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--- math104-s22:notes:lecture_16 [2022/03/09 23:38]
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+++ math104-s22:notes:lecture_16 [2022/03/14 22:58] (current)
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-====== Lecture 16 - 17 ======
+====== Lecture 16 ======
 ===== connectedness =====
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 $\Leftarrow$: suppose $E$ is not connected, thus $E = G \sqcup H$ with $G, H$ open in $E$ (hence both $G, H$ are closed in $E$, since $G = E \RM H$, $H = E\RM G$). Let $G' \In X$ be a closed subset, such that $G' \cap E = G$, then $G' \cap H = \emptyset$. In particular, $\bar G \In G'$, hence $\bar G \cap H = \emptyset$. Similarly $\bar H \cap G=\emptyset$, thus $G,H$ are separated.
-===== Continuous Maps and Compactness, Connectedness =====
-Prop: If $f: X \to Y$ is continuous, and $K \In X$ is compact, then $f(K)$ is compact. \\
-Proof: any open cover of $f(K)$ can be pulled back to be an open cover of $K$, then we can pick a finite subcover in the domain, and the corresponding cover in the target forms a cover of $f(K)$.
-We can also prove using sequential compactness. To see any sequence in $f(K)$ subconverge, we can lift that sequence back to $K$, find a convergent subsequence, and push them back to $f(K)$.
-Lemma: if $f: X \to Y$ is continuous, then for any $E \In X$, $f|_E: E \to Y$ is continuous.
-Lemma: if $f: X \to Y$ is continuous, then $f: X \to f(X)$ is continuous.
-Prop: If $f: X \to Y$ is continuous, and $K \In X$ is connected, then $f(K)$ is connected. \\
-Pf: first, note that map $f: K \to f(K)$ is also continuous. If $f(K)$ is the disjoint union of two non-empty open subsets, $f(K) = U \cap V$, then $K = f^{-1}(U) \cap f^{-1}(V)$ the disjoint union of two non-empty open subsets of $K$.
-Intermediate value theorem: if $[a,b] \In \R$, and $f: \R \to \R$ is continuous, then $f([a,b])$ is also a closed interval. \\
-Proof: since $[a,b]$ is compact, hence $f([a,b])$ is compact, hence closed. Since $[a,b]$ is connected, hence $f([a,b])$ is connected, hence an interval, a closed interval.
-===== Discontinuity =====
-Now we will leave the safe world of continuous functions. We consider more subtle cases of maps.
-Def: Let $f: X \to Y$ be any map, and let $x \in X$ be a point, we say **$f$ is continuous at $x$**, if for any $\epsilon>0$, there exists $\delta>0$, such that $f(B_\delta(x)) \In B_\epsilon(f(x))$.
-Def: Let $E \In X$ and $f: E \to Y$. Suppose $x \in \bar E$. We say $\lim_{p \to x} f(p) = y$ if for any convergent sequence $p_n \to x$ with $p_n \in E$, we have $f(p_n) \to y$.
-note that $x$ may not be in $E$.
-Prop: $f: X \to Y$ is continuous at $x \in X$, if and only if $\lim_{p \to x} f(p) = f(x)$.
-If $f: (a,b) \to \R$ is a function, and $f$ is not continuous at some $x \in (a,b)$, then
-  * if $\lim_{t \to x-} f(t)$ and $\lim_{t \to x+} f(t)$ both exists, but does not equal to $f(x)$, we say this is a simple discontinuity, or first kind discontinuity
-  * otherwise, it is called a second kind.
-===== Uniform Continuity =====
-We say a function $f: X \to Y$ is uniformly continuous, if for any $\epsilon >0$, there exists $\delta > 0$, such that for any pair $x_1, x_2 \in X$ with $d(x_1, x_2)<\delta$, we have $d(f(x_1), f(x_2))< \epsilon$.
-For example, the function $f: (0, 1) \to \R$ $f(x) =1 /x$ is continuous but not uniformly continuous.

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