0
$\begingroup$

Let $(X,d_X)$ and $(Y,d_Y)$ be two metric spaces and let $(D,d_X|D)$ be a metric subspace of $(X,d_X)$. Consider a function $f: D \to Y.$ If D is compact and f is continuous, then f is uniformly continuous.

I have to prove this, I know that this is the Heine-Cantor theorem. My question is: can I use the fact that in metric spaces the subset $D\subseteq X$ is compact and thus is closed and bounded or I do have to use the open cover and the finite subcover?

$\endgroup$
  • $\begingroup$ closed and bounded is not enough. You will need full compactness, so finite covers. $\endgroup$ – Henno Brandsma Oct 9 '14 at 15:02
1
$\begingroup$

It is easiest with an open cover and a finite subcover. Given $\varepsilon > 0$, continuity gives you a ball of size $\delta(x,\varepsilon)$ around each $x \in D$. This is an open cover of $D$. Then compactness lets you extract a finite subcover. What does the subcover do for you?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.