ B_{X}(p, ε), and thus d(x_{j}, p) < ε. Hence the sequence (x_{j}) converges to p.

Conversely, suppose that the sequence (x_{j}) has the property that, given any real number ε satisfying ε > 0, there exists some natural number N such that d(x_{n}, p) < ε whenever n ≥ N. Let U be an open set which contains p. Then there exists some ε > 0 such that B_{X}(p, ε) ⊂ U. But x_{j} → p as j → +∞, and therefore there exists some natural number N such that d(x_{j}, p) < ε for all j ≥ N. If j ≥ N then x_{j} ∈ B_{X}(p, ε) and thus x_{j} ∈ U. Thus the sequence (x_{j}) converges to p, as required.

A sequence of points in a metric space can converge to at most one point of that space. (This is an immediate consequence of Lemma 1.11 below.) However this result does not apply to topological spaces in general: it can happen that a sequence of points in a topological space may convergence to more than one limit. For example, consider the set ℝ of real numbers with the Zariski topology. (The open sets of ℝ in the Zariski topology are the empty set and those subsets of ℝ whose complements are finite.) Let x_{1}, x_{2}, x_{3}, … be the sequence in ℝ defined by x_{j} = j for all natural numbers j. One can readily check that this sequence converges to every real number p with respect to the Zariski topology on ℝ.

The set of real numbers with the Zariski topology is an example of a topological space which is not Hausdorff. We now show that sequences in a Hausdorff space converge to at most one limit.

Lemma 1.11 A sequence x_{1}, x_{2}, x_{3}, … of points in a Hausdorff space X converges to at most one limit.

Proof Suppose that p and q were limits of the sequence (x_{j}), where p ≠ q. Then there would exist open sets U and V such that p ∈ U, q ∈ V and U ∩ V = ∅, since X is a Hausdorff space. But then there would exist natural numbers N_{1} and N_{2} such that x_{j} ∈ U for all j satisfying j ≥ N_{1} and x_{j} ∈ V for all j satisfying j ≥ N_{2}. But then x_{j} ∈ U ∩ V for all j satisfying j ≥ N_{1} and j ≥ N_{2}, which is impossible, since U ∩ V = ∅. This contradiction shows that the sequence (x_{j}) has at most one limit.

Lemma 1.12 Let X be a topological space, and let F be a closed set in X. Let (x_{j} : j ∈ ℕ) be a sequence of points in F. Suppose that the sequence (x_{j}) converges to some point p of X. Then p ∈ F.

Proof Suppose that p were a point belonging to the complement X \F of F. Now X \ F is open (since F is closed). Therefore there would exist some natural number N such that x_{j} ∈ X \ F for all values of j satisfying j ≥ N, contradicting the fact that x_{j} ∈ F for all j. This contradiction shows that p must belong to F, as required.

Lemma 1.13 Let f: X → Y be a continuous function between topological spaces X and Y, and let x_{1}, x_{2}, x_{3}, … be a sequence of points in X which converges to some point p of X. Then the sequence f(x_{1}), f(x_{2}), f(x_{3}), … converges to f(p).

Proof Let V be an open set in Y which contains the point f(p). Then f^{−1}(V) is an open set in X which contains the point p. It follows that there exists some natural number N such that x_{j} ∈ f^{−1}(V) whenever j ≥ N. But then f(x_{j}) ∈ V whenever j ≥ N. We deduce that the sequence f(x_{1}), f(x_{2}), f(x_{3}), … converges to f(p), as required.

1.14 Neighbourhoods, Closures and Interiors

Definition Let X be a topological space, and let x be a point of X. Let N be a subset of X which contains the point x. Then N is said to be a neighbourhood of the point x if and only if there exists an open set U for which x ∈ U and U ⊂ N.

One can readily verify that this definition of neighbourhoods in topological spaces is consistent with that for neighbourhoods in metric spaces.

Lemma 1.14 Let X be a topological space. A subset V of X is open in X if and only if V is a neighbourhood of each point belonging to V.

Proof It follows directly from the definition of neighbourhoods that an open set V is a neighbourhood of any point belonging to V. Conversely, suppose that V is a subset of X which is a neighbourhood of each v ∈ V. Then, given any point v of V, there exists an open set U_{v} such that v ∈ U_{v} and U_{v} ⊂ V. Thus V is an open set, since it is the union of the open sets U_{v} as v ranges over all points of V.

Definition Let X be a topological space and let A be a subset of X. The closure \overline{A} of A in X is defined to be the intersection of all of the closed subsets of X that contain A. The interior A^{0} of A in X is defined to be the union of all of the open subsets of X that are contained in A.

Let X be a topological space and let A be a subset of X. It follows directly from the definition of \overline{A} that the closure \overline{A} of A is uniquely characterized by the following two properties:

(i) the closure \overline{A} of A is a closed set containing A, (ii) if F is any closed set containing A then F contains \overline{A}. Similarly the interior A^{0} of A is uniquely characterized by the following two properties:

(i) the interior A^{0} of A is an open set contained in A, (ii) if U is any open set contained in A then U is contained in A^{0}. Moreover a point x of A belongs to the interior A^{0} of A if and only if A is a neighbourhood of x. Lemma 1.15 Let X be a topological space, and let A be a subset of X. Suppose that a sequence x_{1}, x_{2}, x_{3}, … of points of A converges to some point p of X. Then p belongs to the closure \overline{A} of A. Proof If F is any closed set containing A then x_{j} ∈ F for all j, and therefore p ∈ F, by Lemma 1.12. Therefore p ∈ \overline{A} by definition of \overline{A}.

2 Product Topologies

2.1 The Cartesian Product of Subsets of Euclidean Space

Let X and Y be subsets of ℝ^{m} and ℝ^{n} respectively. We can regard the Cartesian product X × Y of X and Y as a subset of ℝ^{m+n}. If x and y are points of X and Y respectively, with

x = (x_{1}, x_{2}, …, x_{m}), y = (y_{1}, y_{2}, …, y_{n}),

then (x, y) is that point of ℝ^{m+n} with Cartesian coordinates given by

(x, y) = (x_{1}, x_{2}, …, x_{m}, y_{1}, y_{2}, …, y_{n}).

It follows immediately from the definition of the Euclidean distance function that

|(x, y) − (v, w)|^{2} = |x − v|^{2} + |y − w|^{2}.

for all points x and v of X and points y and w of Y.

Lemma 2.1 Let X and Y be subsets of ℝ^{m} and ℝ^{n} respectively. A subset U of X × Y is open in X × Y if and only if, given any point (v, w) of U, there exist positive real numbers δ_{1} > 0 and δ_{2} > 0 such that

{(x, y) ∈ X × Y : |x − v| < δ_{1} and |y − w| < δ_{2}} ⊂ U.

Proof We recall that a subset U of X × Y is open in X × Y if and only if, given any point (v, w) of U, there exists a positive real number δ > 0 such that

{(x, y) ∈ X × Y : |(x, y) − (v, w)| < δ} ⊂ U.

Let (v, w) be a point of U. Suppose that there exists a positive real number δ > 0 such that

{(x, y) ∈ X × Y : |(x, y) − (v, w)| < δ} ⊂ U.

Let δ_{1} = δ_{2} = δ/\sqrt{2}. Then δ_{1}^{2}+δ_{2}^{2} = δ^{2}. Thus if |x−v| < δ_{1} and |y−w| < δ_{2}, then |(x, y) − (v, w)| < δ, and hence (x, y) ∈ U.

Conversely suppose that there exist positive real numbers δ_{1} > 0 and δ_{2} > 0 such that

{(x, y) ∈ X × Y : |x − v| < δ_{1} and |y − w| < δ_{2}} ⊂ U.

Let δ be the minimum of δ_{1} and δ_{2}. If |(x, y) − (v, w)| < δ then |x − v| < δ_{1} and |y − w| < δ_{2}, and hence (x, y) ∈ U.

The result follows.

Lemma 2.2 Let X, Y and Z be subsets of ℝ^{m}, ℝ^{n} and ℝ^{k} respectively. A function f: X × Y → Z is continuous if and only if, given any point (v, w) of X × Y, and given any positive real number ε > 0, there exist positive real numbers δ_{1} > 0 and δ_{2} > 0 such that |f(x, y) − f(v, w)| < ε for all x ∈ X and y ∈ Y satisfying |x − v| < δ_{1} and |y − w| < δ_{2}.

Proof Let f: X × Y → Z be a function satisfying the above criterion. We must show that this function is continuous. Let U be an open set in Z. We show that f^{−1}(U) is open in X × Y.

