Convergence on compact sets is equivalent to uniform convergence. Metric spaces.

Question

Let $ f: E \to M$, where $E$ is some topological space and $(M,d)$ is a metric space. Also let the sequence $ f_n : E \to M$. Assume for each $x\in E$, $d ( f_n(x),f(x) ) \to 0$ as $n\to \infty$, is it true that $\sup_{x\in \Omega} d ( f_n(x),f(x) ) \to 0 $ if $\Omega$ is compact?

Is this just convergence on compacts = uniform convergence ? Does anyone have a reference for this result? What if we also know the limit $f:E \to M$ is continuous.

Answer 1

Not necessarily. Take $E=M=[0,1]$, endowed with the usual topology, and define$$f_n(x)=\begin{cases}1&\text{ if }x=\frac1n\\0&\text{ otherwise.}\end{cases}$$Then, for each $x\in E$, $\lim_{n\to\infty}f_n(x)=0$, but $\sup_{x\in E}|f_n(x)-0|=1.$