Asaf Karagila
I don't have much choice...

## Countable sets of reals

One of the classic results of Sierpinski is that if there are as many countable sets of reals as there are reals, then there is a set which is not Lebesgue measurable. (You can find a wonderful discussion on MathOverflow.)

This is fact is used in the paradoxical decomposition theorems (which I often enjoy bringing up as a counter-argument to bad arguments that the Banach–Tarski paradox implies we need to accept that all sets are measurable as an axiom):

## Preserving Properness

I just posted another problem in the problems page. The prize, by the way, is a bottle of port wine, or equivalent. And I truly hope to make good on that prize.

In another problem there, coming from a work with David Asperó, we asked if an $$\omega_2$$-closed forcing must preserve the property of being proper. Yasou Yoshinobu provided us with a negative answer based on Shelah's "Proper and Improper Forcing" XVII Observation 2.12 (p.826). Take $$\kappa$$ to be uncountable, by forcing with $$\Add(\omega,1)\ast\Col(\omega_1,2^\kappa)$$ and appealing to the gap lemma, $$(2^{<\kappa})$$ is a tree with only $$\aleph_1$$ branches. It can therefore be specialized by a ccc forcing in that model. The iteration of these three forcing (Cohen real, collapse, specialize) is clearly proper. But now by forcing with $$\Add(\kappa,1)$$ we must in fact violate the properness of this forcing, which was defined in the ground model, since the new branch is also generic for the tree and will therefore collapse $$\omega_1$$.