# compact as finite set

• Feb 28th 2009, 04:09 AM
gls
compact as finite set
Hi,
We know if we provide a set X with the discrete topology, then

a subset of X is compact iff it is a finite set (*)

Is there any other topology T, other that the discrete topology, making X a Hausdorff space and satisfying the property (*)?
• Feb 28th 2009, 12:33 PM
Gamma
finite sets are trivially compact, if you take an open cover each point is covered by definition, just take one element of the cover for each point and it is finite since there are a finite number of points.

The converse is just as trivial if you attack it the right way. X has the discrete topology and is compact. Suppose it were infinite for a contradiction. Well each point in X is open under the discrete topology, so take this as your open cover. Try to find a finite subcover of this. Clearly not possible since if you removed one of these open sets from this cover it no longer would cover X thus X cannot be infinite and therefore must be finite.

Hope that helped
• Feb 28th 2009, 11:16 PM
tah
Quote:

Originally Posted by Gamma
finite sets are trivially compact, if you take an open cover each point is covered by definition, just take one element of the cover for each point and it is finite since there are a finite number of points.

The converse is just as trivial if you attack it the right way. X has the discrete topology and is compact. Suppose it were infinite for a contradiction. Well each point in X is open under the discrete topology, so take this as your open cover. Try to find a finite subcover of this. Clearly not possible since if you removed one of these open sets from this cover it no longer would cover X thus X cannot be infinite and therefore must be finite.

Hope that helped

Yes, that is exactly the property (*) that gls said. But the question is: is this property enough characterize the discrete topology or could we construct some Hausdorff topological space satisfying (*) and strictly less finer that the discrete topology ?
• Mar 1st 2009, 03:21 AM
Gamma
oops
wow I totally missed that last line, sorry about that. This is a much more interesting question than I originally thought. Some thoughts:
1) finite set always implies compact
2) X would have to be infinite otherwise Hausdorff is enough to ensure it were discrete topology since finite intersections of open sets are open.
3) I am thinking it is true, since we need to show compact => finite means we have the discrete topology, and the contrapositive says if it is infinite subset then it is not compact, which seems to mean there can be no limit points which would seem to me to mean along with Hausdorff that every point must be isolated and therefore be endowed with the discrete topology.

but it is 5am here and I might be losing my mind. Just thought I should at least offer some actual help since I screwed up and misread the question the first time! Sorry!