Let (v, w) be a point of f^{−1}(U). Then f(v, w) is a point of U. But U is open in Z, and therefore there exists a positive real number ε > 0 such that

{z ∈ Z : |z − f(v, w)| < ε} ⊂ U.

But then there exist real numbers real numbers δ_{1} > 0 and δ_{2} > 0 such that |f(x, y) − f(v, w)| < ε, and hence f(x, y) ∈ U, for all x ∈ X and y ∈ Y satisfying |x − v| < δ_{1} and |y − w| < δ_{2}. Thus

{(x, y) ∈ X × Y : |x − v| < δ_{1} and |y − w| < δ_{2}} ⊂ f^{−1}(U).

We conclude that f^{−1}(U) is open in X × Y for each open set U in Z. Thus the function f: X × Y → Z is continuous.

Conversely suppose that f: X × Y → Z is a continuous function. Let (v, w) be a point of X × Y and let ε > 0 be given. Then

{z ∈ Z : |z − f(v, w)| < ε}

is an open set in Z, and hence its preimage is an open set in X × Y. It follows from Lemma 2.1 that there exist positive real numbers δ_{1} > 0 and δ_{2} > 0 such that

{(x, y) ∈ X × Y : |x − v| < δ_{1} and |y − w| < δ_{2}}

is contained in the preimage of {z ∈ Z : |z − f(v, w)| < ε}. But this means that |f(x, y) − f(v, w)| < ε for all x ∈ X and y ∈ Y satisfying |x − v| < δ_{1} and |y − w| < δ_{2}, as required.

The next result shows how one can describe the collection of open sets of X ×Y in terms of the collections of open sets in X and in Y, without explicit reference to norms or distance functions. This motivates the definition of the product topology on the Cartesian product of two topological spaces.

Proposition 2.3 Let X and Y be subsets of ℝ^{m} and ℝ^{n} respectively. A subset U of X × Y is open in X × Y if and only if, given any point (v, w) of U, there exist an open set V in X and an open set W in Y such that v ∈ V, w ∈ W and V × W ⊂ U.

Proof Let U be open in X × Y. It follows from Lemma 2.1 that there exist positive real numbers δ_{1} > 0 and δ_{2} > 0 such that

{(x, y) ∈ X × Y : |x − v| < δ_{1} and |y − w| < δ_{2}} ⊂ U.

Let

V = {x ∈ X : |x − v| < δ_{1}} and W = {y ∈ Y : |y − w| < δ_{2}}.

Then V is open in X, W is open in Y, v ∈ V, w ∈ W and V × W ⊂ U.

Conversely suppose that U is a subset of X ×Y and that, given any point (v, w) of U, there exist an open set V in X and an open set W in Y such that v ∈ V, w ∈ W and V × W ⊂ U. Then, given any point (v, w) of U, there exist positive real numbers δ_{1} > 0 and δ_{2} > 0 such that

{x ∈ X : |x − v| < δ_{1}} ⊂ V

and

{y ∈ Y : |y − w| < δ_{2}} ⊂ W.

But then

{(x, y) ∈ X × Y : |x − v| < δ_{1} and |y − w| < δ_{2}} ⊂ V × W ⊂ U.

It follows from Lemma 2.1 that U is open in X × Y, as required.

2.2 Product Topologies

The Cartesian product X_{1} × X_{2} × ⋯ × X_{n} of sets X_{1}, X_{2}, …, X_{n} is defined to be the set of all ordered n-tuples (x_{1}, x_{2}, …, x_{n}), where x_{i} ∈ X_{i} for i = 1, 2, …, n.

The sets ℝ^{2} and ℝ^{3} are the Cartesian products ℝ × ℝ and ℝ × ℝ × ℝ respectively.

Cartesian products of sets are employed as the domains of functions of several variables. For example, if X, Y and Z are sets, and if an element f(x, y) of Z is determined for each choice of an element x of X and an element y of Y, then we have a function f: X × Y → Z whose domain is the Cartesian product X × Y of X and Y: this function sends the ordered pair (x, y) to f(x, y) for all x ∈ X and y ∈ Y.

Now suppose that X, Y and Z are topological spaces. We wish to define a notion of continuity for functions f: X × Y → Z from X × Y to Z. In order to do this, we show that the topologies of X and Y together induce in a natural way a topology on X × Y; this topology is referred to as the product topology on X × Y.

First we observe that if V is a subset of X and if W is a subset of Y then V × W is a subset of X × Y: an element of V × W is an ordered pair (v, w) with v ∈ V and w ∈ W, and such an ordered pair belongs to X × Y.

Definition Let X and Y be topological spaces. A subset U of X × Y is said to be open in X × Y (with respect to the product topology) if, given any point (x, y) of U, there exist an open set V in X and an open set W in Y such that x ∈ V, y ∈ W and V × W ⊂ U. The empty set is regarded as an open set in X × Y.

Lemma 2.4 Let X and Y be topological spaces. Then the collection of open sets in X × Y is a topology on X × Y.

Proof The definition of open sets ensures that the empty set and the whole set X × Y are open in X × Y. We must prove that any union or finite intersection of open sets in X × Y is an open set.

Let E be the union of a collection of open sets in X × Y, and let (x, y) be a point of E. Then (x, y) ∈ D for some open set D in the collection. It follows from this that there exists an open set V in X and an open set W in Y such that x ∈ V, y ∈ W and V × W ⊂ D. But then V × W ⊂ E. It follows that E is open in X × Y.

Let U = U_{1} ∩ U_{2} ∩ ⋯ ∩ U_{m}, where U_{1}, U_{2}, …, U_{m} are open sets in X × Y, and let (x, y) be a point of U. Then there exist open sets V_{k} in X and open sets W_{k} in Y for k = 1, 2, …, m such that x ∈ V_{k}, y ∈ W_{k} and V_{k} × W_{k} ⊂ U_{k} for k = 1, 2, …, m. Let

V = V_{1} ∩ V_{2} ∩ ⋯ ∩ V_{m}, W = W_{1} ∩ W_{2} ∩ ⋯ ∩ W_{m}.

Then x ∈ V and y ∈ W. Also V × W ⊂ V_{k} × W_{k} ⊂ U_{k} for k = 1, 2, …, m, hence V × W ⊂ U. It follows that U is open in X × Y, as required.

Let X and Y be topological spaces. The collection of open sets in X × Y defined as described above is referred to as the product topology on X × Y. The definition of the product topology can easily be generalized to Cartesian products of any finite number of topological spaces.

Definition Let X_{1}, X_{2}, …, X_{n} be topological spaces. A subset U of the Cartesian product X_{1} × X_{2} × ⋯ × X_{n} is said to be open (with respect to the product topology) if, given any point p of U, there exist open sets V_{i} in X_{i} for i = 1, 2, …, n such that {p} ⊂ V_{1} × V_{2} × ⋯ × V_{n} ⊂ U.

Lemma 2.5 Let X_{1}, X_{2}, …, X_{n} be topological spaces. Then the collection of open sets in X_{1} × X_{2} × ⋯ × X_{n} is a topology on X_{1} × X_{2} × ⋯ × X_{n}.

Proof Let X = X_{1} ×X_{2} ×⋯×X_{n}. The definition of open sets ensures that the empty set and the whole set X are open in X. We must prove that any union or finite intersection of open sets in X is an open set.

Let E be a union of a collection of open sets in X and let p be a point of E. Then p ∈ D for some open set D in the collection. It follows from this that there exist open sets V_{i} in X_{i} for i = 1, 2, …, n such that

{p} ⊂ V_{1} × V_{2} × ⋯ × V_{n} ⊂ D ⊂ E.

Thus E is open in X.

Let U = U_{1}∩U_{2}∩⋯∩U_{m}, where U_{1}, U_{2}, …, U_{m} are open sets in X, and let p be a point of U. Then there exist open sets V_{ki} in X_{i} for k = 1, 2, …, m and i = 1, 2, …, n such that {p} ⊂ V_{k1} × V_{k2} × ⋯ × V_{kn} ⊂ U_{k} for k = 1, 2, …, m. Let V_{i} = V_{1i} ∩ V_{2i} ∩ ⋯ ∩ V_{mi} for i = 1, 2, …, n. Then

{p} ⊂ V_{1} × V_{2} × ⋯ × V_{n} ⊂ V_{k1} × V_{k2} × ⋯ × V_{kn} ⊂ U_{k}

for k = 1, 2, …, m, and hence {p} ⊂ V_{1} × V_{2} × ⋯ × V_{n} ⊂ U. It follows that U is open in X, as required.

