2 We can take any enumerable set in place of ℕ. Let us take the rationals ℚ.

Now let us associate to each real x ∈ ℝ the set of rationals

S(x) = {r ∈ ℚ : r < x}.

It is clear that set of subsets S(x) ⊂ ℚ is totally ordered; and it is clear that this set of subsets has the same cardinality as ℝ, which we know is uncountable.

3 The monotonic function f(x) is discontinuous at c if and only if

lim_{x→c−} f(x) < lim_{x→c+} f(x),

ie

D(c) = lim_{x→c+} f(x) − lim_{x→c−} f(x) > 0.

4 Let

a(n) = \frac{n^{n}}{e^{n}} .

Then

\frac{a(n + 1)}{a(n)} = (\frac{n + 1}{n})^{n} \frac{n + 1}{e}

= (1 + \frac{1}{n})^{n} \frac{n + 1}{e}.

8 Consider the functional analogue:

\frac{dy}{dx} = \frac{y^{2}}{x^{2}} ,

giving

\frac{1}{y} = C + \frac{1}{x}.

Given the initial condition x = 1, y = c,

C = \frac{1}{c} − 1.

Thus

y = \frac{1}{1/c − 1 + 1/x} = \frac{cx}{x − cx + c}.

Let us set

f(x) = \frac{cx}{x − cx + c}.

Then

f(1) = \frac{c}{1 − c + c} = c,

and

f(n) → \frac{c}{1 − c}

as n → ∞. We want to show that

f(n) ≥ a(n).

We have

f(n + 1) − f(n) = f′(n + θ).

Now

f′(x) = \frac{f(x)^{2}}{x^{2}}

= \frac{c^{2}}{(1 − c)x + c}.

Thus f′(x) is decreasing. Hence

f′(n + θ) ≥ f′(n),

and so

f(n + 1) − f(n) ≥ f′(n) = \frac{f(n)^{2}}{n^{2}} .

Suppose

f(n) ≥ a_{n}.

Then

f(n + 1) ≥ f(n) + \frac{f(n)^{2}}{n^{2}}

≥ a_{n} + \frac{a_{n}^{2}}{n^{2}}

= a_{n+1}.

It follows by induction that

a_{n} ≤ f(n)

for all n. Hence

a_{n} ≤ lim f(n) = \frac{c}{1 − c}.