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Apr 11, 2012 at 1:27 history edited David White CC BY-SA 3.0
Texified, since problem was on front-page anyway
Dec 26, 2009 at 3:15 comment added Joel David Hamkins I'm not sure what the right generalization is. There was something special about omega in my argument, since with the product topology the basic open sets have finite support, and this was what allowed for the diagonalization argument at the end. Perhaps for larger cardinals, there might be a clever workaround...
Dec 25, 2009 at 8:32 comment added Guillermo Mantilla what you wrote is more than perfect,and actually I wonder if your proof can be slightly modify to prove a more general result. If a Hausdorff space $X$ has the property that $X^{\aleph_{\alpha+1}}$ is $\aleph_{\alpha+1}$-compact, does it follows that $X$ is $\aleph_{\alpha}$-compact?
Dec 25, 2009 at 8:24 vote accept Guillermo Mantilla
Dec 24, 2009 at 4:44 comment added Joel David Hamkins Oh, I see you wanted either a reference or a counterexample, whereas I gave a proof. I'm sorry that I don't know a reference, but since it appears to be true, it must have been known classically, so surely there is a reference. What a fun problem!
Dec 24, 2009 at 3:10 history edited Joel David Hamkins CC BY-SA 2.5
fixed typo.
Dec 24, 2009 at 3:02 history answered Joel David Hamkins CC BY-SA 2.5