Lemma 2.6 Let X_{1}, X_{2}, …, X_{n} and Z be topological spaces. Then a function f: X_{1} × X_{2} × ⋯ × X_{n} → Z is continuous if and only if, given any point p of X_{1} × X_{2} × ⋯ × X_{n}, and given any open set U in Z containing f(p), there exist open sets V_{i} in X_{i} for i = 1, 2, …, n such that p ∈ V_{1} × V_{2} ⋯ × V_{n} and f(V_{1} × V_{2} × ⋯ × V_{n}) ⊂ U.

Proof Let V_{i} be an open set in X_{i} for i = 1, 2, …, n, and let U be an open set in Z. Then V_{1}×V_{2}×⋯×V_{n} ⊂ f^{−1}(U) if and only if f(V_{1}×V_{2}×⋯×V_{n}) ⊂ U. It follows that f^{−1}(U) is open in the product topology on X_{1}×X_{2}×⋯×X_{n} if and only if, given any point p of X_{1} ×X_{2} ×⋯×X_{n} satisfying f(p) ∈ U, there exist open sets V_{i} in X_{i} for i = 1, 2, …, n such that f(V_{1} ×V_{2} ×⋯×V_{n}) ⊂ U. The required result now follows from the definition of continuity.

Let X_{1}, X_{2}, …, X_{n} be topological spaces, and let V_{i} be an open set in X_{i} for i = 1, 2, …, n. It follows directly from the definition of the product topology that V_{1} × V_{2} × ⋯ × V_{n} is open in X_{1} × X_{2} × ⋯ × X_{n}.

Theorem 2.7 Let X = X_{1} × X_{2} × ⋯ × X_{n}, where X_{1}, X_{2}, …, X_{n} are topological spaces and X is given the product topology, and for each i, let p_{i}: X → X_{i} denote the projection function which sends (x_{1}, x_{2}, …, x_{n}) ∈ X to x_{i}. Then the functions p_{1}, p_{2}, …, p_{n} are continuous. Moreover a function f: Z → X mapping a topological space Z into X is continuous if and only if p_{i} ∘ f: Z → X_{i} is continuous for i = 1, 2, …, n.

Proof Let V be an open set in X_{i}. Then

p_{i}^{−1}(V) = X_{1} × ⋯ × X_{i−1} × V × X_{i+1} × ⋯ × X_{n},

and therefore p_{i}^{−1}(V) is open in X. Thus p_{i}: X → X_{i} is continuous for all i.

Let f: Z → X be continuous. Then, for each i, p_{i} ∘ f: Z → X_{i} is a composition of continuous functions, and is thus itself continuous.

Conversely suppose that f: Z → X is a function with the property that p_{i} ∘ f is continuous for all i. Let U be an open set in X. We must show that f^{−1}(U) is open in Z.

Let z be a point of f^{−1}(U), and let f(z) = (u_{1}, u_{2}, …, u_{n}). Now U is open in X, and therefore there exist open sets V_{1}, V_{2}, …, V_{n} in X_{1}, X_{2}, …, X_{n} respectively such that u_{i} ∈ V_{i} for all i and V_{1} × V_{2} × ⋯ × V_{n} ⊂ U. Let

N_{z} = f_{1}^{−1}(V_{1}) ∩ f_{2}^{−1}(V_{2}) ∩ ⋯ ∩ f_{n}^{−1}(V_{n}),

where f_{i} = p_{i} ∘ f for i = 1, 2, …, n. Now f_{i}^{−1}(V_{i}) is an open subset of Z for i = 1, 2, …, n, since V_{i} is open in X_{i} and f_{i}: Z → X_{i} is continuous. Thus N_{z}, being a finite intersection of open sets, is itself open in Z. Moreover

f(N_{z}) ⊂ V_{1} × V_{2} × ⋯ × V_{n} ⊂ U,

so that N_{z} ⊂ f^{−1}(U). It follows that f^{−1}(U) is the union of the open sets N_{z} as z ranges over all points of f^{−1}(U). Therefore f^{−1}(U) is open in Z. This shows that f: Z → X is continuous, as required.

Proposition 2.8 The usual topology on ℝ^{n} coincides with the product topology on ℝ^{n} obtained on regarding ℝ^{n} as the Cartesian product ℝ×ℝ×⋯×ℝ of n copies of the real line ℝ.

Proof We must show that a subset U of ℝ^{n} is open with respect to the usual topology if and only if it is open with respect to the product topology.

Let U be a subset of ℝ^{n} that is open with respect to the usual topology, and let u ∈ U. Then there exists some δ > 0 such that B(u, δ) ⊂ U, where

B(u, δ) = {x ∈ ℝ^{n} : |x − u| < δ}.

Let I_{1}, I_{2}, …, I_{n} be the open intervals in ℝ defined by

I_{i} = \{t ∈ ℝ : u_{i} − \frac{δ}{\sqrt{n}} < t < u_{i} + \frac{δ}{\sqrt{n}}\} (i = 1, 2, …, n),

Then I_{1}, I_{2}, …, I_{n} are open sets in ℝ. Moreover

{u} ⊂ I_{1} × I_{2} × ⋯ × I_{n} ⊂ B(u, δ) ⊂ U,

since

|x − u|^{2} = ∑_{i=1}^{n}(x_{i} − u_{i})^{2} < n (\frac{δ}{\sqrt{n}})^{2} = δ^{2}

for all x ∈ I_{1} × I_{2} × ⋯ × I_{n}. This shows that any subset U of ℝ^{n} that is open with respect to the usual topology on ℝ^{n} is also open with respect to the product topology on ℝ^{n}.

Conversely suppose that U is a subset of ℝ^{n} that is open with respect to the product topology on ℝ^{n}, and let u ∈ U. Then there exist open sets V_{1}, V_{2}, …, V_{n} in ℝ containing u_{1}, u_{2}, …, u_{n} respectively such that V_{1} × V_{2} × ⋯ × V_{n} ⊂ U. Now we can find δ_{1}, δ_{2}, …, δ_{n} such that δ_{i} > 0 and (u_{i} − δ_{i}, u_{i} + δ_{i}) ⊂ V_{i} for all i. Let δ > 0 be the minimum of δ_{1}, δ_{2}, …, …, δ_{n}. Then

B(u, δ) ⊂ V_{1} × V_{2} × ⋯ V_{n} ⊂ U,

for if x ∈ B(u, δ) then |x_{i} − u_{i}| < δ_{i} for i = 1, 2, …, n. This shows that any subset U of ℝ^{n} that is open with respect to the product topology on ℝ^{n} is also open with respect to the usual topology on ℝ^{n}.

The following result is now an immediate corollary of Proposition 2.8 and Theorem 2.7.

Corollary 2.9 Let X be a topological space and let f: X → ℝ^{n} be a function from X to ℝ^{n}. Let us write

f(x) = (f_{1}(x), f_{2}(x), …, f_{n}(x))

for all x ∈ X, where the components f_{1}, f_{2}, …, f_{n} of f are functions from X to ℝ. The function f is continuous if and only if its components f_{1}, f_{2}, …, f_{n} are all continuous.

Let f: X → ℝ and g: X → ℝ be continuous real-valued functions on some topological space X. We claim that f +g, f −g and f.g are continuous. Now it is a straightforward exercise to verify that the sum and product functions s: ℝ^{2} → ℝ and p: ℝ^{2} → ℝ defined by s(x, y) = x + y and p(x, y) = xy are continuous, and f + g = s ∘ h and f.g = p ∘ h, where h: X → ℝ^{2} is defined by h(x) = (f(x), g(x)). Moreover it follows from Corollary 2.9 that the function h is continuous, and compositions of continuous functions are continuous. Therefore f + g and f.g are continuous, as claimed. Also −g is continuous, and f − g = f + (−g), and therefore f − g is continuous. If in addition the continuous function g is non-zero everywhere on X then 1/g is continuous (since 1/g is the composition of g with the reciprocal function t ↦ 1/t), and therefore f/g is continuous.

Lemma 2.10 The Cartesian product X_{1} × X_{2} × … X_{n} of Hausdorff spaces X_{1}, X_{2}, …, X_{n} is Hausdorff.

Proof Let X = X_{1} × X_{2} × …, X_{n}, and let u and v be distinct points of X, where u = (x_{1}, x_{2}, …, x_{n}) and v = (y_{1}, y_{2}, …, y_{n}). Then x_{i} ≠ y_{i} for some integer i between 1 and n. But then there exist open sets U and V in X_{i} such that x_{i} ∈ U, y_{i} ∈ V and U ∩ V = ∅ (since X_{i} is a Hausdorff space). Let p_{i}: X → X_{i} denote the projection function. Then p_{i}^{−1}(U) and p_{i}^{−1}(V) are open sets in X, since p_{i} is continuous. Moreover u ∈ p_{i}^{−1}(U), v ∈ p_{i}^{−1}(V), and p_{i}^{−1}(U) ∩ p_{i}^{−1}(V) = ∅. Thus X is Hausdorff, as required.

