Weierstrass compactness states that each infinite set has a limit point. Why Infinite set which is Dedekind finite with discrete metric not Weierstrass compact.
2026-04-03 07:27:24.1775201244
Infinite set which is Dedekind finite and Weierstrass compactness
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Recall that $x$ is a limit point of $A$ if and only if every neighborhood of $x$ meets $A$ on an infinite set. So in any case a discrete space will never be Weierstrass compact.
Simply because $\{x\}$ is a neighborhood of $x$. So no infinite set has any limit point. So the only way a discrete space can be Weierstrass is that it is finite, and the requirement is satisfied vacuously.
(The terminology is actually an accumulation point in some places, e.g. in Herrlich's "The Axiom of Choice", p.35)