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Is there a partition of the real numbers into infinitely many closed subsets so that no infinite union of these subsets (other than the whole set of real numbers) is closed?

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I'd think that if the answer is positive it lies within Cantor set sort of sets. –  Asaf Karagila Feb 27 '12 at 22:47
    
Note that it is impossible to write $\mathbb R$ as the disjoint union of countably many (but $\ge 2$) nonempty closed sets. So you'll need an uncountable partition. –  Robert Israel Feb 27 '12 at 23:42
    
Another observation is that almost all the sets needs to be very close to one another at some point, if you had infinitely many sets which are separated by pairwise disjoint open sets then their union would be closed. –  Asaf Karagila Feb 28 '12 at 0:14

1 Answer 1

Perhaps it's worthwhile to state a weaker result I was able to come up with. There is a partition of $\mathbb R$ into sets of cardinality $2$ such that no uncountable union of these sets (other than $\mathbb R$) is closed.

The cardinality of the set $\cal F$ of closed proper subsets of $\mathbb R$ is that of the continuum, so we can well-order it by the first ordinal of cardinality $\cal c$. Thus we have a well-ordering $\prec$ of $\cal F$ in which for each $A \in \cal F$, the set of $B \in \cal F$ such that $B \prec A$ has cardinality strictly less than $\cal c$. Now using transfinite recursion, for each $A \in \cal F$, take $P_A$ as follows. If $A \backslash \bigcup_{B \prec A} P_B$ is nonempty, choose one of its members to be in $P_A$. Also choose one or two members of $A^c \backslash \bigcup_{B \prec A} P_B$ to be in $P_A$, so that $P_A$ has two members. Note that the nonempty open set $A^c$ has cardinality $\cal c$, which is larger than the cardinality of $\bigcup_{B \prec A} P_B$, so this is always possible. Any uncountable closed subset of $\mathbb R$ has cardinality $\cal c$, so if $A$ is uncountable $P_A$ will contain a member of $A$.

By construction the $P_A$ are all disjoint. Their union is all of $\mathbb R$, since any $x$ will be in $P_{\{x\}}$ if it is not in some $P_B$ for $B \prec \{x\}$. And no uncountable closed set $C$ can be a union of the sets $P_A$: such a union would have to include $P_C$, because one member of $P_C$ is in $C$, but $P_C$ also contains a member of $C^c$.

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Thanks for presenting this clever construction. It took some time for me to understand why $P_C$ must contain an element of $C$ in the case of an uncountable closed $C$, but of course it follows from the classical result that such $C$ must be of cardinality continuum. However, as far as I understand this construction leaves room for closed countably infinite unions, so for the time being I would like to respectfully leave the question still open for more conclusive answers. –  LostInMath Feb 29 '12 at 23:55
    
Yes, as I said, this is a weaker result than what you want, but perhaps of some interest. –  Robert Israel Mar 1 '12 at 2:17
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Also perhaps of interest, let me point out this. The partition ${\cal P} = \{P_A: A \in \cal F\}$ corresponds to a involution $f: {\mathbb R} \to {\mathbb R}$ such that ${\cal P} = \{\{x,f(x)\}: x \in {\mathbb R}\}$. But $f$ is non-measurable. In fact, suppose $\mu$ is a nonatomic positive regular Borel measure on $\mathbb R$ and $S$ some $\mu$-measurable set with $0 < \mu(S) < \infty$. Suppose $g \cdot I_S$ is $\mu$-measurable (where $I_S$ is the indicator function of $S$). ... continued –  Robert Israel Mar 1 '12 at 2:42
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Then by Lusin's theorem there is a continuous function $h$ on $\mathbb R$ such that $g = h$ on a set of positive $\mu$-measure. This set has an uncountable compact subset $K$ of positive $\mu$-measure. Then $g(K) = h(K)$ is compact, and the uncountable closed set $K \cup g(K) = \bigcup_{x \in K} \{x, g(x)\}$ is the union of uncountably many members of the partition corresponding to $g$. –  Robert Israel Mar 1 '12 at 2:47

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