3 Identification Maps and Quotient Topologies

3.1 Cut and Paste Constructions

Suppose we start out with a square of paper. If we join together two opposite edges of this square we obtain a cylinder. The boundary of the cylinder consists of two circles. If we join together the two boundary circles we obtain a torus (which corresponds to the surface of a doughnut).

Let the square be represented by the set [0, 1] × [0, 1] consisting of all ordered pairs (s, t) where s and t are real numbers between 0 and 1. There is an equivalence relation on the square [0, 1] × [0, 1], where points (s, t) and (u, v) of the square are related if and only if at least one of the following conditions is satisfied:

• s = u and t = v;

• s = 0, u = 1 and t = v;

• s = 1, u = 0 and t = v;

• t = 0, v = 1 and s = u;

• t = 1, v = 0 and s = u;

• (s, t) and (u, v) both belong to {(0, 0), (0, 1), (1, 0), (1, 1)}.

Note that if 0 < s < 1 and 0 < t < 1 then the equivalence class of the point (s, t) is the set {(s, t)} consisting of that point. If s = 0 or 1 and if 0 < t < 1 then the equivalence class of (s, t) is the set {(0, t), (1, t)}. Similarly if t = 0 or 1 and if 0 < s < 1 then the equivalence class of (s, t) is the set {(s, 0), (s, 1)}. The equivalence class of each corner of the square is the set (0, 0), (1, 0), (0, 1), (1, 1) consisting of all four corners. Thus each equivalence class contains either one point in the interior of the square, or two points on opposite edges of the square, or four points at the four corners of the square. Let T^{2} denote the set of these equivalence classes. We have a map q: [0, 1] × [0, 1] → T^{2} which sends each point (s, t) of the square to its equivalence class. Each element of the set T^{2} is the image of one, two or four points of the square. The elements of T^{2} represent points on the torus obtained from the square by first joining together two opposite sides of the square to form a cylinder and then joining together the boundary circles of this cylinder as described above. We say that the torus T^{2} is obtained from the square [0, 1] × [0, 1] by identifying the points (0, t) and (1, t) for all t ∈ [0, 1] and identifying the points (s, 0) and (s, 1) for all s ∈ [0, 1].

The topology on the square [0, 1]×[0, 1] induces a corresponding topology on the set T^{2}, where a subset U of T^{2} is open in T^{2} if and only if q^{−1}(U) is open in the square [0, 1] × [0, 1]. (The fact that these open sets in T^{2} constitute a topology on the set T^{2} is a consequence of Lemma 3.1.) The function q: [0, 1] × [0, 1] → T^{2} is then a continuous surjection. We say that the topological space T^{2} is the identification space obtained from the square [0, 1] × [0, 1] by identifying points on the sides to the square as described above. The continuous map q from the square to the torus is an example of an identification map, and the topology on the torus T^{2} is referred to as the quotient topology on T^{2} induced by the identification map q: [0, 1] × [0, 1] → T^{2}.

Another well-known identification space obtained from the square is the Klein bottle (Kleinsche Flasche). The Klein bottle K^{2} is obtained from the square [0, 1] × [0, 1] by identifying (0, t) with (1, 1 − t) for all t ∈ [0, 1] and identifying (s, 0) with (s, 1) for all s ∈ [0, 1]. These identifications correspond to an equivalence relation on the square, where points (s, t) and (u, v) of the square are equivalent if and only if one of the following conditions is satisfied:

• s = u and t = v;

• s = 0, u = 1 and t = 1 − v;

• s = 1, u = 0 and t = 1 − v;

• t = 0, v = 1 and s = u;

• t = 1, v = 0 and s = u;

• (s, t) and (u, v) both belong to {(0, 0), (0, 1), (1, 0), (1, 1)}.

The corresponding set of equivalence classes is the Klein bottle K^{2}. Thus each point of the Klein bottle K^{2} represents an equivalence class consisting of either one point in the interior of the square, or two points (0, t) and (1, 1 − t) with 0 < t < 1 on opposite edges of the square, or two points (s, 0) and (s, 1) with 0 < s < 1 on opposite edges of the square, or the four corners of the square. There is a surjection r: [0, 1] × [0, 1] → K^{2} from the square to the Klein bottle that sends each point of the square to its equivalence class. The identifications used to construct the Klein bottle ensure that r(0, t) = r(1, 1 − t) for all t ∈ [0, 1] and r(s, 0) = r(s, 1) for all s ∈ [0, 1]. One can construct a quotient topology on the Klein bottle K^{2}, where a subset U of K^{2} is open in K^{2} if and only if its preimage r^{−1}(U) is open in the square [0, 1] × [0, 1].

3.2 Identification Maps and Quotient Topologies

Definition Let X and Y be topological spaces and let q: X → Y be a function from X to Y. The function q is said to be an identification map if and only if the following conditions are satisfied:

• the function q: X → Y is surjective,

• a subset U of Y is open in Y if and only if q^{−1}(U) is open in X.

It follows directly from the definition that any identification map is continuous. Moreover, in order to show that a continuous surjection q: X → Y is an identification map, it suffices to prove that if V is a subset of Y with the property that q^{−1}(V) is open in X then V is open in Y.

Example Let S^{1} denote the unit circle {(x, y) ∈ ℝ^{2} : x^{2}+y^{2} = 1} in ℝ^{2}, and let q: [0, 1] → S^{1} be the continuous map defined by q(t) = (cos 2πt, sin 2πt) for all t ∈ [0, 1]. We show that q: [0, 1] → S^{1} is an identification map. This map is continuous and surjective. It remains to show that if V is a subset of S^{1} with the property that q^{−1}(V) is open in [0, 1] then V is open in S^{1}.

Note that |q(s) − q(t)| = 2| sin π(s − t)| for all s, t ∈ [0, 1] satisfying |s − t| ≤ \frac{1}{2}. Let V be a subset of S^{1} with the property that q^{−1}(V) is open in [0, 1], and let v be an element of V. We show that there exists ε > 0 such that all points u of S^{1} satisfying |u − v| < ε belong to V. We consider separately the cases when v = (1, 0) and when v ≠ (1, 0).

Suppose that v = (1, 0). Then (1, 0) ∈ V, and hence 0 ∈ q^{−1}(V) and 1 ∈ q^{−1}(V). But q^{−1}(V) is open in [0, 1]. It follows that there exists a real number δ satisfying 0 < δ < \frac{1}{2} such that [0, δ) ⊂ q^{−1}(V) and (1 − δ, 1] ∈ q^{−1}(V). Let ε = 2 sin πδ. Now if −π ≤ θ ≤ π then the Euclidean distance between the points (1, 0) and (cos θ, sin θ) is 2 sin \frac{1}{2}|θ|. Moreover, this distance increases monotonically as |θ| increases from 0 to π. Thus any point on the unit circle S^{1} whose distance from (1, 0) is less than ε must be of the form (cos θ, sin θ), where |θ| < 2πδ. Thus if u ∈ S^{1} satisfies |u − v| < ε then u = q(s) for some s ∈ [0, 1] satisfying either 0 ≤ s < δ or 1 − δ < s ≤ 1. But then s ∈ q^{−1}(V), and hence u ∈ V.

Next suppose that v ≠ (1, 0). Then v = q(t) for some real number t satisfying 0 < t < 1. But q^{−1}(V) is open in [0, 1], and t ∈ q^{−1}(V). It follows that (t − δ, t + δ) ⊂ q^{−1}(V) for some real number δ satisfying δ > 0. Let ε = 2 sin πδ. If u ∈ S^{1} satisfies |u − v| < ε then u = q(s) for some s ∈ (t − δ, t + δ). But then s ∈ q^{−1}(V), and hence u ∈ V.

We have thus shown that if V is a subset of S^{1} with the property that q^{−1}(V) is open in [0, 1] then there exists ε > 0 such that u ∈ V for all elements u of S^{1} satisfying |u − v| < ε. It follows from this that V is open in S^{1}. Thus the continuous surjection q: [0, 1] → S^{1} is an identification map.

Lemma 3.1 Let X be a topological space, let Y be a set, and let q: X → Y be a surjection. Then there is a unique topology on Y for which the function q: X → Y is an identification map.

