Hello, everyone. I want to discuss about this theorem, which I read in "Principles of Mathematical Analysis" by Rudin, to enrich my knowledge about equicontinuity. I think "equicontinuity" in the theorem means uniform equicontinuity, is it right?
Then, the theorem said "$K$ is a compact metric space", why it must be a compact metric space? What will happen if $K$ is not a compact metric space? Will it be equicontinuous at $x$ (not uniformly equicontinuous) if $K$ is not compact?
Thanks for any help.
Yes, equi-continuity means uniform equi-continuity.
Let $K$ be the real line with the usual metric, $f_n(x)=x^{2}+\frac 1 n$ and $f(x)=x^{2}$. The $f_n \to f$ uniformly but $(f_n)$ is not equi-continuous: In the inequality $|f_n(x)-f_n(y)| <\epsilon$ for $|x-y| <\delta$ take $x=N+\delta /2$ and $y=N$. You will see that the inequality fails for large enough $N$.
Of course equi-continuity at a point holds because a singleton set is compact.