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Assume A is an open set and B is a closed set. Determine if the following sets are definitely
open, definitely closed, both, or neither.
(e) A̅^c ∩ A̅^c̅

(e) Neither in general. Note that A̅^c =/= A̅^c̅ consider how A = {1/n : n ∈ N} has A̅^c̅ = R
A̅^c =/= R.

This is actually definitely open. As A is open, A^c is closed, thus its closure A̅^c̅ is itself, i.e. A̅^c̅ = A^c.
Next, a set is always a subset of it's own closure. Thus A ⊆ A̅. Taking complements reverses the containment: A̅^c ⊆ A^c. Thus A̅^c ⊆ A̅^c̅
Thus A̅^c ∩ A̅^c̅ = A̅^c which is open, as A̅ is closed.

Hey :)
In your solution to 2.5.5:

(a2, a3, . . .) Is a convergent subsequence, so obviously...

but actually (a2, a3, . . .) isn't necessary convergent, because all you know about (an) is that it is just bounded, not necessarily convergent.

For part b), the negation should be "For every real number x > 0 there exists an n ∈ N such that x >= 1/n" (the current version has x < 1/n).

Just wanted to let you know that (b) is wrong. (a_n+2)^2 - 4 = a_n^2 + 4a_n, not a_n^2 + 2a_n

Hey!

There is a mistake in the proof where the inequality x_n^2 \ge 2 is applied for the first time. Since you have the x_n^2 in the denominator, the application of inequality gives you inequality in wrong direction, so that instead of

    \frac 14\left(\frac{(x_n^2 + 2)^2}{x_n^2}\right)
    \ge \frac 14\left(\frac{(x_n^2 + 2)^2}{2}\right)

there should be

    \frac 14\left(\frac{(x_n^2 + 2)^2}{x_n^2}\right)
    \le \frac 14\left(\frac{(x_n^2 + 2)^2}{2}\right)

which is true, but unhelpful.

Here is an idea to prove this:

    x_{n+1}^2 = \frac 14\left(x_n + \frac{2}{x_n}\right)^2
    = \frac 14 x_n^2 + 1 + \frac{1}{x_n^2}
    = \frac 14 x_n^2 - 1 + \frac{1}{x_n^2} + 2
    = \left( \frac{x_n}{2} - \frac{1}{x_n} \right)^2 + 2
    \ge 2

Let N₁ be large enough that n > m ≥ N₁ implies a_n > a_m − ϵ.

Since (a_n) is bounded we can set s = sup{a_n} applying Lemma 1.3.8 tells us there exists
an N > N₁ such that a_N > s − ϵ.

This is false and the proof falls apart at the point of assumption that there is N > N₁ satisfying the quasi-increasing condition and Lemma 1.3.8 for the same given ϵ. This makes the assumption that quasi-increasing implies (eventually-)monotonically increasing. See the sequence (a_n) = (-1)ⁿ / n analyzed in 2.6.6a as a direct counter-example to the offered proof.

Thus we can't assume that (a_n) converges sup{a_n} in general. Convergence to the limit superior, the existence of which is guaranteed by (a_n) being bounded, is true but perhaps circular reasoning.

It's better to apply Bolzano-Weierstrass and show that the convergent subsequence (a_nk) and quasi-increasingness of (a_n) together imply (a_n) is itself convergent.

Your solutions have helped and will help me so much.
But I think I found two errors in your solution.
Could you check them?

Exercise 2.3.4
(a_n) -> 0 but your answer is when a_n -> inf
Exercise 2.3.7 (c)
if b_n = 1/n, (b_n) -> 0 and (1/b_n) diverges.

Hi Ulisse,

Thank you for your solutions. They are really helpful!
But I think there are some problems for exercise 2.6.6(b) (c).
The sequence you mentioned in 2.6.6(b), i.e. a_n = (2, 1/2, 3, 1/3, ...) is not quasi-increasing.

