Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-06
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Mazur theorem: weak and norm closure agree for convex sets

Statement

For a convex subset C of a normed space X, its norm closure equals its weak closure. Here the weak topology is generated by the basic neighbourhoods of x

{y:fj(yx)<ε (1jn)}

for finitely many fjX and some ε>0.

Facts & Assumptions

Given: A convex set CX and D=C.

[F1]

Disjoint convex sets, one open, are strictly separated by a nonzero continuous functional (Separation of disjoint convex sets when one is open).

Proof

technique · direct
1.1

Every fX is norm-continuous, so every displayed basic weak neighbourhood is norm-open. Thus every weak-open set is norm-open, and DCweak.

given
1.2

If C=, both closures are empty. Otherwise D is nonempty; if xD, choose r>0 with xD+B(0,r). This latter set is open and convex, so [F1] yields f0 with Ref(d+b)<Ref(x) for dD, b<r.

givenF1choose
2.1

Taking suprema over bB(0,r) gives supdDRef(d)+rfRef(x). The weak neighbourhood {y:f(yx)<rf/2} therefore misses D, hence misses C.

step 1.2algebra
3.1

Thus x is not in the weak closure whenever it is not in D, proving the reverse inclusion and equality.

step 1.1step 2.1

Depends on

Used by

Dependency tree · two levels

10 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