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--- math104-s22:notes:lecture_17 [2022/03/14 23:05]
pzhou created
+++ math104-s22:notes:lecture_17 [2022/03/14 23:06] (current)
pzhou
@@ Line 16: / Line 16: @@
 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.
+===== 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.
+Theorem: if  $f: X \to Y$  is continuous, and  $X$  is compact, then  $f$  is uniformly continuous.
+Proof: Fix $\epsilon>0 $. For each$ x \in X $, let$ r_x > 0 $be small enough such that$ f(B_{2r_x}(x)) \In B_{\epsilon/2}(f(x)) $. Let$ B_x = B_{r_x}(x) $. The,$ \{B_x: x \in X\} $forms an open cover of$ X $. Pass to finite subcover,$ X = \cup_{i=1}^N B_{x_i} $. Then let$ \delta = \min \{r_{x_i}\} $. Then, suppose$ x $and$ x' $has distance less than$ \delta $, then$ x $is contained in some$ B_{r_{x_i}}(x_i)$, and  $x' \in B_{2 r_{x_i}}(x_i)$ , thus $f(x), f(x') \in B_{\epsilon/2}(f(x_i))$, thus  $f(x)$  and $f(x') $has distance less than$ \epsilon$.
 ===== Discontinuity =====
@@ Line 33: / Line 43: @@
   * otherwise, it is called a second kind.
-===== Uniform Continuity =====
+===== Sequences of functions =====
-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.
-Theorem: if  $f: X \to Y$  is continuous, and  $X$  is compact, then  $f$  is uniformly continuous.
-Proof: Fix $\epsilon>0 $. For each$ x \in X $, let$ r_x > 0 $be small enough such that$ f(B_{2r_x}(x)) \In B_{\epsilon/2}(f(x)) $. Let$ B_x = B_{r_x}(x) $. The,$ \{B_x: x \in X\} $forms an open cover of$ X $. Pass to finite subcover,$ X = \cup_{i=1}^N B_{x_i} $. Then let$ \delta = \min \{r_{x_i}\} $. Then, suppose$ x $and$ x' $has distance less than$ \delta $, then$ x $is contained in some$ B_{r_{x_i}}(x_i)$, and  $x' \in B_{2 r_{x_i}}(x_i)$ , thus $f(x), f(x') \in B_{\epsilon/2}(f(x_i))$, thus  $f(x)$  and $f(x') $has distance less than$ \epsilon$.

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