Proof Let τ be the collection consisting of all subsets U of Y for which q^{−1}(U) is open in X. Now q^{−1}(∅) = ∅, and q^{−1}(Y) = X, so that ∅ ∈ τ and Y ∈ τ. If {V_{α} : α ∈ A} is any collection of subsets of Y indexed by a set A, then it is a straightforward exercise to verify that

⋃_{α∈A} q^{−1}(V_{α}) = q^{−1} (⋃_{α∈A} V_{α}) , ⋂_{α∈A} q^{−1}(V_{α}) = q^{−1} (⋂_{α∈A} V_{α})

(i.e., given any collection of subsets of Y, the union of the preimages of the sets is the preimage of the union of those sets, and the intersection of the preimages of the sets is the preimage of the intersection of those sets). It follows easily from this that unions and finite intersections of sets belonging to τ must themselves belong to τ. Thus τ is a topology on Y, and the function q: X → Y is an identification map with respect to the topology τ. Clearly τ is the unique topology on Y for which the function q: X → Y is an identification map.

Let X be a topological space, let Y be a set, and let q: X → Y be a surjection. The unique topology on Y for which the function q is an identification map is referred to as the quotient topology (or identification topology) on Y.

Let ∼ be an equivalence relation on a topological space X. If Y is the corresponding set of equivalence classes of elements of X then there is a surjection q: X → Y that sends each element of X to its equivalence class. Lemma 3.1 ensures that there is a well-defined quotient topology on Y, where a subset U of Y is open in Y if and only if q^{−1}(U) is open in X. (Appropriate equivalence relations on the square yield the torus and the Klein bottle, as discussed above.)

Lemma 3.2 Let X and Y be topological spaces and let q: X → Y be an identification map. Let Z be a topological space, and let f: Y → Z be a function from Y to Z. Then the function f is continuous if and only if the composition function f ∘ q: X → Z is continuous.

Proof Suppose that f is continuous. Then the composition function f ∘ q is a composition of continuous functions and hence is itself continuous.

Conversely suppose that f ∘ q is continuous. Let U be an open set in Z. Then q^{−1}(f^{−1}(U)) is open in X (since f ∘ q is continuous), and hence f^{−1}(U) is open in Y (since the function q is an identification map). Therefore the function f is continuous, as required.

Example Let S^{1} be the unit circle in ℝ^{2}, and let q: [0, 1] → S^{1} be the map that sends t ∈ [0, 1] to (cos 2πt, sin 2πt). Then q: [0, 1] → S^{1} is an identification map, and therefore a function f: S^{1} → Z from S^{1} to some topological space Z is continuous if and only if f ∘q: [0, 1] → Z is continuous.

Example The Klein bottle K^{2} is the identification space obtained from the square [0, 1] × [0, 1] by identifying (0, t) with (1, 1 − t) for all t ∈ [0, 1] and identifying (s, 0) with (s, 1) for all s ∈ [0, 1]. Let q: [0, 1] × [0, 1] → K^{2} be the identification map determined by these identifications. Let Z be a topological space. A function g: [0, 1]×[0, 1] → Z mapping the square into Z which satisfies g(0, t) = g(1, 1 − t) for all t ∈ [0, 1] and g(s, 0) = g(s, 1) for all s ∈ [0, 1], determines a corresponding function f: K^{2} → Z, where g = f ∘ q. It follows from Lemma 3.2 that the function f: K^{2} → Z is continuous if and only if g: [0, 1] × [0, 1] → Z is continuous.

Example Let S^{n} be the n-sphere, consisting of all points x in ℝ^{n+1} satisfying |x| = 1. Let ℝP^{n} be the set of all lines in ℝ^{n+1} passing through the origin (i.e., ℝP^{n} is the set of all one-dimensional vector subspaces of ℝ^{n+1}). Let q: S^{n} → ℝP^{n} denote the function which sends a point x of S^{n} to the element of ℝP^{n} represented by the line in ℝ^{n+1} that passes through both x and the origin. Note that each element of ℝP^{n} is the image (under q) of exactly two antipodal points x and −x of S^{n}. The function q induces a corresponding quotient topology on ℝP^{n} such that q: S^{n} → ℝP^{n} is an identification map. The set ℝP^{n}, with this topology, is referred to as real projective n-space. In particular ℝP^{2} is referred to as the real projective plane. It follows from Lemma 3.2 that a function f: ℝP^{n} → Z from ℝP^{n} to any topological space Z is continuous if and only if the composition function f ∘ q: S^{n} → Z is continuous.

4 Compactness

4.1 Compact Topological Spaces

Let X be a topological space, and let A be a subset of X. A collection of subsets of X in X is said to cover A if and only if every point of A belongs to at least one of these subsets. In particular, an open cover of X is collection of open sets in X that covers X.

If U and V are open covers of some topological space X then V is said to be a subcover of U if and only if every open set belonging to V also belongs to U.

Definition A topological space X is said to be compact if and only if every open cover of X possesses a finite subcover.

Lemma 4.1 Let X be a topological space. A subset A of X is compact (with respect to the subspace topology on A) if and only if, given any collection U of open sets in X covering A, there exists a finite collection V_{1}, V_{2}, …, V_{r} of open sets belonging to U such that A ⊂ V_{1} ∪ V_{2} ∪ ⋯ ∪ V_{r}.

Proof A subset B of A is open in A (with respect to the subspace topology on A) if and only if B = A ∩ V for some open set V in X. The desired result therefore follows directly from the definition of compactness.

We now show that any closed bounded interval in the real line is compact. This result is known as the Heine-Borel Theorem. The proof of this theorem uses the least upper bound principle which states that, given any non-empty set S of real numbers which is bounded above, there exists a least upper bound (or supremum) sup S for the set S.

Theorem 4.2 (Heine-Borel) Let a and b be real numbers satisfying a < b. Then the closed bounded interval [a, b] is a compact subset of ℝ.

Proof Let U be a collection of open sets in ℝ with the property that each point of the interval [a, b] belongs to at least one of these open sets. We must show that [a, b] is covered by finitely many of these open sets.

Let S be the set of all τ ∈ [a, b] with the property that [a, τ] is covered by some finite collection of open sets belonging to U, and let s = sup S. Now s ∈ W for some open set W belonging to U. Moreover W is open in ℝ, and therefore there exists some δ > 0 such that (s − δ, s + δ) ⊂ W. Moreover s − δ is not an upper bound for the set S, hence there exists some τ ∈ S satisfying τ > s − δ. It follows from the definition of S that [a, τ] is covered by some finite collection V_{1}, V_{2}, …, V_{r} of open sets belonging to U.

Let t ∈ [a, b] satisfy τ ≤ t < s + δ. Then

[a, t] ⊂ [a, τ] ∪ (s − δ, s + δ) ⊂ V_{1} ∪ V_{2} ∪ ⋯ ∪ V_{r} ∪ W,

and thus t ∈ S. In particular s ∈ S, and moreover s = b, since otherwise s would not be an upper bound of the set S. Thus b ∈ S, and therefore [a, b] is covered by a finite collection of open sets belonging to U, as required.

Lemma 4.3 Let A be a closed subset of some compact topological space X. Then A is compact.

Proof Let U be any collection of open sets in X covering A. On adjoining the open set X \ A to U, we obtain an open cover of X. This open cover of X possesses a finite subcover, since X is compact. Moreover A is covered by the open sets in the collection U that belong to this finite subcover. It follows from Lemma 4.1 that A is compact, as required.

Lemma 4.4 Let f: X → Y be a continuous function between topological spaces X and Y, and let A be a compact subset of X. Then f(A) is a compact subset of Y.

Proof Let V be a collection of open sets in Y which covers f(A). Then A is covered by the collection of all open sets of the form f^{−1}(V) for some V ∈ V. It follows from the compactness of A that there exists a finite collection V_{1}, V_{2}, …, V_{k} of open sets belonging to V such that

A ⊂ f^{−1}(V_{1}) ∪ f^{−1}(V_{2}) ∪ ⋯ ∪ f^{−1}(V_{k}).

But then f(A) ⊂ V_{1} ∪ V_{2} ∪ ⋯ ∪ V_{k}. This shows that f(A) is compact.

Lemma 4.5 Let f: X → ℝ be a continuous real-valued function on a compact topological space X. Then f is bounded above and below on X.

Proof The range f(X) of the function f is covered by some finite collection I_{1}, I_{2}, …, I_{k} of open intervals of the form (−m, m), where m ∈ ℕ, since f(X) is compact (Lemma 4.4) and ℝ is covered by the collection of all intervals of this form. It follows that f(X) ⊂ (−M, M), where (−M, M) is the largest of the intervals I_{1}, I_{2}, …, I_{k}. Thus the function f is bounded above and below on X, as required.

