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Let $B$ be a compact subset of a metric space $X$. Let $x \in X \setminus B$. Prove that there exists disjoint open subsets $U,V$ of $X$ such that $x \in U$ and $B \subseteq V$.

Is this an existence proof? If yes, do I get to define $U,V$?

mathdaddy
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  • you cannot choose such an $\varepsilon$. Because, $B$ is compact (and therefore closed) so set of all limit points of $B$ is a subset of $B$ and therefore , there may exist $ b\in B:\forall \varepsilon>0 B_\varepsilon(b)\cap B\neq \phi$ – Martund Dec 20 '18 at 02:39
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    Am I missing something? If $b\in B$ and $B_\varepsilon(b)$ means the ball centered at $b$ of radius $\varepsilon$, then $B_\varepsilon(b)\cap B\neq\varnothing$ no matter how big or small $\varepsilon>0$. This is because $b\in B_\varepsilon(b)$ and $b\in B$, so it is in the intersection. – Clayton Dec 20 '18 at 03:00

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You cannot say for all $b∈B$, for if $b$ is a limit point then definitely $B(b,r)\setminus\{b\}\cap B\neq\varnothing\;\forall\;r>0.$

Since every metric space is a regular space, so there definitely exist neighbourhoods that comply with the requirements in the question.

Instead choose $\delta=\inf\{d(x,y)\::\:y\in B\}$. Consider $U=B\left(x,\dfrac{\delta}{2}\right)$ and $\displaystyle V=\bigcup_{b\in B} B\left(b,\frac{\delta}{2}\right)$

Yadati Kiran
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$B$ is compact so it is closed. Its complement is open so for any $x\in B^c$ there exists $\epsilon\gt 0$ such that the open ball of radius $\epsilon$ around $x$ is a subset of $B^c$.

Let $U$ be the open ball of radius $\frac{\epsilon}{2}$ around $x$ and let $V=\{y:d(x,y)\gt\frac{\epsilon}{2}\}$. Then $x\in U$ and $B\subset\{y:d(x,y)\ge\epsilon\}\subset\{y:d(x,y)\gt\frac{\epsilon}{2}\}=V$.

John Douma
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