Is a subspace of a supercompact space also supercompact?

Question

Of course, we assume that subspace is closed

Answer 1

NO.

Supercompact means that there is a subbase for the topology such that any cover by subbasic elements has a subcover with exactly two elements. Any supercompact space is compact by the Alexander sub-base theorem (which is sometimes used to prove Tychonoff's theorem).

The closed unit interval is supercompact, but the open unit interval is not. It's not clear to me immediately whether closed subspaces of supercompact spaces need be supercompact.

Added:

It turns out that the Stone-Cech compactification of the naturals is not supercompact. Since it can be embedded as a closed subspace of a product of intervals and since products of supercompact spaces are supercompact, this gives an example of a closed subspace of a supercomapct space whichis not supercompact.

Answer 2

Yes. A subspace has the same properties as that of the space it is a subspace of so yes it will be supercompact whatever that is.

Answer 3

I'm not sure about the answer to this question, so Paul may be right, but his reasoning is not. Compactness is not carried to subspaces, that is easy to show. I would imagine that Supercompactness would also not carry, but since it is stricter maybe it is, as I said, I don't know.

Answer 4

Please define supercompact.