I would appreciate if you gave my proof a look over, i'm getting a bit stuck towards the end. thank you very much. first time writing in latex by the way, so apologies beforehand.

claim: let $f: M \to N$ be a uniformly continuous function. Show that $f$ maps bounded sets to bounded sets.

proof: If $ m \subset M$ is bounded, then $ d(p,q) < B $ for fixed p and all q $ \in $ m.

let $F (m) $ be the image of m.

by definition of uniform continuity,

For all $ \epsilon > 0$ there exists $\delta > 0$ such that for all p,q such that $d(p,q) < \delta $ $\implies $ $d(f(p),f(q)) < \epsilon$

edit: cannot let $\delta = m$

so now i'm thinking that if we hold p and vary q, we will cover M with a set of open balls. Because M is bounded, that means this set of open balls is finite. and we can use that to similarly bound $f(m)$ with a finite set of open balls by the continuity definition.

*** this strategy would be assuming the set were "totally bounded", meaning it can be covered by a finite number of open balls of any radius. thanks for pointing out the term, user251257

edit: it's two general metric spaces, we cannot assume it's R.

  • $\begingroup$ what is $M$? ${}$ $\endgroup$
    – user251257
    Apr 6 '16 at 2:08
  • 2
    $\begingroup$ You can't say "let $B=\delta$": $\delta$ depends on $\epsilon$. $\endgroup$
    – spinoza
    Apr 6 '16 at 2:09
  • $\begingroup$ M and N are metric spaces. $\endgroup$
    – mac5
    Apr 6 '16 at 2:11
  • $\begingroup$ thank you spinoza. i thought that was might be wrong $\endgroup$
    – mac5
    Apr 6 '16 at 2:12
  • 3
    $\begingroup$ It is wrong. Consider the identity function from the integers endowed with the discrete metric to the integers endowed with the usual Euclidean metric. $\endgroup$
    – user251257
    Apr 6 '16 at 2:27

In general this is not true for arbitrary metric spaces but if $M=N=\mathbb R$ with the standard metric, you can argue as follows:

$f$ is uniformly continuous on $m\subseteq M$, so it extends to a (uniformly) continuous function on $\overline m$, which is also bounded, hence compact. Of course, $f(m)\subseteq f(\overline m)$, and the latter is compact hence bounded,so that $f(m)$ is bounded as well.


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