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](https://math.stackexchange.com/questions/114170/infinite-closed-partition-of-the-real-numbers-with-a-certain-property). 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.



Infinite closed partition of the real line with no closed infinite unions
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.


