Exercise 21.8

Question

Let $X$ be a topological space and let $Y$ be a metric space.
  Let $f_n: X 	o Y$ 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 $f(x)$.

Dan Whitman · Accepted Answer

\def\a{\alpha} \def\b{\beta} \def\d{\delta} \def\e{\epsilon} \def { ho} \begin{proof} Suppose that $(f_n)$ converges to $f$ uniformly and let $d$ denote the metric for $Y$. Consider any $\e > 0$. Since $(f_n)$ converges to $f$ uniformly there is an $N_1 \in {{\mathbb{Z}}_+}$ where $d(f_n(x'),f(x')) < \e/2$ for any $n > N_1$ and $x' \in X$. We also know from the uniform limit theorem (Theorem~21.6) that $f$ is continuous since each $f_n$ is continuous. It then follows from Theorem~21.3 that the sequence $(f(x_n))$ converges to $f(x)$ since $x_n o x$. Hence there is an $N_2 \in {{\mathbb{Z}}_+}$ such that $d(f(x_n),f(x)) < \e/2$ for all $n \geq N_2$. So set $N = \max\left\{N_1+1, N_2 ight\}$ and consider any $n \geq N$. Then of course $n \geq N \geq N_1+1 > N_1$ and $x_n \in X$ so that \begin{gather*} d(f_n(x_n), f(x_n)) < \e/2 \,. \end{gather*} We also have that $n \geq N \geq N_2$ so that \begin{gather*} d(f(x_n),f(x)) < \e/2 \,. \end{gather*} We then have \begin{gather*} d(f_n(x_n),f(x)) \leq d(f_n(x_n), f(x_n)) + d(f(x_n), f(x)) < \frac{\e}{2} + \frac{\e}{2} = \e \end{gather*} since $d$ is a metric, and so $f_n(x_n) \in B_d(f(x),\e)$. Since $n \geq N$ and $\e > 0$ were arbitrary, we have shown that the sequence $(f_n(x_n))$ converges to $f(x)$ in the metric space $Y$ by definition. \end{proof}

Exercise 21.8

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Exercise 21.8

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