Let $X$ be a compact, metrizable space, and let $Y\subseteq X$ be a closed subset. Is is always possible to find a decreasing sequence $(Y_m)_{m\geq 1}$ of closed subsets of $X$, s.t. $\mathrm{int}(Y_m)\neq\emptyset$ and $Y=\bigcap\limits_{m=1}^{\infty} Y_m$?

If $Y$ is a subset of finitely many points, I know that the answer is yes.

I also know that if $X$ is not compact, this fails.

  • $\begingroup$ I don't think you need compactness. Can you find a mistake in my answer? $\endgroup$ – Kavi Rama Murthy Jul 25 '19 at 9:15

$Y=\emptyset$ is a counterexample in any compact space, since the intersection of a decreasing sequence of nonempty closed sets is nonempty.

If we assume $Y\ne\emptyset$ then compactness is not needed: if $Y$ is a nonempty closed set in a metric space $X$, then $Y$ is the intersection of a decreasing sequence of closed sets with nonempty interiors.

Proof. Choose a point $p\in Y$. Let $Y_n=Y\cup\{x\in X:d(x,p)\le\frac1n\}$. Then $Y_n$ is a closed set with nonempty interior (since $p$ is an interior point of $Y_n$), and $\bigcap_{n=1}^\infty Y_n=Y$.

That is a rather trivial and uninteresting statement. Here's a better one:

Proposition. In a metric space $X$, any closed set $Y$ is the intersection of a decreasing sequence of closed sets $Y_n$ such that $Y\subseteq\operatorname{int}(Y_n)$.

Proof. Let $Y_n=\{x\in X:d(x,Y)\le\frac1n\}$. Then $\bigcap_{n=1}^\infty Y_n=Y$ because $Y$ is closed, and $Y\subseteq\{x\in X:d(x,Y)\lt\frac1n\}\subseteq\operatorname{int}(Y_n)$.


For every integer $n$, there exists a finite number of elements $y_{i_1},..,y_{i_n}$ such that $y_{i_j}\in Y$ and $Y_n=B(y_{i_1},{1\over n})\cup...\cup B(y_{i_n},{1\over n})$ contains $Y$ since $Y$ is compact.

$Z=\cap_{n>0}Y_n=Y$. Suppose this is not true. There exists $z\in Z$ not in $Y$. For every $n>0$, $z\in B(y_{i^z_n},{1\over n})$. Since $Y$ is compact we can extract a sequence from $(y_{i^z_n})$ which converges towards $y\in Y$. $d(y,z)\leq d(y,y_{i^z_n})+d(y_{i^z_n},z)$. This implies that $y=z$ contradiction.

To make the sequence $Y_n$ a decreasing sequence, replace it by $Z_n=\cap_{i=1}^{i=n}Y_i$.

  • $\begingroup$ Compactness of $X$ is not necesasry for this result. $\endgroup$ – Kavi Rama Murthy Jul 25 '19 at 9:21


Space $X$ has a finite underlying set and is equipped with discrete topology.

Let $Y=\varnothing$.

Then by decreasing $Y_m$ the equality $\bigcap_{m=1}^{\infty}Y_n=Y=\varnothing$ implies that $Y_m=\varnothing$ for $m$ large enough.

Then also $\mathsf{int}(Y_m)=\varnothing$.

  • $\begingroup$ Okay, Thanks. Good point- I forgot to assume that $Y\neq \emptyset$. $\endgroup$ – User3231 Jul 25 '19 at 9:11

The conclusion holds in any metric space. Compactness is not necessary. (I will assume that $Y$ is non-empty). Let $W_n =\{x: d(x,Y) <\frac 1 n\}$. Let $Y_n=\overset {-} {W_n}$. Then $W_n$ is a nonempty open set for each $n$ , so $Y_n$ is a closed set with non-empty interior. Since $Y_n \subset \{x: d(x,Y) \leq \frac 1 n\}$ it is clear that $\cap_n Y_n =Y$.

Any point of $Y$ is an interior point of $W_n$ because it belongs to $W_n$ and $W_n$ is open (by continuity of the map $ x \to d(x,Y)$).

  • $\begingroup$ How do we know $Y_n$ is a decreasing sequence? $\endgroup$ – 5xum Jul 25 '19 at 9:06
  • $\begingroup$ How do we know $W_n$ is decreasing? What if $X$ has the discrete metric? It could be that $W_{n}=W_{n+1}$ for some $n$ in such a case $\endgroup$ – 5xum Jul 25 '19 at 9:11
  • $\begingroup$ All you proved is that $W_{n+1}\subseteq W_n$, but I think OP wants a strictly decreasing sequence, i.e. $Y_{n+1}\neq Y_n$. Otherwise, we can simply take $Y_n=Y$ and be done with it. Also, you didn't prove that the interior of your $T_n$ is nonempty. $\endgroup$ – 5xum Jul 25 '19 at 9:16
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    $\begingroup$ Then why not simply take $Y_n=Y$? Also, how do you know your $Y_n$ have a nonempty interior? $\endgroup$ – 5xum Jul 25 '19 at 9:18
  • $\begingroup$ @5xum Read the last part of my answer. $\endgroup$ – Kavi Rama Murthy Jul 25 '19 at 12:16

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