# Is a locally countable family of open subsets of an separable space countable?

Definition A topological space $(X, \mathfrak{T} )$ is said to be an separable space if $X$ contains a contable subset $D \subseteq X$ such that $D$ is dense in $X$

Definition Let $(X, \mathfrak{T} )$ be a topological space and $\mathfrak{F}$ be a family of open subsets of $X$. The family $\mathfrak{F}$ is said to be locally countable if it satisfies the following property: for every $x \in X$, there exists an open subset $x \in U(x) \subseteq X$ such that there are only countably many members in $\mathfrak{T}$ which intersect $U(x)$.

Question Let $(X,\mathfrak{T})$ be an separable space and $\mathfrak{F}$ be a locally countable family of open subsets of $X$. Is $\mathfrak{F}$ a countable family?

I want to prove that $\mathfrak{F}$ is a countable family.

My Approach For convenience, we may assume that $\emptyset \notin \mathfrak{F}$. Because $(X,\mathfrak{T})$ is an analytic space, $X$ contains a countable subsets $D = \lbrace x_n \ | \ n \in \mathbb{N} \rbrace$ in $X$ such that $D \subseteq X \subseteq \overline{D}$. Therefore, for every $U \in \mathfrak{F}$, there is a smallest $n \in \mathbb{N}$ such that $x_n \in U$. Because $\mathfrak{F}$ is a locally countable family, there are only countable many members $V \in \mathfrak{F}$ containing $x_n$. In other words, we may separate $\mathfrak{F}$ to be a disjoint union of the form $\mathfrak{F} = \bigsqcup_{n \in \mathbb{N}} \mathfrak{F}_n$, where $\mathfrak{F}_n$ is the family of all $U \in \mathfrak{F}$ so that $n$ is the smallest positive integer such that $x_n \in U$. By the preceding arguments, because each $\mathfrak{F}_n$ is a countable family, there is an injective map $\iota_n : \mathfrak{F}_n \rightarrow \mathbb{N}$. We define $\iota : \mathfrak{F} \rightarrow \mathbb{N} \times \mathbb{N}$ to be the map $U \mapsto (n,\iota_n(U))$ whenever $U \in \mathfrak{F}_n$. Obviously, the map $\iota$ is injective. Because $\mathbb{N} \times \mathbb{N}$ is countable, it shows that $\mathfrak{F}$ is also countable.

Is my proof correct? Thank you!

• On the whole this seems correct. If you already know that any countable union of countable sets is countable, it might be easier to just define $\mathfrak{F}_n = \{ U \in \mathfrak{F} : x_n \in U \}$. You will still get that each $\mathfrak{F}_n$ is countable and $\bigcup_n \mathfrak{F}_n = \mathfrak{F}$ (it just won't be a disjoint union). – the thesaurus saurus Sep 4 '15 at 17:29
• In addition to the simplification noted by @hereisnowhy, you might want to note that you don’t $\mathfrak{F}$ to be locally countable: it need only be point countable. – Brian M. Scott Sep 4 '15 at 20:08
• @BrianM.Scott . What is an analytic space? – DanielWainfleet Jan 20 '17 at 23:04

## 1 Answer

Your error is assuming that $\cup F\supset D.$ So there may not exist $U\in F$ such that $x_n\in U.$

Here is a proof: Let $F$ be an uncountable open family. For each $f\in F$ \ $\{\phi\}$ let $G(f)\in D\cap f.$ This is possible because $D$ is dense and $F$ is an open family. Then $G$ maps $F$ \ $\{\phi \}$, which is uncountable, into the countable set $D.$ So for some $d\in D$ the set $G^{-1}\{d\}$ is uncountable .That is, $d$ belongs to uncountably many members of $F$, so any nbhd of $d$ intersects uncountably many members of $F.$ So $F$ is not locally countable.

• As Brian M. Scott said in a comment, this shows that $F$ is not even point-countable. – DanielWainfleet Jan 20 '17 at 23:11