8
$\begingroup$

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?

$\endgroup$
9
$\begingroup$

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 $a\in A$: $x\notin B(a,\frac{1}{n}) \Longrightarrow x\notin \bigcup_{a\in A} B(a,\frac{1}{n}) \Longrightarrow x\notin U_n \Longrightarrow x\notin \bigcap U_n$.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – ALannister Sep 21 '15 at 16:05
  • 2
    $\begingroup$ @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 :) $\endgroup$ – Ludolila Sep 21 '15 at 16:14
  • $\begingroup$ 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? $\endgroup$ – ALannister Sep 21 '15 at 16:17
  • $\begingroup$ 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. $\endgroup$ – Ludolila Sep 21 '15 at 16:22
  • $\begingroup$ also, how does $A^{c} \notin \bigcap U_{n}$ necessarily imply that $\bigcap U_{n} \subseteq A$? $\endgroup$ – ALannister Sep 21 '15 at 20:10
4
$\begingroup$

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 $\bigcap_{\epsilon > 0} U_\epsilon(A)$?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – hardmath Sep 25 '13 at 13:54
  • 2
    $\begingroup$ As $\mathop{\rm dist}(\cdot, A) \colon X \to [0,\infty)$ is continuous ... $\endgroup$ – martini Sep 25 '13 at 13:56
  • $\begingroup$ True, but more elementary is that a union of open sets is open. $\endgroup$ – hardmath Sep 25 '13 at 13:59
  • 1
    $\begingroup$ 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. $\endgroup$ – martini Sep 25 '13 at 14:02
1
$\begingroup$

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 $x\in A$ $\Rightarrow$ $\cap_{n=1}^{\infty}D(A,\varepsilon_{n})\subset A$.

$\endgroup$
  • $\begingroup$ l-2: Isn't the $D(x,\epsilon)$ meant to be a ball centered at x ? $\endgroup$ – zebullon Jan 13 '15 at 1:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.