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Is there a partition of the real line into infinitely many closed subsets so that no infinite union of these subsets (except the whole space) is closed?

This question was asked also at math.stackexchange.com. While an interesting construction of a partition with no closed uncountable unions was given, there has not yet been a conclusive answer.

Two observations:

  1. No infinite subcollection of the partition can be locally finite since the union of a locally finite collection of closed sets is closed.
  2. The partition must be uncountable since the real line cannot be partitioned into countably many (and $\geq 2$) closed subsets by a theorem of Sierpiński.
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1 Answer 1

up vote 3 down vote accepted

Let $\mathcal{F}$ be this partition. As noted above it must be uncountable. We may as well assume it lives on the interval $(0,1)$ and add the set $ \lbrace 0, 1\rbrace$ to each member to obtain a family of closed subsets of $[0,1]$. Now $\mathcal{F}$ is an uncountable subset of the space of closed subsets of $[0,1]$ endowed with the Hausdorff metric. The latter space is separable metric, so $\mathcal{F}$ contains a non-trivial sequence $\langle F_n:n\in\mathbb{N}\rangle$ that converges to a point $F$ of $\mathcal{F}$. The union $F\cup\bigcup_n F_n$ is closed: if $x$ is outside the union, in particular outside $F$, let $\epsilon=\frac12 d(x,F)$. Then there is an $N$ such that $F_n\subseteq B(F, \epsilon)$ for $n\ge N$. This now easily implies that $x$ is not in the closure of the union.

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After a bit of ingenuity to consider the space of closed subsets it becomes so easy! Thank you very much. By the way, if I understand it correctly, one should perhaps add $0$ and $1$ to each element of $\mathcal{F}$ to make sure they are closed in $[0,1]$ too. –  LostInMath Mar 18 '12 at 3:05
    
Indeed, I'll edit the answer to reflect your suggestion. –  KP Hart Mar 18 '12 at 9:04

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