Problem of the Week #101 - May 5th, 2014

[Chris L T521]

Here's this week's problem!

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Problem: 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 $f(x)$.

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[Chris L T521]

This week's problem was correctly answered by Ackbach and girdav.

You can find Ackbach's solution below.

[sp]Let $\varepsilon>0$.

Assume $X$ is a topological space, and $Y$ is a metric space with metric $d$. Let $f_n:X \to Y$ be a sequence of continuous functions converging uniformly to $f$. Let $\{x_n\}$ be a sequence of points of $X$ converging to $x\in X$. We wish to show that $\{f_{n}(x_{n})\}$ converges to $f(x)$.

Since $f_n$ converges uniformly to $f$, by definition, $(\forall \varepsilon>0)(\exists N_{f}\in \mathbb{Z}^{+})$ s.t. $(\forall x\in X)$ and $(\forall n\ge N_{f})$, $d(f_{n}(x),f(x))<\dfrac{\varepsilon}{2}$.

Since the $f_n$ are continuous, for any sequence $\{y_n\}\to y$, all in $X$, it must be that $\{f_n(y_n)\}\to f_n(y)$. This latter statement means that there exists $N_{c}$ such that for all $n\ge N_{c}$, it must be that $d(f_n(x_n),f_n(x))<\dfrac{\varepsilon}{2}$.

Let $N=\max(N_f,N_c)$, and assume $n\ge N$. It follows that
$$d(f_n(x_n),f(x))\le d(f_n(x_n),f_n(x))+d(f_n(x),f(x)) \le \frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.$$

This proves that $\{f_n(x_n)\}\to f(x)$, as required.[/sp]