Cofinite Topology: Borel Algebra vs. Power Set

Question

Being curious...

Are there uncountable spaces such that any uncountable subset has countable complement: $$\#\Omega>\aleph_0:\quad\#A>\aleph_0\implies\#A^\complement\leq\aleph_0\quad(A\subseteq\Omega)$$

If so then for these the Borel algebra induced by the cofinite topology is already the power set itself.

I might be wrong, but isn't this like asking whether there is an uncountable space that is actually smaller than $\mathbb{R}$ since $\mathbb{R}$ is 'too big' for this property to hold? If so, it might be equivalent to the continuum hypothesis. — j4GGy, Dec 10 '14 at 06:24

score 1 · Accepted Answer · edited Apr 13 '17 at 12:20

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Every uncountable set can be written as the union of two disjoint uncountable sets.

See, for example, Does any uncountable set contain two disjoint uncountable sets?

edited Apr 13 '17 at 12:20

Community

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answered Dec 10 '14 at 06:26

MPW

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Great, thanks. :D – freishahiri Dec 10 '14 at 06:34
@Freeze_S: I think you can say that every infinite set is the union of two disjoint copies of itself. And so you can say that a nonempty set is infinite if and only it is the union of two disjoint copies of itself. – MPW Dec 10 '14 at 06:48
Nitpick: this does use the axiom of choice. Without AC such a set might exist! – Henno Brandsma Dec 10 '14 at 07:56

Cofinite Topology: Borel Algebra vs. Power Set

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