Closed sets in a topological space

[logarithmic]

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?

 

[Hurkyl]

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...

 

[logarithmic]

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]

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.

 

[logarithmic]

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]

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]

?? 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]

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.