For example, given ε = 1/2, we can not find a N, such that whenever n > m ≥ N, a_n > a_m − ε .
This is because, say let m = 2k+1, n = 2k+2,
then a_{2k+1} = (k+2), a_{2k+2} = 1/(k+2).
So a_n - a_m = a_{2k+2} - a_{2k+1} = 1/(k+2) - (k+2) < -ε = -1/2.
So a_n < a_m - ε.
Which does not satisfy the definition of quasi-increasing, i.e. "whenever n > m ≥ N it follows that a_n > a_m − ε".

Hi,

Thank you so much for the solutions! They have been really helpful as a reference.

I just wanted to point out a possible error in the proof for Exercise 1.6.10 (c) - specifically, the step where any antichain of P(N) is expressed as the countable union of all A_L, where A_L is every subset of N in A with size L for some L in N. This only accounts for finite-sized elements of P(N), but not for infinite-sized elements of P(N) that can possibly be in A. One example would be the set of all even natural numbers, which doesn't belong in any A_L, but can exist as an element of an antichain.

• Fix \epsilon > 0. Since \lim_{n \rightarrow \infty} a_{mn} = b_m for all m \in \mathbb{N}, there exists N_m \in \mathbb{N} such that for all n \geq N_m, |a_{mn} - b_m| < \epsilon.
• Notably, the choice of lower bound N_m depends on m; i.e. it may be different for different m. Because of this, I am not following how we can choose a sufficiently large lower bound N when proving the limit \lim_{m \rightarrow \infty} b_m = a.
• You seem to be implicitly taking a maximum over the sequence N_1, N_2, \dots, N_m, \dots, but that requires showing that the sequence is bounded (e.g. by showing that it converges). However, I am not sure how to do that.

Let me know if you want me to clarify my concern in any way.

Since it is not necessary that f(c) and g(c) exist (a limit point of a function is not necessarily in the domain of the function), they can not be used in the proof.

You should substitute f(c) and g(c) by L and M resp., using the notations in part a of the proof.

f(x) = f ′(x) = 0. We have
...
therefore f ′(x) = lim f '_n(x) everywhere.

This misses that f '_n(0) = 1 for all n ∈ ℕ. Therefore lim_n->∞ f '_n(0) = 1.

Thus f '(x) = lim f '_n(x) on ℝ \ {0} is the solution.

In Exercise 1.3.10 (b), we are asked to show that the Cut Property implies Axiom of Completeness. In your answer, you first construct set A and B, and then show that the c found by the Cut Property is the lowest upper bound of A. However, I think the argument in step (ii) is not rigorous because B is not a set of all upper bounds of E, but of all numbers that are strictly larger than elements in E. Therefore c <= b does not imply that c is the lowest upper bound.

In my opinion, the solution is wrong.
I think that :

The formula in (a) does not extend to the infinite case.

Take A_{n} = [n, n +1], then A_{n} is obviously nonempty and bounded above, but the infinite union becomes [0, + infty[ which is not bounded above .

Missing the case where c = 0.
It should be "there exists a ∈ A such that s′/c < a meaning s′ < ca thus s′ cannot be an upper bound on cA" (the direction of the inequality sign is incorrect in the current version)

Hi Ulisse,

I think this question asks 3 questions for each of (a), (b), (c).
E.g. for (a), the 3 statements are
(a.1) Every finite set has a maximum.
(a.2) Every compact set has a maximum.
(a.3) Every closed set has a maximum.

It seems you did not answer for the "finite" case.

Also for (b), the "sum" of two closed sets might not be closed. Here is a counterexample.

The question is to disprove 'For all colleges in the United States, there exists a professor who gives every student
a grade of either A or B.'

The answer that is given is 'Find a collage in the United States with a professor who has given a grade other then
an A or B.', but this does not seem correct to me, you would only check if there exists a professor that only not only grades A or B. But the statement would not be disproven if a college has a professor that only grades A or B, but also a professor which gives other grades. Wouldn't you need to find a college in the United States with no professors that only grade A or B.