Closed sets in a topological space

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SUMMARY

In a topological space (X, τ), the assertion that any closed subset of A is also a closed subset of B, where A ⊆ B, is false if A is not closed in B. The discussion highlights that while A is closed in itself, it does not guarantee closure in B unless A is explicitly stated to be closed. The concept of compactness does not rectify this issue, as compact sets are not necessarily closed, as illustrated by the Zariski topology on R. The example of the interval (0,1) demonstrates that compact sets can exist without being closed.

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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?
 
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logarithmic said:
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...
 
Hurkyl said:
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.
 
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.
 
HallsofIvy said:
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?
 
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).
 
?? 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.
 
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.
 

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