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I saw the following statement in some notes without proof, and I have been trying to verify it (using finite open covers and triangle inequalities) without success, so I would like to ask for some suggestions.

Let $X$ be a compact metric space. For a subset $Y$ of $X$ and $\varepsilon>0$, let $N_\varepsilon(Y)=\bigcup_{y\in Y}B_\varepsilon(y)$, where $B_\varepsilon(y)$ denotes the open ball of radius $\varepsilon$ about $y$. Let $A$ and $B$ be closed subsets of $X$. Then for any $\varepsilon>0$, there exists $\delta>0$ such that $N_\delta(A)\cap N_\delta(B)\subseteq N_\varepsilon(A\cap B)$.

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  • $\begingroup$ Have you tried using the fact that closed substes of compact metric space are compact? I think it will be used somewhere after you consider the set $N_{\delta} \left( A \right) \cap N_{\delta} \left( B \right)$. Even I am stuck there! $\endgroup$ – Aniruddha Deshmukh Jul 10 '18 at 16:40
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Let's set up some nested family of compact sets: $$ E_k = \overline{N_{1/k}(A)\cap N_{1/k}(B)} \setminus N_\varepsilon(A\cap B) $$ We want to show some $E_k$ is empty. If not, then there exists $x\in \bigcap_k E_k$. By the construction of $E_k$ we have $\operatorname{dist}(x, A)\le 1/k$ for every $k$, hence $\operatorname{dist}(x, A)=0$, hence $x\in A$ (here we use the fact that $A$ is closed). Similarly $x\in B$. But then $x\in A\cap B$ which is definitely disjoint from any $E_k$ by construction. A contradiction.

Just as an aside: the property of being closed is important: the statement would fail for $A=[0, 1]$ and $B=(1, 2]$ on the real line.

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  • $\begingroup$ I was also wondering whether something similar might be true when $X$ is a proper metric space (closed balls are compact). $\endgroup$ – cyc Jul 14 '18 at 12:51
  • $\begingroup$ No. For example, in $\mathbb R$ take $A=\mathbb N$ and $B=\{n+1/n:n\in\mathbb{N}, n>1\}$. Then $N_\delta(A)\cap N_\delta(B)$ is never empty but $ N_\varepsilon(A\cap B)$ is empty. $\endgroup$ – user357151 Jul 14 '18 at 14:46

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