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Dec 18, 2010 at 21:38 comment added Daniel Litt This certainly only works for normal spaces.
Dec 18, 2010 at 14:14 comment added Greg Graviton Found a slightly different equivalence relation that only mentions the topology of $X$. The trick is too choose $\sim$ large enough to identify convergent ultrafilters while making it small enough to separate ultrafilters that can be distinguished by continuous functions into compact spaces.
Dec 18, 2010 at 14:10 history edited Greg Graviton CC BY-SA 2.5
New equivalence relation
Dec 18, 2010 at 9:09 comment added Greg Graviton What do you mean? The quantification over all compact spaces $C$ is fairly natural and the proof goes through. But you seem to be unhappy with it, probably because it's "too indirect", i.e. it would be nicer if only the space $X$ were mentioned? It's clear that any two convergent ultrafilters need to be equal in the quotient; the problem is to identify ultrafilter which are not convergent in $X$ but do converge in all compact images of $X$.
Dec 18, 2010 at 1:50 comment added Qiaochu Yuan Yes, this is essentially what I was thinking. To prove continuity and the universal property it should be easier to prove that all the relevant maps preserve convergence of ultrafilters. But the essential problem, it seems to me, is the definition of \sim, and I will edit my question accordingly.
Dec 17, 2010 at 11:13 history edited Greg Graviton CC BY-SA 2.5
Language
Dec 17, 2010 at 11:09 comment added Greg Graviton Fair enough. I have completely rewritten my answer.
Dec 17, 2010 at 11:08 history edited Greg Graviton CC BY-SA 2.5
Proper answer
Dec 16, 2010 at 18:52 comment added Qiaochu Yuan Right, but this presupposes the existence of the Stone-Čech compactification in general. I'm trying to use this argument to construct the Stone-Čech compactification in general.
Dec 16, 2010 at 16:59 history edited Greg Graviton CC BY-SA 2.5
Fix type of inclusion
Dec 16, 2010 at 11:26 history answered Greg Graviton CC BY-SA 2.5