TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Banach–Alaoglu

Statement

Assume the ultrafilter lemma. If $X$ is a real or complex normed space, then its closed dual unit ball $B_{X^{*}}$ is compact in the weak-star topology $σ (X^{*}, X)$ . Completeness of $X$ is not required.

Facts & Assumptions

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

[F1]

Evaluation is a weak-star homeomorphism of $B_{X^{*}}$ onto a closed subspace of $\prod_{x \in X} {z : ∣ z ∣ \leq ∥ x ∥}$ (Dual ball as a closed subset of a product).

[F2]

Assuming the ultrafilter lemma, an arbitrary product of compact Hausdorff spaces is compact (Assuming the ultrafilter lemma, an arbitrary product of compact Hausdorff spaces is compact).

[F3]

Closed and bounded subsets of $R^{n}$ are compact, and in particular closed bounded intervals in $R$ are compact (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

[F4]

A closed subspace of a compact topological space is compact (A closed subspace of a compact space is compact, and a finite union of compact subspaces is compact).

Proof

technique · direct

1.1

For each $x \in X$ , the disk $D_{x} = {z \in K : ∣ z ∣ \leq ∥ x ∥}$ is compact and Hausdorff: for $K = R$ it is a closed bounded interval, and for $K = C$ it is the closed Euclidean disk in $R^{2}$ . This finite-dimensional fact uses no choice; when $x = 0$ , $D_{x} = {0}$ .

2.1

The product $P = \prod_{x \in X} D_{x}$ is compact by compact-Hausdorff Tychonoff. This is the unique step that uses the assumed ultrafilter lemma.

F2step 1.1

3.1

By [F1], evaluation carries $B_{X^{*}}$ homeomorphically onto a closed subspace $C$ of $P$ . The subspace $C$ is compact by [F4].

F1F4step 2.1

4.1

An open cover of $B_{X^{*}}$ transports under the homeomorphism to an open cover of $C$ ; a finite subcover of $C$ pulls back to a finite subcover of the ball. Therefore $B_{X^{*}}$ is weak-star compact. No step used completeness of $X$ ; if $X = 0$ , both spaces in [F1] are singletons.

F1step 3.1∎

Depends on

Used by

A separable predual has weak-star sequentially compact dual ball Corollary
Dual unit ball has extreme points Corollary
Weak-star compact does not imply weak-star sequentially compact Counterexample
Weak-star compactness of probability measures Example
Countable compactness closes in the bidual Lemma
Banach–Alaoglu versus sequential Alaoglu Remark
Banach–Dieudonné linear-subspace criterion Theorem
Eberlein–Šmulian theorem Theorem
Reflexive iff unit ball weakly compact Theorem
Weak-star compactness of polar sets Theorem

Dependency tree · two levels

46 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Bühler–Salamon, Functional Analysis (standard reference, not scraped)
Gerald Teschl, Topics in Real and Functional Analysis (standard reference, not scraped)