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I'm reading the proof of "if there's a continuous function $f:X\to Y$ where $X$ is a compact metric space and $Y$ is a metric space, then $f$ is uniformly continuous on $X$."

The proof proceeds thus: Take any $\epsilon\in\Bbb{R}$. For any point $p\in Y$, there exists a $\delta_p\in\Bbb{R}$ such that $|f(x)-f(p)|<\epsilon$ for $|x-p|<\delta_p$. Take the cover $\bigcup_{p\in X} B(p,\displaystyle{\frac{\delta_p}{2}})$. As $X$ is compact, it contains a finite subcover $\bigcup_{i=1}^n B(x_i,\displaystyle{\frac{\delta_{x_i}}{2}})$. Let $m=\min\{\displaystyle{\frac{\delta_{x_1}}{2},\frac{\delta_{x_2}}{2},\dots,\frac{\delta_{x_n}}{2}}\}$. The proof goes on to prove that $\forall x,y\in X$, $d(f(x),f(y))<\epsilon$ for $d(x,y)<m$.

The point of the proof is that $m$ is fixed for every particular $\epsilon$. But is $m$ really unique? There can be multiple finite subcovers for a particular cover. Hence, $m$ is set to vary. If $m$ has an upper bound, then this still makes sense. However, if there are are an infinite subcovers, then it is possible $m$ has no upper bound. I'm getting confused here.

Thanks in advance!

2 Answers2

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Your interpretation is slightly wrong. To prove uniform continuity you need only find some $m$ for $\epsilon$. The proof shows that the $m$ chosen by the expression given works. This is all that matters. Had a different subcover been chosen you might have got a different $m$; but it would still be an $m$ which worked. You already knew $m$ would not be unique because e.g. any smaller choice would also work.

not all wrong
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Well $m$ makes it such that as long as $d(x,y)$ then $|f(x)-f(p)|<\epsilon$ in $X$. Possibly, if you took a different subcover, you would have found a different "$m$". However, what matters is that you found an $m$ that works. Sure, you can choose something smaller than $m$ and it would still work, but who cares =]?

LASV
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