Section 21: Problem 8 Solution

Working problems is a crucial part of learning mathematics. No one can learn topology merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.

James R. Munkres

Let $X$ be a topological space and let $Y$ be a metric space. Let $f_{n}:X\rightarrow Y$ be a sequence of continuous functions. Let $x_{n}$ be a sequence of points of $X$ converging to $x$ . Show that if the sequence $(f_{n})$ converges uniformly to $f$ , then $(f_{n}(x_{n}))$ converges to $f(x)$ .

$f$ is continuous, by Theorem 21.6, therefore, for every $\epsilon>0$ , there is an open neighborhood $U$ of $x$ such that $f(U)\subset B_{d_{Y}}(f(x),\tfrac{\epsilon}{2})$ . Let $N$ be such that for $n>N$ , $x_{n}\in U$ and $d_{Y}(f_{n}(x),f(x))<\tfrac{\epsilon}{2}$ for all $x\in X$ . Then, by the triangle inequality, $d_{Y}(f_{n}(x_{n}),f(x))\le$ $d_{Y}(f_{n}(x_{n}),f(x_{n}))+d_{Y}(f(x_{n}),f(x))<\epsilon$ for $n>N$ .

This is of course not true if the convergence is not uniform. For example, $x^{n}\rightarrow\begin{cases} 0 & x<1\\ 1 & x=1 \end{cases}$ on $[0,1]$ , but $(1-\tfrac{1}{n})^{n}\rightarrow\tfrac{1}{e}\neq1$ . But this is also not always true even if all $f_{n}$ and $f$ are continuous while the convergence is not uniform. For example, $\min\{|nx-1|,1\}\rightarrow1$ , but $\min\{|n\tfrac{1}{n}-1|,1\}\rightarrow0\neq1$ . See also the function $f$ of Exercise 9, which converges to $0$ but $f(\tfrac{1}{n})\rightarrow1$ .

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Section 21: Problem 8 Solution

Section 21: Problem 8 Solution