# Compactness is topology-independent ?

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Title says it all. If (X,T) is a compact topological space wrt T, is (X,T') compact wrt T' ?? My answer would be 'yes', but I don't know how to prove it...Any thoughts from the experts ?

T and T' of course arbitrary.

quasar987
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On X:=R u {∞} (the set R plus some other point) consider:
a) the topology that makes X into a circle (there is an obvious bijection btw X and S1. Use that to define the topology on X that make X homeomorphic to a circle)
b) the topology generated by the usual open sets of R plus the singleton {∞}

Under a), X is compact, but not under b) since {{∞},(-1/n,1/n)} is a covering withouth a finite subcover.

No the topology very much matters.

The silliest example is that (ℝ,T) is compact if T is the topology where the only open sets are ℝ and ∅.

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I see...The Alexandroff compactification. http://en.wikipedia.org/wiki/Alexandroff_extension.

But essentially R is non-compact, irrespective of the topology on it ?? Ok, I didn't see post 3, so that settles it. I assume the same conclusion goes for connectedness as well. What about path properties, like multiple-connectedness ?

I see...The Alexandroff compactification. http://en.wikipedia.org/wiki/Alexandroff_extension.

But essentially R is non-compact, irrespective of the topology on it ??

No, there are topologies on R in which R is compact. For example, the trivial/indiscrete topology.

The only spaces which are compact regardless of the topology are the finite spaces. ALL other spaces have topologies which make them noncompact: the discrete topology. And they also have topologies which make them compact: the trivial topology.

What about path properties, like multiple-connectedness ?

All these things are crucially dependent on the topology!!

If you disregard the topology, then you can only talk about the size of the underlying set. All other things have to do with the topology.

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I understand it now. It's clear.

Bacle2