Closed sets in a topological space

**logarithmic** · May26-09, 10:31 AM

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

**Hurkyl** · May26-09, 10:36 AM

Originally Posted by logarithmic

If $LaTeX Code: A\\subseteq B$ are both subsets of a topological space $LaTeX Code: (X,\\tau)$ , 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** · May26-09, 10:50 AM

Originally Posted by Hurkyl

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** · May26-09, 11:40 AM

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** · May26-09, 11:51 AM

Originally Posted by 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.

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** · May26-09, 11:59 AM

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** · May26-09, 02:46 PM

?? 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** · May26-09, 02:57 PM

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.