# Closed sets in a topological space

If $$A\subseteq B$$ are both subsets of a topological space $$(X,\tau)$$, is it true that any closed subset of A is also a closed subset of B?

## Answers and Replies

Hurkyl
Staff Emeritus
Science Advisor
Gold Member
If $$A\subseteq B$$ are both subsets of a topological space $$(X,\tau)$$, is it true that any closed subset of A is also a closed subset of B?

Note that A is a closed subset of A....

Note that A is a closed subset of A....
Hmm, I'm not sure. I never said A was closed. A is in B which is in X. And some other set in A, maybe call it U, is closed. I think the answer would be U is closed in X but I don't quite see why.

HallsofIvy
Science Advisor
Homework Helper
Hurkyl's point is that any topological space is both open and closed as a subset of itself. If A is NOT closed as a subset of topological space B, since it IS closed as a subset of itself, the statement "any closed subset of A is also a closed subset of B" is false.

Hurkyl's point is that any topological space is both open and closed as a subset of itself. If A is NOT closed as a subset of topological space B, since it IS closed as a subset of itself, the statement "any closed subset of A is also a closed subset of B" is false.
Ahh i see. Thanks. But is that the only problem here? If we insist that A is compact in B, then that fixes the problem and the statement is true, right?

matt grime
Science Advisor
Homework Helper
Um, just insisting that A is closed is all you need, nothing to do with compactness. In fact compactness won't help you at all - compact does not imply closed (e.g. the Zariski topology on R).

HallsofIvy
Science Advisor
Homework Helper
?? I thought compact did imply closed!

Oh, I see. I started to give the proof and then realized I was saying "given points p and q construct neighborhoods about p and q that do not intersect". That's not possible in some topological spaces.

matt grime
Science Advisor
Homework Helper
If you don't know what the Zariski topology is (and Halls does but forget, temporarily) consider the topology on R given by:

U is open if and only if U contains the interval (0,1) - the set (0,1) is in this and is certainly compact, but not closed.