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CorollaryStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14
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A separable predual has weak-star sequentially compact dual ball

Statement

Assume the ultrafilter lemma. If X is a separable real or complex normed space, then every sequence in BX has a subsequence converging in the weak-star topology. Completeness of X is not required.

Facts & Assumptions

Given: The ultrafilter lemma and a separable real or complex normed space X.

[F1]

A separable space has an at most countable dense subset (Separability: the existence of an at most countable dense subset).

[F2]

A fixed dense sequence metrizes the weak-star topology on every norm-bounded subset of the dual (Dual ball weak-star metrizable for a separable predual).

[F3]

Under the ultrafilter lemma, BX is weak-star compact, without completeness of X (Banach–Alaoglu).

Proof

technique · direct
1.1

Fix an at most countable dense set DX. It is nonempty because 0X and the empty set is not dense in a nonempty space. If D is countably infinite, a witnessing bijection ND is a dense sequence. If D is finite, a witnessing finite list can be repeated periodically (and its first entry repeated after the list ends) to give a sequence with range D. Thus X has a fixed dense sequence; no countable family of choices was made.

F1given
1.2

The same ball is weak-star compact by Banach–Alaoglu; the ultrafilter lemma is used at this step through [F3].

F3
2.1

Applying [F2] to the norm-bounded set BX gives a metric inducing precisely its relative weak-star topology.

F2step 1.1
3.1

By steps 2.1 and 1.2 the ball is a compact metric space, hence countably compact by [F4] and sequentially compact by [F5]. Equivalently, every sequence in it has a weak-star convergent subsequence.

F4F5step 2.1step 1.2

Depends on

Used by

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Sources