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Under the ultrafilter lemma, an ultrafilter algebra maps each ultrafilter to its unique limit

Statement

Assume UL/BPI. For an ultrafilter algebra ξ:βXX with its induced topology, every ultrafilter U converges to exactly one point, namely ξ(U).

Facts & Assumptions

Given: UL/BPI, an ultrafilter algebra ξ:βXX, and an ultrafilter U on X.

[L2]

A filter converges to p when every neighbourhood of p belongs to the filter (Convergence and cluster points of a filter on a topological space).

[L3]

The ultrafilter extension principle says that every filter on a set is contained in an ultrafilter on that set (The ultrafilter extension principle (UL/BPI)).

Proof

technique · direct
1.1

If an induced-open neighbourhood O contains ξ(U), the definition of induced-open gives OU. Thus U converges to ξ(U) by [L2].

givenL2
1.2

Let x be any limit of U. For each AU, every neighbourhood of x meets A, so xA=ξ[A^] by [L1].

L1L2
2.1

On βX, the family {A^:AU}{ξ1[{x}]} has the finite-intersection property by step 1.2. Extend it by [L3] to an ultrafilter W on βX.

step 1.2L3choose
3.1

The inclusions forced by step 2.1 and maximality give μX(W)=U and β(ξ)(W)=ηX(x). The algebra laws therefore give ξ(U)=ξμX(W)=ξβ(ξ)(W)=ξηX(x)=x.

step 2.1algebra
4.1

Step 1.1 supplies the limit ξ(U) and step 3.1 identifies every other limit with it, proving existence and uniqueness. If X=, no ultrafilter exists and the assertion is vacuous.

step 1.1step 3.1

Depends on

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