From 0a6dc8cf69402d3da2dbf6ebacf3b271284832dc Mon Sep 17 00:00:00 2001
From: Trance-0 <60459821+Trance-0@users.noreply.github.com>
Date: Wed, 19 Feb 2025 12:48:37 -0600
Subject: [PATCH] Update Math4121_L15.md

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-# Lecture 15
\ No newline at end of file
+# Math 4121 Lecture 15
+
+## Continue on patches for Riemann integral
+
+### Hankel's conjecture
+
+If $f$ is pointwise discontinuous (set of points of continuity is dense), then $f\in\mathcal{R}$.
+
+Why the conjecture is believable:
+
+- $\{w(f)\leq \sigma\}=\bigcup_{a\in D} I_a$, so then $S_{\sigma}$ contains no intervals, so the outer content _maybe_ zero.
+
+However, it turns out that the complement of the set of points of continuity can be large.
+
+#### Definition: Accumulation point
+
+Given a set $S$, $x$ is an accumulation point of $S$ if every open interval containing $x$ also contains infinitely many points of $S$.
+
+The derived set of $S$, denoted $S'$, is the set of all accumulation points of $S$.
+
+#### Lemma
+
+$$
+c_e(S)\leq c_e(S')
+$$
+
+Proof:
+
+Given $\epsilon > 0$, we can find open intervals $I_1, I_2, \ldots$ such that $S'\subseteq \bigcup_{i=1}^{n} I_i$ and
+
+$$
+\sum_{i=1}^{n} \ell(I_i) \leq c_e(S') + \frac{\epsilon}{2}
+$$
+
+So $S\setminus \bigcup_{i=1}^{n} I_i$ contains only finitely many points, say $N$, so we can cover $S\setminus \bigcup_{i=1}^{n} I_i$ by $N$ intervals of length $\frac{\epsilon}{2N}$. We call these intervals $J_1, J_2, \ldots, J_N$. Then $S$ is covered by the intervals $C=\bigcup_{i=1}^{n} I_i \cup \bigcup_{i=1}^{N} J_i$. and $\ell(C)\leq c_e(S') + \frac{\epsilon}{2} + \frac{N\epsilon}{2N}=c_e(S')+\epsilon$.
+
+So $c_e(S)\leq \ell(C)\leq c_e(S')+\epsilon$.
+
+EOP
+
+#### Corollary: sef of first species
+
+There exists $n\in \mathbb{N}$ such that $S^{(n)'}=\phi$.
+
+If $S$ is of the first species, then $c_e(S)=0$. This leads to further credence to Hankel's conjecture, but it begs the question is the complement of $\bigcup_{a\in D} I_a$ indeed of first species?
+
+#### Theorem (Baire Category Theorem)
+
+Every open set in $\mathbb{R}$ is a countable union of disjoint open intervals.
+
+Proof:
+
+Let $U$ be open and for each $t\in U$, set $a=\inf \{x:(x,t]\subseteq U\}$ and $b=\sup \{x:[t,x)\subseteq U\}$.
+
+We define $I(t)=(a,b)$, $U\subseteq \bigcup_{t\in U} I(t)$.
+
+Notice that  if $I(t)\cap I(s)\neq \emptyset$, then there exists rational $r\in U$ such that $I(t)\cap I(s)\subseteq (r,r)$.
+
+> Take a dense, contable set in $[0,1]$, say $D=\{a_n\}_{n=1}^{\infty}$. Let $\epsilon>0$ and define $I_n=(a_n-\frac{\epsilon}{2^{n+1}}, a_n+\frac{\epsilon}{2^{n+1}})$. Can the entire interval $[0,1]$ be placed inside a union of intervals whose lengths add up to $\epsilon$?
+
+$$
+\sum_{n=1}^{\infty} \ell(I_n) = \sum_{n=1}^{\infty} \frac{2\epsilon}{2^{n+1}} = \epsilon
+$$
+
+_Harnack believed that the complement of a countable union of intervals is another countable union of intervals._
+
+Borel found the flaw in Harnack's argument. In fact, $[0,1]\setminus \bigcup_{n=1}^{\infty} I_n$ is uncountable.
+
+#### Theorem (Heine-Borel)
+
+If $\{U_\alpha\}_{\alpha\in A}$ is a collection of open sets (countable or uncountable) such that
+
+$$
+[a,b]\subseteq \bigcup_{\alpha\in A} U_\alpha
+$$
+
+then there exists a finite or countable subcollection $\{U_{\alpha_n}\}_{n=1}^{\infty}$ such that
+
+$$
+[a,b]\subseteq \bigcup_{n=1}^{\infty} U_{\alpha_n}
+$$
+
+Continue on next lecture.