Proposition 4.6 Let f: X → ℝ be a continuous real-valued function on a compact topological space X. Then there exist points u and v of X such that f(u) ≤ f(x) ≤ f(v) for all x ∈ X.

Proof Let m = inf{f(x) : x ∈ X} and M = sup{f(x) : x ∈ X}. There must exist v ∈ X satisfying f(v) = M, for if f(x) < M for all x ∈ X then the function x ↦ 1/(M − f(x)) would be a continuous real-valued function on X that was not bounded above, contradicting Lemma 4.5. Similarly there must exist u ∈ X satisfying f(u) = m, since otherwise the function x ↦ 1/(f(x)−m) would be a continuous function on X that was not bounded above, again contradicting Lemma 4.5. But then f(u) ≤ f(x) ≤ f(v) for all x ∈ X, as required.

Proposition 4.7 Let A be a compact subset of a metric space X. Then A is closed in X.

Proof Let p be a point of X that does not belong to A, and let f(x) = d(x, p), where d is the distance function on X. It follows from Proposition 4.6 that there is a point q of A such that f(a) ≥ f(q) for all a ∈ A, since A is compact. Now f(q) > 0, since q ≠ p. Let δ satisfy 0 < δ ≤ f(q). Then the open ball of radius δ about the point p is contained in the complement of A, since f(x) < f(q) for all points x of this open ball. It follows that the complement of A is an open set in X, and thus A itself is closed in X.

Proposition 4.8 Let X be a Hausdorff topological space, and let K be a compact subset of X. Let x be a point of X \ K. Then there exist open sets V and W in X such that x ∈ V, K ⊂ W and V ∩ W = ∅.

Proof For each point y ∈ K there exist open sets V_{x,y} and W_{x,y} such that x ∈ V_{x,y}, y ∈ W_{x,y} and V_{x,y} ∩ W_{x,y} = ∅ (since X is a Hausdorff space). But then there exists a finite set {y_{1}, y_{2}, …, y_{r}} of points of K such that K is contained in W_{x,y_{1}} ∪ W_{x,y_{2}} ∪ ⋯ ∪ W_{x,y_{r}}, since K is compact. Define

V = V_{x,y_{1}} ∩ V_{x,y_{2}} ∩ ⋯ ∩ V_{x,y_{r}}, W = W_{x,y_{1}} ∪ W_{x,y_{2}} ∪ ⋯ ∪ W_{x,y_{r}}.

Then V and W are open sets, x ∈ V, K ⊂ W and V ∩W = ∅, as required.

Corollary 4.9 A compact subset of a Hausdorff topological space is closed.

Proof Let K be a compact subset of a Hausdorff topological space X. It follows immediately from Proposition 4.8 that, for each x ∈ X \ K, there exists an open set V_{x} such that x ∈ V_{x} and V_{x} ∩ K = ∅. But then X \ K is equal to the union of the open sets V_{x} as x ranges over all points of X \ K, and any set that is a union of open sets is itself an open set. We conclude that X \ K is open, and thus K is closed.

Proposition 4.10 Let X be a Hausdorff topological space, and let K_{1} and K_{2} be compact subsets of X, where K_{1} ∩ K_{2} = ∅. Then there exist open sets U_{1} and U_{2} such that K_{1} ⊂ U_{1}, K_{2} ⊂ U_{2} and U_{1} ∩ U_{2} = ∅.

Proof It follows from Proposition 4.8 that, for each point x of K_{1}, there exist open sets V_{x} and W_{x} such that x ∈ V_{x}, K_{2} ⊂ W_{x} and V_{x} ∩ W_{x} = ∅. But then there exists a finite set {x_{1}, x_{2}, …, x_{r}} of points of K_{1} such that

K_{1} ⊂ V_{x_{1}} ∪ V_{x_{2}} ∪ ⋯ ∪ V_{x_{r}},

since K_{1} is compact. Define

U_{1} = V_{x_{1}} ∪ V_{x_{2}} ∪ ⋯ ∪ V_{x_{r}}, U_{2} = W_{x_{1}} ∩ W_{x_{2}} ∩ ⋯ ∩ W_{x_{r}}.

Then U_{1} and U_{2} are open sets, K_{1} ⊂ U_{1}, K_{2} ⊂ U_{2} and U_{1} ∩ U_{2} = ∅, as required.

Lemma 4.11 Let f: X → Y be a continuous function from a compact topological space X to a Hausdorff space Y. Then f(K) is closed in Y for every closed set K in X.

Proof If K is a closed set in X, then K is compact (Lemma 4.3), and therefore f(K) is compact (Lemma 4.4). But any compact subset of a Hausdorff space is closed (Corollary 4.9). Thus f(K) is closed in Y, as required.

Remark If the Hausdorff space Y in Lemma 4.11 is a metric space, then Proposition 4.7 may be used in place of Corollary 4.9 in the proof of the lemma.

Theorem 4.12 A continuous bijection f: X → Y from a compact topological space X to a Hausdorff space Y is a homeomorphism.

Proof Let g: Y → X be the inverse of the bijection f: X → Y. If U is open in X then X \ U is closed in X, and hence f(X \ U) is closed in Y, by Lemma 4.11. But f(X \ U) = g^{−1}(X \ U) = Y \ g^{−1}(U). It follows that g^{−1}(U) is open in Y for every open set U in X. Therefore g: Y → X is continuous, and thus f: X → Y is a homeomorphism.

We recall that a function f: X → Y from a topological space X to a topological space Y is said to be an identification map if it is surjective and satisfies the following condition: a subset U of Y is open in Y if and only if f^{−1}(U) is open in X.

Proposition 4.13 A continuous surjection f: X → Y from a compact topological space X to a Hausdorff space Y is an identification map.

Proof Let U be a subset of Y. We claim that Y \ U = f(K), where K = X \ f^{−1}(U). Clearly f(K) ⊂ Y \ U. Also, given any y ∈ Y \ U, there exists x ∈ X satisfying y = f(x), since f: X → Y is surjective. Moreover x ∈ K, since f(x) ∉ U. Thus Y \ U ⊂ f(K), and hence Y \ U = f(K), as claimed.

We must show that the set U is open in Y if and only if f^{−1}(U) is open in X. First suppose that f^{−1}(U) is open in X. Then K is closed in X, and hence f(K) is closed in Y, by Lemma 4.11. It follows that U is open in Y. Conversely if U is open in Y then f^{−1}(Y) is open in X, since f: X → Y is continuous. Thus the surjection f: X → Y is an identification map.

Example Let S^{1} be the unit circle in ℝ^{2}, defined by S^{1} = {(x, y) ∈ ℝ^{2} : x^{2} + y^{2} = 1}, and let q: [0, 1] → S^{1} be defined by q(t) = (cos 2πt, sin 2πt) for all t ∈ [0, 1]. It has been shown that the map q is an identification map. This also follows directly from the fact that q: [0, 1] → S^{1} is a continuous surjection from the compact space [0, 1] to the Hausdorff space S^{1}.

We shall show that a finite Cartesian product of compact spaces is compact. To prove this, we apply the following result, known as the Tube Lemma.

Lemma 4.14 Let X and Y be topological spaces, let K be a compact subset of Y, and U be an open set in X × Y. Let V = {x ∈ X : {x} × K ⊂ U}. Then V is an open set in X.

Proof Let x ∈ V. For each y ∈ K there exist open subsets D_{y} and E_{y} of X and Y respectively such that (x, y) ∈ D_{y} × E_{y} and D_{y} × E_{y} ⊂ U. Now there exists a finite set {y_{1}, y_{2}, …, y_{k}} of points of K such that K ⊂ E_{y_{1}} ∪ E_{y_{2}} ∪ ⋯ ∪ E_{y_{k}}, since K is compact. Set N_{x} = D_{y_{1}} ∩ D_{y_{2}} ∩ ⋯ ∩ D_{y_{k}}. Then N_{x} is an open set in X. Moreover

N_{x} × K ⊂ ⋃_{i=1}^{k}(N_{x} × E_{y_{i}}) ⊂ ⋃_{i=1}^{k}(D_{y_{i}} × E_{y_{i}}) ⊂ U,

so that N_{x} ⊂ V. It follows that V is the union of the open sets N_{x} for all x ∈ V. Thus V is itself an open set in X, as required.

Theorem 4.15 A Cartesian product of a finite number of compact spaces is itself compact.

Proof It suffices to prove that the product of two compact topological spaces X and Y is compact, since the general result then follows easily by induction on the number of compact spaces in the product.

