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In reply to "Re: free ultrafilter", posted by mars on Aug 30, 2002:
>In reply to "free ultrafilter", posted by Vidhyanath on Aug 30, 2002:
>>Is it really impossible to construct a free ultrafilter? If so, why?
>No, it's possible. Construct the Frechet filter on the integers,
>which is the filter containing all cofinite sets of integers.
>Now apply theorem, every filter can be extended to an ultrafilter.
True enough. But this does not qualify as a construction,
only a proof of existence.
And without Choice (AC, or in fact a weaker variant of it) one cannot prove that
there are any free ultrafilters.
Sketch of proof: there is a model of set theory (ZF) with dependent choice
(which implies countable choice), such that all subsets of R are measurable.
If we have a free ultrafilter on N we can define (explicitly, if the ultrafilter
is given explicitly as well) a non-measurable subset of R.
So no free ultrafilters on N can exist in this model.
So from ZF alone (even with countable choice) we cannot find a free ultrafilter on N.
No hope for an explicit construction, I'm afraid..
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