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I'm trying to prove the following statement: "Let $\Omega \subseteq \Bbb{C}$ is open and $K\subset \Omega$ be compact. Prove that there exists $\epsilon >0$ independent of $z\in K$ where $D_\epsilon(z)\subset \Omega$ for all $z\in K$"

I wanted to use a general definition of open cover compactness and do a similar proof to showing that compact subspace of a metric space is bounded. However, I haven't had much success.

I was told that sequential compactness may be easier to use in this proof however, I don't know where to start with that. I understand the definition is similar to the Bolzano-Weierstrass Theorem: $K\subseteq\mathbb{C}$ is sequentially compact if every $\{z_{n}\}\subseteq K$ has a convergent subsequence where the limit is in $K$.

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Suppose that there is no such $\varepsilon>0$. Then, for each $n\in\mathbb N$, there is a $z_n\in K$ such that $D_{\frac1n}(z_n)\not\subset\Omega$. Since $K$ is compact, you can assume without loss of generality that $(z_n)_{n\in\mathbb N}$ converges to some $z_0\in K$. Since $z_0\in\Omega$ and $\Omega$ is an open set, there is a $r>0$ such that $D_r(z_0)\subset\Omega$. Take $n$ so large that $\frac1n<\frac r2$ and that $d(z_n,z_0)<\frac r2$. Then$$D_{\frac1n}(z_n)\subset D_r(z_0)\subset\Omega,$$which is impossible, since $D_{\frac1n}(z_n)\not\subset\Omega$.

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  • $\begingroup$ I have a very elementary question. How do we know that the disk $D_{\frac{1}{n}(z_n)\notin \Omega$? As in, how do we know that the radius of $\frac{1}{n}$ is sufficient? $\endgroup$ – Ya G Aug 31 '18 at 21:24
  • $\begingroup$ Because we are assuming that, for each $\varepsilon>0$, there is a $z\in K$ such $D_\varepsilon(z)\not\subset\Omega$. So, in particular, if $n\in\mathbb N$, then, since $\frac1n>0$, there is a $z_n\in K$ such that $D_{\frac1n}(z_n)\not\subset\Omega$. $\endgroup$ – José Carlos Santos Aug 31 '18 at 21:36
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Hint:

Assume the contrary: For every $\varepsilon>0$ you can find $z,w$, such that $|z-w|<\varepsilon$, $z\in K$, $w\not\in \Omega$.

Then you can extract a sequence in $K$, so that you can apply Bolzano-Weierstrass.

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For each $z \in K$ find $\varepsilon_z$ such that $D_{\varepsilon_z}(z) \subseteq \Omega$ by $K \subseteq \Omega$ and $\Omega$ being open.

Now let $\delta > 0$ be the Lebesgue number for the open cover $\{B_{\varepsilon_z}(z): z \in K\}$ of $K$ and show it is as required.

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For the non-trivial case $K\ne \emptyset \ne \Omega^c=\Bbb C \backslash \Omega.$

For $x\in K$ let $f(z)=\inf \{|z-x|: x\in \Omega^c\}.$ Then $f$ is continuous on $K$ and $K$ is compact so $M=\min \{f(z):z\in K\}$ exists.

And $M>0.$ Because $f(z)=0\implies z\in \overline {\Omega^c}=\Omega^c$ (which is absurd because $z\in \Omega).$

Now $\forall z\in K\;(\;D_{M/2}(z)\subset \Omega\;).$

More strongly the closure of $\bigcup_{z\in K}D_{M/2}(z)$ is a subset of $\Omega.$

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  • $\begingroup$ In any metric space $(X,d)$, if $\emptyset \ne Y\subset X$ then the function $f(z)=\{\inf d(z,y):y\in Y\}$ is continuous on $X$ and hence continuous on any $K\subset X$. $\endgroup$ – DanielWainfleet Sep 1 '18 at 3:23

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