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I was trying to prove that any continuous function from a compact metric space to any other metric space is uniformly continuous.

I proved it as follows:

Let $f\colon X\to Y$ be continuous and $X$ be compact. For a given $r>0$, for each $x$, there exists $s(x)>0$ by continuity of $f$ at $x$. Since $X$ is compact, the open cover $\{B(x,s(x))\mid x\in X\}$ admits a finite subcover say $\{B(x_i,s(x_i))\}_{i=1}^n$. If we take $s=\min(s(x_i))$, then we are through.

But this proof is given to be wrong. I just can't understand what am I doing wrong. Please help me.


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Well this proof is obviously incomplete as it stands, since you just assert that your $s$ works with no justification. Why do you think it works? –  Chris Eagle Jan 31 '13 at 13:31
Your "proof", as presented here, is very poorly worded and lacks lots of explanation. You should also try to enhance your presentation and write it using LaTeX. In the FAQ section you can find directions about this. –  DonAntonio Jan 31 '13 at 13:38
there exists $s(x)>0$ by continuity of $f$ at $x$... Good to know! But say, there exists $s(x)$ such that WHAT holds? –  Did Jan 31 '13 at 14:04
sorry!! i need to be more specific. so my strategy was to take s as defined above. now given any x,y belonging to X such that d(x,y)<s we will have d(f(x),f(y))<r, by the continuity of f and hence the function is uniformly continuous. –  wanderer Jan 31 '13 at 16:21

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