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[parent] every filter is contained in an ultrafilter (Theorem)

Let $X$ be a set and $\mathcal{F}$ be a filter on $X$ . Then there exists an ultrafilter $\mathcal{U}$ on $X$ which is finer than $\mathcal{F}$ .

An importance consequnce of this theorem is the existence of free ultrafilters on infinite sets. According to the theorem, there must exist an ultrafilter which is finer than the cofinite filter. Since the cofinite filter is free, every filter finer than it must also be free, and hence there exists a free ultafilter.

Also note that this theorem requires the axiom of choice.




"every filter is contained in an ultrafilter" is owned by rspuzio. [ full author list (2) ]
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proof that every filter is contained in an ultrafilter (Proof) by rspuzio
proof of every filter is contained in an ultrafilter (alternate proof) (Proof) by brunoloff
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Cross-references: axiom of choice, cofinite filter, infinite sets, free ultrafilters, theorem, finer, ultrafilter, filter
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This is version 3 of every filter is contained in an ultrafilter, born on 2004-10-06, modified 2006-01-26.
Object id is 6304, canonical name is EveryFilterIsContainedInAnUltrafilter.
Accessed 1567 times total.

Classification:
AMS MSC54A20 (General topology :: Generalities :: Convergence in general topology )

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