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

Posts tagged open problems

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):

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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 \(\omega_1\).

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Open Problems!

I've decided to have a list of open problems on my site. I am no Erdős, nor Hilbert, nor Knuth.

But I want my own problems page, and it's my site. So to celebreate the new website, I created just that. For the first couple of problems, I've chosen to focus on the axiom of choice. And I don't think that I have much choice, but to keep that interest running. But I can promise that this is not the only type of problems that I will add there.

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