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December 11

Unique normal ultrafilter

So I'm supposed to know the answer to this, I suppose, but I don't seem to :-)

"Everyone knows" that, in $L[U]$ , Gödel's constructible universe relative to an ultrafilter $U$ on some measurable cardinal $\kappa$ , there is only a single normal ultrafilter, namely $U$ itself. See for example John R. Steel's monograph here, at Theorem 1.7.

So I guess that must mean that the product measure $U\times U$ , meaning you fix some identification between $\kappa \times \kappa$ and $\kappa$ and then say a set has measure 1 if measure 1 many of its vertical sections have measure 1, must not be normal. (Unless it's somehow just equal to $U$ but I don't think it is.)

But is there some direct way to see that? Say, a ~~continuous~~ function $f:\kappa \to \kappa$ with $\forall \alpha f(\alpha )<\alpha$ such that ~~the set of fixed points of $f$ is not in the ultrafilter~~ no singleton has a preimage under $f$ that's in the ultrafilter? I haven't been able to come up with it. --Trovatore (talk) 06:01, 11 December 2024 (UTC)[reply]