Is it possible to decompose $\mathbf R$, or in general, any uncountable Polish space, into $\mathfrak c$-many disjoint closed subsets such that the union of any infinite subfamily is dense?

If it is, what additional constraints (if any) have to be put on the space to allow this (I suspect connectedness/zero dimensionality might play an important role there, if it is possible at all)?

I tried to construct such a decomposition using the Bernstein set (or a family of those), but without much success.

This question is motivated by another question (which, as far as I can see, is equivalent to my question in case of $\mathbf R^2$).

edit: As for the last remark, it seems that I was under the false impression that for quotient maps, the closure of preimage is the preimage of closure, which is not true in general.


Such a partition does not exist for any first countable space, so certainly not for Polish spaces.

Let $P$ be a collection of closed sets, such that the union of every infinite subset is everywhere dense. If $x$ is a point with a countable neighbourhood base, then $x \in F$ for all but countably many $F \in P$.

Proof: Let $\{B_n\}_{n \in \mathbb{N}}$ be a decreasing neighbourhood base of $x$. Define $d: P \to [-\infty, +\infty]$ by $d(F) = \sup \{ n \mid F \cap B_n \neq \emptyset \}$. Because the union of every infinite subset of $P$ must intersect every $B_n$, we have that $\{ F \in P \mid d(F) \le n \}$ is finite for every $n \lt \infty$. If follows that $d(F) = \infty$ for all but countably many $F\in P$, and for all these $F$, since $F$ is closed, $x \in F$.

  • $\begingroup$ Thanks. I knew it shouldn't be too complicated. :) I think it would be clearer if you wrote $\lbrace F\vert d(F)\leq n\rbrace$, though, since that's easier to see and that's what we really want. $\endgroup$ – tomasz Jun 24 '12 at 2:09
  • $\begingroup$ I originally had it that way. I don't know why I changed it, but it's changed back now. $\endgroup$ – Niels J. Diepeveen Jun 24 '12 at 10:31

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