# How many open sets are needed to separate all pairs of disjoint closed sets?

Let $\mathscr{U}$ be a collection of open subsets of $\mathbb{R}$ which separates closed sets. That is, for every pair of disjoint closed sets $A,B \subset \mathbb{R}$, there exist disjoint $U,V \in \mathscr{U}$ such that $A \subset U$ and $B \subset V$. Is it obvious that $\mathscr{U}$ needs to have $\mathfrak{c}$ elements?

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Every subset of $\mathbb Z$ is closed. Let as assign to each $A\subseteq\mathbb Z$ disjoint open sets $U_A$, $U_{\mathbb Z\setminus A}$ such that $A\subseteq U_A$ and $\mathbb Z\setminus A\subseteq U_{\mathbb Z\setminus A}$. Then the map $A\mapsto U_A$ is an injective map from $\mathcal P(\mathbb Z)$ to $\mathscr U$.

Proof of injectivity: Let $B\ne A$, w.l.o.g $B\subsetneqq A$. Let $a\in A\setminus B$. Then $a\in U_A$, since $a\in A$. Also $a\in\mathbb Z\setminus B$ implies $a\notin U_B$. Thus $U_A\ne U_B$.

EDIT: The above answers the original question. I just want to add some thoughts what can change if we work with a compact subspace of $\mathbb R$ instead of the whole real line.

So let us ask the same question about $[0,1]$ instead of $\mathbb R$. Since $[0,1]$ is compact, we cannot find infinite closed discrete subspace in it, so the above trick will not work.

We have a countable basis $\mathcal B=\{(p,q)\cap[0,1]; p<q, p,q\in\mathbb Q\}$ for this space. Using compactness we can show that any two closed subsets can be separated by sets $U$, $V$ such that each of them is a finite intersection of finite unions or a finite union of finite intersections of basic sets from $\mathcal B$. (See bellow.) I.e., every separating set is of the form $\bigcap_{i=1}^k \bigcup_{j=1}^l B_{ij}$ or of the form $\bigcup_{i=1}^k \bigcap_{j=1}^l B_{ij}$, where each $B_{ij}$ is in $\mathcal B$. The system $\mathscr U$ consisting of all such sets is countable system of open sets. (The set of all finite sequences of element of given countable set is again countable.) So in this space, a set of size $\aleph_0$ is sufficient.

Basically the same proof would work in any second countable compact space.

We know that any two points can be separated by basic sets from $\mathcal B$. Suppose we are given two disjoint closed sets $A$ and $B$. Closed subsets of compact space are compact.

We can show that for any $x\notin A$ there exist sets $U_x$ and $V_x$ such that:

• $A\subseteq U_x$, $x\in V_x$ and $U_x\cap V_x = \emptyset$ (i.e., these sets separate $U$, $V$)
• $U_x$ is a finite union of sets from $\mathcal B$
• $V_x$ is a finite intersection of sets from $\mathcal B$

The proof of this fact is exactly the same as in this answer with the only exception that we have to work with basic sets only.

Now we can repeat basically the same argument again. For every $b\in B$ we have sets $U_b$, $V_b$ with the above properties. Now as $V_b$ is an open cover of $B$, thus there exists finite subcover $\{V_b; b\in F\}$. The sets $U:=\bigcap_{b\in F} U_b$ and $V:=\bigcup_{b\in F} V_b$ separate $A$ and $B$ and they belong to the family $\mathscr U$ described above.

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Perfect, thank you Martin. – Mike F Nov 18 '11 at 22:41