Point-set topology, a short review

Here, we work through (in review) selected exercises from Chapter 2 of Charles Pugh's Real Mathematical Analysis, in review for a second course on real analysis I am planning to take this coming (junior) fall. At the end, we have a discussion of the Riemann integral and its limitations, in anticipation of the Lebesgue integral, which is introduced in the aforementioned course.

(7) Write the limit of the sequence as \(p \). Fix some arbitrary \( \epsilon > 0 \) and find \( k \in \mathbb{N} \) such that \( |p_j - p| < \epsilon \ \forall j > k\) and split \( (p_n) \) into two sub-sequences \( (p_0, \cdots, p_k) \) and \( (p_{k+1},p_{k+2}, \cdots) \). Then see that we know all points in the second sequence are within \( \epsilon \) of \( p\) by construction since \( (p_n) \) converges to \(p \). Then see that since the first sequence is finite, we can find a maximum distance of any of those points from \( p \), and call this distance \( d \in \mathbb{R} \). Then we can let \( \delta = \text{max}(d, \epsilon)+1 \) to get that \(p_i \in B_\delta(p) \) for all \( p_i \in (p_n) \), as required.

(10) Consider, WLOG (the same idea goes for decreasing sequences) an increasing sequence, \( (p_n) \in \mathbb{R} \). Assume it is bounded. Then by the LUB property of the real numbers, it has a supremum, call it \( p \). To see this is in fact the limit of the sequence, fix an \( \epsilon > 0 \) and see that eventually \( (p_n) \) must be \( \epsilon-\text{close} \) to the LUB because if it was never that close, you could find a tighter upper bound, \( p' \), for the sequence. This would contradict the fact that \( p\) is the least upper bound. Therefore \( (p_n) \) converges to \( p \), as required.

(11)(a) Let \( (x_n) \in \mathbb{R} \) be a sequence, and we seek to find a monotone subsequence. Call a natural number \( x_n \) "nice" if \(m > n \implies x_m > x_n \). That is, if every number after \( x_n \) is strictly larger than it. Then see that either our sequence \( (x_n) \) has finitely many "nice" numbers or not.

If it does, take some \( x_k, k > n \) for \(x_n \) the final "nice" number. Then we are guaranteed that there exists some \( x_{k+1} \) that is smaller than or equal to \( x_k \) because \( x_k \) is not nice. And same with \( x_{k+1} \) inductively, to generate an infinite sequence of non-increasing numbers.

If \( (x_n) \) has infinitely many "nice" numbers, then those form a non-decreasing sequence since each is greater than the last, by definition of "nice." Therefore, either way, our sequence \( (x_n) \) has a monotone subsequence, as required.

(b) Every subsequence of a bounded sequence is also bounded, and so if we take a subsequence of \( (x_n) \) that is monotone, we can use (10) to see that subsequence converges, and thus \( (x_n) \) has a convergent subsequence.

(c) The fact that every bounded sequence has a convergent subsequence that we just proved above is exactly equivalent to Bolzano-Weierstrass for \( n=1 \).

(d) Heine-Borel states that sequentially compact \( \iff \) closed and bounded. A metric space \( (X, d) \) is sequentially compact if every sequence within it converges to a limit in the same metric space. BW just gives that every bounded sequence in the reals has a convergent subsequence, which isn't enough to establish sequential compactness of the reals due to the boundedness condition (in fact, the reals do not comprise a compact set), so this does not help us get a proof of Heine-Borel.

(13) Fix arbitrary \( \epsilon > 0 \) and \( x_0 \in M \). We seek a \( \delta \) such that \(|x-x_0| < \delta \implies |f| < \epsilon \)







[TODO. The Riemann Integral and its limitations]

[TODO. Theorems to recap: Rolle, Darboux, IVT, EVT, MVT, Bolzano-Weierstrass, Heine-Borel, Arzela-Ascoli, Inverse Function Theorem, Implicit Function Theorem]