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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 =====
@@ Line 41: / Line 41: @@
  $\Leftarrow$ : 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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