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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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