# Closed sets in a topological space

1. May 26, 2009

### 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?

2. May 26, 2009

### Hurkyl

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

3. May 26, 2009

### logarithmic

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.

4. May 26, 2009

### HallsofIvy

Staff Emeritus
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.

5. May 26, 2009

### logarithmic

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?

6. May 26, 2009

### 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).

7. May 26, 2009

### HallsofIvy

Staff Emeritus
?? 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.

8. May 26, 2009

### 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.