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Let E be a compact metric space, {$U_i$},i∈I a collection of open sets whose union is E. Show that there exists a real number $\epsilon$ > 0 such that any closed ball in E of radius $\epsilon$ > 0 is entirely contained in at least one open set $U_i$

Here are my thoughts:

"Suppose not,

Then for all $\epsilon$>0, $\exists$p$\in$E such that the closed ball $B_{\epsilon}(p)$ $\cap$${U_i}^{C}$$\neq$ $\emptyset$

Now, since p$\in$E, p$\in$$U_i$ for some i$\in$I, where $U_i$ is an open set from the open cover of E.

$\implies$ p$\in$$closure$($U_i$) and p$\in$$closure$(${U_i}^{C}$)

So p is in the boundary of $U_i$ and p$\in$$U_i$ so this contradicts $U_i$ being open."

I realize there is a way to prove this with sequences, but if I show that for a single point p$\in$E, there is no such closed ball, will I have successfully completed the proof? My professor and the grader say that this is not a valid proof and I recently failed my first quiz.

Jungleshrimp
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1 Answers1

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Your mistake come from the fact that $i$ such that $B_{\epsilon}(p)$ $\cap$${U_i}^{C}$$\neq$ $\emptyset$ is not necessarily the same $i$ for which $p\in U_i$, so you should have said there exists $j$ such that $p\in U_j$.

Hint: take integers $n>0$, suppose that $B_{1\over n}(p_n)$ is not contained in $U_i$ for every $i\in I$, since $E$ is compact, the sequence $p_n$ has a limit point $limp_{i_n}=p\in U_{i_0}$, there exists $N$ such that $B_{1\over N}(p)\subset U_{i_0}$ deduce that for $p_{i_n}$ enough big, $B_{1\over{p_{i_n}}}(p_{i_n})\subset U_{i_0}$.

Tsemo Aristide
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