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3 votes

On the Menger property and the Alexandroff duplicate

The reference is here, provided you cannot find anything more classical. Every closed subset of a Menger set is Menger. Thus if $A(X)$ is Menger, then its closed subset $X\times\{0\}\cong X$ is Menger …
Steven Clontz's user avatar
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Almost compact sets

I don't specifically recall "almost compact" in the literature, but it's quite natural as it relates to "almost Menger" and "almost Lindelof". In general, relative compactness is not equivalent to a s …
Steven Clontz's user avatar