Let U be an open cover of X × Y. We must show that this open cover possesses a finite subcover.

Let x be a point of X. The set {x}×Y is a compact subset of X ×Y, since it is the image of the compact space Y under the continuous map from Y to X × Y which sends y ∈ Y to (x, y), and the image of any compact set under a continuous map is itself compact (Lemma 4.4). Therefore there exists a finite collection U_{1}, U_{2}, …, U_{r} of open sets belonging to the open cover U such that {x} × Y is contained in U_{1} ∪ U_{2} ∪ ⋯ ∪ U_{r}. Let V_{x} denote the set of all points x′ of X for which {x′} × Y is contained in U_{1} ∪ U_{2} ∪ ⋯ ∪ U_{r}. Then x ∈ V_{x}, and Lemma 4.14 ensures That V_{x} is an open set in X. Note that V_{x} × Y is covered by finitely many of the open sets belonging to the open cover U.

Now {V_{x} : x ∈ X} is an open cover of the space X. It follows from the compactness of X that there exists a finite set {x_{1}, x_{2}, …, x_{r}} of points of X such that X = V_{x_{1}} ∪ V_{x_{2}} ∪ ⋯ ∪ V_{x_{r}}. Now X × Y is the union of the sets V_{x_{j}} × Y for j = 1, 2, …, r, and each of these sets can be covered by a finite collection of open sets belonging to the open cover U. On combining these finite collections, we obtain a finite collection of open sets belonging to U which covers X × Y. This shows that X × Y is compact.

Theorem 4.16 Let K be a subset of ℝ^{n}. Then K is compact if and only if K is both closed and bounded.

Proof Suppose that K is compact. Then K is closed, since ℝ^{n} is Hausdorff, and a compact subset of a Hausdorff space is closed (by Corollary 4.9). For each natural number m, let B_{m} be the open ball of radius m about the origin, given by B_{m} = {x ∈ ℝ^{n} : |x| < m}. Then {B_{m} : m ∈ ℕ} is an open cover of ℝ^{n}. It follows from the compactness of K that there exist natural numbers m_{1}, m_{2}, …, m_{k} such that K ⊂ B_{m_{1}} ∪ B_{m_{2}} ∪ ⋯ ∪ B_{m_{k}}. But then K ⊂ B_{M}, where M is the maximum of m_{1}, m_{2}, …, m_{k}, and thus K is bounded.

Conversely suppose that K is both closed and bounded. Then there exists some real number L such that K is contained within the closed cube C given by

C = {(x_{1}, x_{2}, …, x_{n}) ∈ ℝ^{n} : −L ≤ x_{j} ≤ L for j = 1, 2, …, n}.

Now the closed interval [−L, L] is compact by the Heine-Borel Theorem (Theorem 4.2), and C is the Cartesian product of n copies of the compact set [−L, L]. It follows from Theorem 4.15 that C is compact. But K is a closed subset of C, and a closed subset of a compact topological space is itself compact, by Lemma 4.3. Thus K is compact, as required.

4.2 The Lebesgue Lemma and Uniform Continuity

Definition Let X be a metric space with distance function d. A subset A of X is said to be bounded if there exists a non-negative real number K such that d(x, y) ≤ K for all x, y ∈ A. The smallest real number K with this property is referred to as the diameter of A, and is denoted by diam A. (Note that diam A is the supremum of the values of d(x, y) as x and y range over all points of A.)

Lemma 4.17 (Lebesgue Lemma) Let (X, d) be a compact metric space. Let U be an open cover of X. Then there exists a positive real number δ such that every subset of X whose diameter is less than δ is contained wholly within one of the open sets belonging to the open cover U.

Proof Every point of X is contained in at least one of the open sets belonging to the open cover U. It follows from this that, for each point x of X, there exists some δ_{x} > 0 such that the open ball B(x, 2δ_{x}) of radius 2δ_{x} about the point x is contained wholly within one of the open sets belonging to the open cover U. But then the collection consisting of the open balls B(x, δ_{x}) of radius δ_{x} about the points x of X forms an open cover of the compact space X. Therefore there exists a finite set x_{1}, x_{2}, …, x_{r} of points of X such that

B(x_{1}, δ_{1}) ∪ B(x_{2}, δ_{2}) ∪ ⋯ ∪ B(x_{r}, δ_{r}) = X,

where δ_{i} = δ_{x_{i}} for i = 1, 2, …, r. Let δ > 0 be given by

δ = minimum(δ_{1}, δ_{2}, …, δ_{r}).

Suppose that A is a subset of X whose diameter is less than δ. Let u be a point of A. Then u belongs to B(x_{i}, δ_{i}) for some integer i between 1 and r. But then it follows that A ⊂ B(x_{i}, 2δ_{i}), since, for each point v of A,

d(v, x_{i}) ≤ d(v, u) + d(u, x_{i}) < δ + δ_{i} ≤ 2δ_{i}.

But B(x_{i}, 2δ_{i}) is contained wholly within one of the open sets belonging to the open cover U. Thus A is contained wholly within one of the open sets belonging to U, as required.

Let U be an open cover of a compact metric space X. A Lebesgue number for the open cover U is a positive real number δ such that every subset of X whose diameter is less than δ is contained wholly within one of the open sets belonging to the open cover U. The Lebesgue Lemma thus states that there exists a Lebesgue number for every open cover of a compact metric space.

Let X and Y be metric spaces with distance functions d_{X} and d_{Y} respectively, and let f: X → Y be a function from X to Y. The function f is said to be uniformly continuous on X if and only if, given ε > 0, there exists some δ > 0 such that d_{Y} (f(x), f(x′)) < ε for all points x and x′ of X satisfying d_{X}(x, x′) < δ. (The value of δ should be independent of both x and x′.)

Theorem 4.18 Let X and Y be metric spaces. Suppose that X is compact. Then every continuous function from X to Y is uniformly continuous.

Proof Let d_{X} and d_{Y} denote the distance functions for the metric spaces X and Y respectively. Let f: X → Y be a continuous function from X to Y. We must show that f is uniformly continuous.

Let ε > 0 be given. For each y ∈ Y, define

V_{y} = \{x ∈ X : d_{Y} (f(x), y) < \frac{1}{2}ε\}.

Note that V_{y} = f^{−1} (B_{Y} (y, \frac{1}{2}ε)), where B_{Y} (y, \frac{1}{2}ε) denotes the open ball of radius \frac{1}{2}ε about y in Y. Now the open ball B_{Y} (y, \frac{1}{2}ε) is an open set in Y, and f is continuous. Therefore V_{y} is open in X for all y ∈ Y. Note that x ∈ V_{f(x)} for all x ∈ X.

Now {V_{y} : y ∈ Y } is an open cover of the compact metric space X. It follows from the Lebesgue Lemma (Lemma 4.17) that there exists some δ > 0 such that every subset of X whose diameter is less than δ is a subset of some set V_{y}. Let x and x′ be points of X satisfying d_{X}(x, x′) < δ. The diameter of the set {x, x′} is d_{X}(x, x′), which is less than δ. Therefore there exists some y ∈ Y such that x ∈ V_{y} and x′ ∈ V_{y}. But then d_{Y} (f(x), y) < \frac{1}{2}ε and d_{Y} (f(x′), y) < \frac{1}{2}ε, and hence

d_{Y} (f(x), f(x′)) ≤ d_{Y} (f(x), y) + d_{Y} (y, f(x′)) < ε.

This shows that f: X → Y is uniformly continuous, as required.

Let K be a closed bounded subset of ℝ^{n}. It follows from Theorem 4.16) and Theorem 4.18 that any continuous function f: K → ℝ^{k} is uniformly continuous.

5 Connectedness

5.1 Connected Topological Spaces

Definition A topological space X is said to be connected if the empty set ∅ and the whole space X are the only subsets of X that are both open and closed.

Lemma 5.1 A topological space X is connected if and only if it has the following property: if U and V are non-empty open sets in X such that X = U ∪ V, then U ∩ V is non-empty.

Proof If U is a subset of X that is both open and closed, and if V = X \ U, then U and V are both open, U ∪ V = X and U ∩ V = ∅. Conversely if U and V are open subsets of X satisfying U ∪ V = X and U ∩ V = ∅, then U = X \V, and hence U is both open and closed. Thus a topological space X is connected if and only if there do not exist non-empty open sets U and V such that U ∪ V = X and U ∩ V = ∅. The result follows.

