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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 …
1
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Accepted
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 …