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Let's take a metric space. Then any closed set can be written as a countable intersection of open sets.

How can I prove that?

user96849
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4 Answers4

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Let $A\subseteq X$ be closed. For all $n\in \mathbb N$ define $$U_n=\bigcup _{a\in A} B(a,\frac{1}{n}).$$ $U_n$ is open as a union of open balls. We prove that $A=\bigcap _{n\in \mathbb N} U_n$.

Clearly $A \subseteq \bigcap _{n\in \mathbb N} U_n$.

To prove $A \supseteq \bigcap _{n\in \mathbb N} U_n$ we take $x\notin A$ and show that $x\notin \bigcap U_n$.

Since $A$ is closed, $A^C$ is open, therefore $\exists n \in \mathbb N$ such that $B(x,\frac{1}{n}) \cap A=\emptyset$. That is, for all $a\in A$: $a\notin B(x,\frac{1}{n})$; and thus for all $a\in A$: $x\notin B(a,\frac{1}{n}) \Longrightarrow x\notin \bigcup_{a\in A} B(a,\frac{1}{n}) \Longrightarrow x\notin U_n \Longrightarrow x\notin \bigcap U_n$.

Ludolila
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  • it is not clear to me why $A \subseteq \bigcap_{n \in \mathbb{N}}U_{n}$, could you please explain this part in more detail? –  Sep 21 '15 at 16:05
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    @JessyCat: well, let's take an element $a\in A$ and show that it's in $\bigcap U_n$. Note that for every $n\in \mathbb N$ we have $a\in B(a,\frac{1}{n})$, and thus $a\in U_n$ for every $n\in \mathbb N$. This means that $a\in \bigcap U_n$. Hope it helped :) – Ludolila Sep 21 '15 at 16:14
  • let's say we were talking about $\mathbb{R}$ here, instead of a generalized metric space. What would an element of $A$ look like then? The problem would reduce to showing that a closed interval in $\mathbb{R}$ equals the countable intersection of open intervals in $\mathbb{R}$, so I would like to see in more detail what an element of a closed interval in $\mathbb{R}$ looks like. Is it just a point? –  Sep 21 '15 at 16:17
  • Not every closed set in $\mathbb R$ is an interval, so we can't reduce the problem this way. But we can prove, in particular, that every closed interval is a countable intersection of open intervals. For this purpose, the elements of the intervals in question are indeed points. – Ludolila Sep 21 '15 at 16:22
  • also, how does $A^{c} \notin \bigcap U_{n}$ necessarily imply that $\bigcap U_{n} \subseteq A$? –  Sep 21 '15 at 20:10
  • it doesn't, and I didn't say it does... – Ludolila Sep 24 '15 at 13:29
  • @Ludolila can I ask you to explain me one thing: why do we need to define the union of open balls? – Daniil Yefimov Nov 23 '16 at 12:34
  • @DanielYefimov I'm not sure I understand your question. We want $A$ to be a countable intersection of open sets. We define these open sets, $U_n$, as unions of (open) neighborhoods of elements of $A$. In a metric space, we can work with open balls, instead of some general neighborhoods... – Ludolila Nov 23 '16 at 13:22
  • @Ludolila Thank you, now I got it:) Just forgot, that every open set is a union of neighborhoods of points of open set. – Daniil Yefimov Nov 24 '16 at 08:35
  • @Ludolila Why choosing $1/n$ as a radius I don't see where did we use it. We could simply use $n$ instead or any sequence? – palio Oct 19 '22 at 09:53
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Hint: Let $A\subseteq X$ be closed. For $\epsilon > 0$ let $$ U_\epsilon(A) = \{x \in X \mid \def\dist{\mathop{\rm dist}}\dist(x,A) < \epsilon \}$$ where $\dist(x, A) := \inf_{y \in A} d(x,y)$. What can you say about the sets $U_\epsilon(A)$, what is $\bigcap_{\epsilon > 0} U_\epsilon(A)$?

martini
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  • I think this approach makes it a little difficult to see that the $U_\epsilon$ are open. Defining $U_\epsilon$ to be the union of open balls of radius $\epsilon$ around points of $A$ would make its openness immediate. – hardmath Sep 25 '13 at 13:54
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    As $\mathop{\rm dist}(\cdot, A) \colon X \to [0,\infty)$ is continuous ... – martini Sep 25 '13 at 13:56
  • True, but more elementary is that a union of open sets is open. – hardmath Sep 25 '13 at 13:59
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    You are right, but on the other hand, the way I used above is how I imagine the set $U_\epsilon(A)$ in the first place, namely as $A$ with a small border around it, not as a union of balls. – martini Sep 25 '13 at 14:02
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Let $D(A,\varepsilon)=\left\{ y\in M | d(A,y)<\varepsilon \right\}$, where $d(A,y)=\inf \left\{ d(z,y)|z\in A \right\}$. This set is open .Define a sequence of $\varepsilon$ as $\varepsilon_{n}=\frac{1}{n}$. Claim $A=\cap_{n=1}^{\infty}D(A,\varepsilon_{n})$. Proof: If $x\in A$, then $d(A,x)=0$ and so $x \in D(A,\varepsilon_{n})$ for all $n\in \mathbb{N}$ $\Rightarrow$ $A\subset \cap_{n=1}^{\infty}D(A,\varepsilon_{n})$. Conversely if $x\in \cap_{n=1}^{\infty}D(A,\varepsilon_{n})$ then $x\in D(A,\varepsilon_{n})$ for all $n\in \mathbb{N}$ $\Rightarrow$ $\forall \varepsilon>0$ $D(x,\varepsilon)\cap A\setminus \left\{ x \right\}\neq \emptyset$ and so $x$ is an accumulation point of $A$. $A$ is, however, closed so contains all its accumulation points, so $x\in A$ $\Rightarrow$ $\cap_{n=1}^{\infty}D(A,\varepsilon_{n})\subset A$.

Mael
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About the openness of $G_n=\{x\in M|d(x,F)<\frac 1n\}$, where $n\in \Bbb N, (M,d)$ our metric space and $F\subset M$ a closed set; let $y\in G_n$ and $ε:=\frac 1n -d(y,F)>0$. If $z\in B^d_ε(y)$, we have $d(z,F)\leq d(z,x)$, for all $x\in G_n$; so $d(z,F)\leq d(z,x)\leq d(z,y)+d(y,x)\implies d(z,F)\leq d(z,y)+d(y,F)<ε+d(y,F)=\frac 1n$, thus $z\in G_n$, which implies that $B^d_ε(y)\subset G_n$, so $G_n$ is open.

SK_
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