Let ℤ be the set of integers with the usual topology (i.e., the subspace topology on ℤ induced by the usual topology on ℝ). Then {n} is open for all n ∈ ℤ, since

\{n\} = ℤ ∩ \{t ∈ ℝ : |t − n| < \frac{1}{2}\}.

It follows that every subset of ℤ is open (since it is a union of sets consisting of a single element, and any union of open sets is open). It follows that a function f: X → ℤ on a topological space X is continuous if and only if f^{−1}(V) is open in X for any subset V of ℤ. We use this fact in the proof of the next theorem.

Proposition 5.2 A topological space X is connected if and only if every continuous function f: X → ℤ from X to the set ℤ of integers is constant.

Proof Suppose that X is connected. Let f: X → ℤ be a continuous function. Choose n ∈ f(X), and let

U = {x ∈ X : f(x) = n}, V = {x ∈ X : f(x) ≠ n}.

Then U and V are the preimages of the open subsets {n} and ℤ \ {n} of ℤ, and therefore both U and V are open in X. Moreover U ∩ V = ∅, and X = U ∪ V. It follows that V = X \ U, and thus U is both open and closed. Moreover U is non-empty, since n ∈ f(X). It follows from the connectedness of X that U = X, so that f: X → ℤ is constant, with value n.

Conversely suppose that every continuous function f: X → ℤ is constant. Let S be a subset of X which is both open and closed. Let f: X → ℤ be defined by

f(x) = \{ \begin{matrix}1 & if x ∈ S; \\ 0 & if x ∉ S. \end{matrix}

Now the preimage of any subset of ℤ under f is one of the open sets ∅, S, X \ S and X. Therefore the function f is continuous. But then the function f is constant, so that either S = ∅ or S = X. This shows that X is connected.

Lemma 5.3 The closed interval [a, b] is connected, for all real numbers a and b satisfying a ≤ b.

Proof Let f: [a, b] → ℤ be a continuous integer-valued function on [a, b]. We show that f is constant on [a, b]. Indeed suppose that f were not constant. Then f(τ) ≠ f(a) for some τ ∈ [a, b]. But the Intermediate Value Theorem would then ensure that, given any real number c between f(a) and f(τ), there would exist some t ∈ [a, τ] for which f(t) = c, and this is clearly impossible, since f is integer-valued. Thus f must be constant on [a, b]. We now deduce from Proposition 5.2 that [a, b] is connected.

Example Let X = {(x, y) ∈ ℝ^{2} : x ≠ 0}. The topological space X is not connected. Indeed if f: X → ℤ is defined by

f(x, y) = \{ \begin{matrix}1 & if x > 0, \\ −1 & if x < 0, \end{matrix}

then f is continuous on X but is not constant.

A concept closely related to that of connectedness is path-connectedness. Let x_{0} and x_{1} be points in a topological space X. A path in X from x_{0} to x_{1} is defined to be a continuous function γ: [0, 1] → X such that γ(0) = x_{0} and γ(1) = x_{1}. A topological space X is said to be path-connected if and only if, given any two points x_{0} and x_{1} of X, there exists a path in X from x_{0} to x_{1}.

Proposition 5.4 Every path-connected topological space is connected.

Proof Let X be a path-connected topological space, and let f: X → ℤ be a continuous integer-valued function on X. If x_{0} and x_{1} are any two points of X then there exists a path γ: [0, 1] → X such that γ(0) = x_{0} and γ(1) = x_{1}. But then f ∘ γ: [0, 1] → ℤ is a continuous integer-valued function on [0, 1]. But [0, 1] is connected (Lemma 5.3), therefore f ∘ γ is constant (Proposition 5.2). It follows that f(x_{0}) = f(x_{1}). Thus every continuous integer-valued function on X is constant. Therefore X is connected, by Proposition 5.2.

The topological spaces ℝ, ℂ and ℝ^{n} are all path-connected. Indeed, given any two points of one of these spaces, the straight line segment joining these two points is a continuous path from one point to the other. Also the n-sphere S^{n} is path-connected for all n > 0. We conclude that these topological spaces are connected.

Let A be a subset of a topological space X. Using Lemma 5.1 and the definition of the subspace topology, we see that A is connected if and only if the following condition is satisfied:

• if U and V are open sets in X such that A∩U and A∩V are non-empty and A ⊂ U ∪ V then A ∩ U ∩ V is also non-empty.

Lemma 5.5 Let X be a topological space and let A be a connected subset of X. Then the closure \overline{A} of A is connected.

Proof It follows from the definition of the closure of A that \overline{A} ⊂ F for any closed subset F of X for which A ⊂ F. On taking F to be the complement of some open set U, we deduce that \overline{A} ∩ U = ∅ for any open set U for which A ∩ U = ∅. Thus if U is an open set in X and if \overline{A} ∩ U is non-empty then A ∩ U must also be non-empty.

Now let U and V be open sets in X such that \overline{A} ∩ U and \overline{A} ∩ V are non-empty and \overline{A} ⊂ U ∪ V. Then A ∩ U and A ∩ V are non-empty, and A ⊂ U ∪ V. But A is connected. Therefore A ∩ U ∩ V is non-empty, and thus \overline{A} ∩ U ∩ V is non-empty. This shows that \overline{A} is connected.

Lemma 5.6 Let f: X → Y be a continuous function between topological spaces X and Y, and let A be a connected subset of X. Then f(A) is connected.

Proof Let g: f(A) → ℤ be any continuous integer-valued function on f(A). Then g ∘ f: A → ℤ is a continuous integer-valued function on A. It follows from Proposition 5.2 that g ∘ f is constant on A. Therefore g is constant on f(A). We deduce from Proposition 5.2 that f(A) is connected.

Lemma 5.7 The Cartesian product X × Y of connected topological spaces X and Y is itself connected.

Proof Let f: X×Y → ℤ be a continuous integer-valued function from X×Y to Z. Choose x_{0} ∈ X and y_{0} ∈ Y. The function x ↦ f(x, y_{0}) is continuous on X, and is thus constant. Therefore f(x, y_{0}) = f(x_{0}, y_{0}) for all x ∈ X. Now fix x. The function y ↦ f(x, y) is continuous on Y, and is thus constant. Therefore

f(x, y) = f(x, y_{0}) = f(x_{0}, y_{0})

for all x ∈ X and y ∈ Y. We deduce from Proposition 5.2 that X × Y is connected.

We deduce immediately that a finite Cartesian product of connected topological spaces is connected.

Proposition 5.8 Let X be a topological space. For each x ∈ X, let S_{x} be the union of all connected subsets of X that contain x. Then

(i) S_{x} is connected,

(ii) S_{x} is closed,

(iii) if x, y ∈ X, then either S_{x} = S_{y}, or else S_{x} ∩ S_{y} = ∅.

Proof Let f: S_{x} → ℤ be a continuous integer-valued function on S_{x}, for some x ∈ X. Let y be any point of S_{x}. Then, by definition of S_{x}, there exists some connected set A containing both x and y. But then f is constant on A, and thus f(x) = f(y). This shows that the function f is constant on S_{x}. We deduce that S_{x} is connected. This proves (i). Moreover the closure \overline{S_{x}} is connected, by Lemma 5.5. Therefore \overline{S_{x}} ⊂ S_{x}. This shows that S_{x} is closed, proving (ii).

Finally, suppose that x and y are points of X for which S_{x} ∩ S_{y} ≠ ∅. Let f: S_{x} ∪ S_{y} → ℤ be any continuous integer-valued function on S_{x} ∪ S_{y}. Then f is constant on both S_{x} and S_{y}. Moreover the value of f on S_{x} must agree with that on S_{y}, since S_{x} ∩ S_{y} is non-empty. We deduce that f is constant on S_{x} ∪ S_{y}. Thus S_{x} ∪ S_{y} is a connected set containing both x and y, and thus S_{x} ∪ S_{y} ⊂ S_{x} and S_{x} ∪ S_{y} ⊂ S_{y}, by definition of S_{x} and S_{y}. We conclude that S_{x} = S_{y}. This proves (iii).

Given any topological space X, the connected subsets S_{x} of X defined as in the statement of Proposition 5.8 are referred to as the connected components of X. We see from Proposition 5.8, part (iii) that the topological space X is the disjoint union of its connected components.

Example The connected components of {(x, y) ∈ ℝ^{2} : x ≠ 0} are

{(x, y) ∈ ℝ^{2} : x > 0} and {(x, y) ∈ ℝ^{2} : x < 0}.

Example The connected components of

\{t ∈ ℝ : |t − n| < \frac{1}{2} for some integer n\}.

are the sets J_{n} for all n ∈ ℤ, where J_{n} = (n − \frac{1}{2}, n + \frac{1}{2}).