I need some help to prove that if $\mathcal{A}$ is a chain of closed subsets in a separable metric space then there is a countable subfamily $\mathcal{A}'\subseteq\mathcal{A}$ such that $\bigcup\mathcal{A}'=\bigcup\mathcal{A}$.

My attempt was to pick a countable dense subset $D$ and for each $d\in D\cap\bigcup\mathcal{A}$ choose one $A_d\in\mathcal{A}$ so that $d\in A_d$. Then the set of $A_d$'s is countable but I think that the union may not be the whole $\bigcup\mathcal{A}$. Is there any way to fix this idea or is there some better way to prove the statement?

  • 3
    $\begingroup$ Just to point something out, it is entirely possible that the dense set does not meet $\bigcup\mathcal A$. For example, let $D=\mathbb Q$, and take $A_i = \{\pi + n\mid n<i\}$, finite sets are closed, and $\mathcal A=\{A_n\mid n\in\mathbb N\}$ is clearly a chain. However there is not a single rational point in $\bigcup\mathcal A$. $\endgroup$ – Asaf Karagila Sep 20 '11 at 18:43

A separable metric space has a countable basis of its topology. I.e., there is a countable family of open sets such that each open set $O$ is a union of sets from the open family. Let $\mathcal B$ denote such a countable basis of the topology of your space.

First we choose a maximal wellordered subchain $\mathcal A'$ of $\mathcal A$. You can use Zorn's Lemma to do this. (By maximal I mean that you cannot add an element on top of the wellordered chain.) Because of the maximality, $\bigcup\mathcal A'=\bigcup\mathcal A$.

Now let $O\in\mathcal A'$ and let $U\in\mathcal A'$ be the successor of $O$ in the wellordering. $O$ is a proper subset of $U$. Hence there is $V_O\in\mathcal B$ such that $V_O\subseteq U$ but $V_O\not\subseteq O$. Note that the map $O\mapsto V_O$ is 1-1 and has a countable range. It follows that $\mathcal A'$ is countable.

Edit: As LostinMath pointed out in his comment, it is not clear that such a set $V_O$ always exists. However, since the elements of $\mathcal A$ are closed, their complements are open and hence, for each $O\in\mathcal A'$ and successor $U\in\mathcal A'$, there is $V_O\in\mathcal B$ such that $V_O$ is disjoint from $O$ and has non-empty intersection with $U$. These conditions guarantee that the map $O\mapsto V_O$ is 1-1 and the proof goes through as above.

  • $\begingroup$ I think the conditions $V_O\subseteq U$ and $V_O\not\subseteq O$ should be replaced by $V_O\cap U\neq\emptyset$ and $V_O\cap O=\emptyset$. $\endgroup$ – LostInMath Sep 21 '11 at 18:20
  • $\begingroup$ Yes you are right. For some reason I was under the wrong impression that the elements of $\mathcal A$ are open. I edited my answer accordingly. $\endgroup$ – Stefan Geschke Sep 22 '11 at 18:40

A slightly different version of the same basic idea as Stefan’s proof may prove more intuitive for some. Let $M$ be the metric space. I too first want to get a maximal well-ordered subchain $\mathcal{A}'$ of $\mathcal{A}$; I’ll comment on that below, but for now assume that I have one. Just as in Stefan’s proof, its maximality ensures that $$\bigcup\mathcal{A}'=\bigcup\mathcal{A}.$$

Now for each $A \in \mathcal{A}'$ let $A^+$ be the successor of $A$ in the well-ordered chain $\mathcal{A}'$, choose a point $x_A \in A^+\setminus A$, and let $X = \{x_A:A \in \mathcal{A}'\}$.) For convenience write $x_A \preceq x_B$ iff $A \subseteq B$; $X$ is well-ordered by $\preceq$.

A separable metric space is hereditarily separable (for instance because it’s second countable), so $X$ has a countable dense subset $D$: every open set in the space that meets $X$ must also meet $D$. If there were some $x_A \in X$ such that $x \prec x_A$ for every $x \in D$, then $M\setminus A$ would be an open set containing $x_A$ and disjoint from $D$. This is impossible, so $D$ must be cofinal (unbounded) in $X$ with respect to $\preceq$. This clearly implies that $$\bigcup\limits_{x_A \in D} A = \bigcup \mathcal{A}',$$ so $\{A \in \mathcal{A}':x_A \in D\}$ is the desired countable subset of $\mathcal{A}$.

Concerning the construction of $\mathcal{A}'$: If you use Zorn’s lemma to get $\mathcal{A}'$, note that you want to order the family of well-ordered subchains of $\mathcal{A}$ by extension, not by simple inclusion: $\mathscr{C} \preceq \mathscr{C}'$ iff $\mathscr{C}$ is an initial segment of $\mathscr{C}'$. For those familiar with ordinals it may be simpler to construct $\mathcal{A}'$ recursively. Suppose that $\alpha$ is an ordinal, and that we’ve chosen $A_\xi \in \mathcal{A}$ for each $\xi < \alpha$ in such a way that $A_\xi \subsetneq A_\eta$ whenever $\xi < \eta < \alpha$. If $$\bigcup\mathcal{A} \setminus \bigcup\limits_{\xi<\alpha}A_\xi \ne \varnothing,$$ there is an $A_\alpha \in \mathcal{A}$ such that $$A_\alpha \setminus \bigcup\limits_{\xi<\alpha}A_\xi \ne \varnothing,$$ and the recursion continues; if not, we’re done, and $\mathcal{A}' = \{A_\xi:\xi<\alpha\}$ is the desired subchain.

  • $\begingroup$ I think it is very clever to use hereditary separability here. I like this argument. $\endgroup$ – Stefan Geschke Sep 27 '11 at 12:18

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