# Closed subsets of metrizable compact spaces can be viewed as intersection of closed with non-empty interior

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.

• I don't think you need compactness. Can you find a mistake in my answer? – 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 $$Z_n=\cap_{i=1}^{i=n}Y_i$$.

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

Counterexample:

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

• Okay, Thanks. Good point- I forgot to assume that $Y\neq \emptyset$. – 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 $$x \to d(x,Y)$$).

• How do we know $Y_n$ is a decreasing sequence? – 5xum Jul 25 '19 at 9:06
• 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 – 5xum Jul 25 '19 at 9:11
• 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. – 5xum Jul 25 '19 at 9:16
• Then why not simply take $Y_n=Y$? Also, how do you know your $Y_n$ have a nonempty interior? – 5xum Jul 25 '19 at 9:18
• @5xum Read the last part of my answer. – Kavi Rama Murthy Jul 25 '19 at 12:16