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

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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

1.1F3

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 $x = 0$ , $D_{x} = {0}$ .

2.1F2step 1.1

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.

3.1F1F4step 2.1

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

4.1F1step 3.1∎

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.

Banach–Alaoglu

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources