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Compactness is topology-independent ?

  1. Nov 26, 2011 #1

    dextercioby

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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.
     
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  3. Nov 26, 2011 #2

    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.
     
  4. Nov 27, 2011 #3
    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 ∅.
     
  5. Nov 27, 2011 #4

    dextercioby

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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 ?
     
  6. Nov 27, 2011 #5

    micromass

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    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.
     
  7. Nov 27, 2011 #6

    micromass

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    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.
     
  8. Nov 27, 2011 #7

    dextercioby

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    I understand it now. It's clear.
     
  9. Nov 28, 2011 #8

    Bacle2

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    Actually if (X,T) is compact and T'≤T then (X,T') is also compact; same for connectedness--if you cannot find a disconnection in T, and T'≤ T, you will not find one in T'-- but you can make no conclusions the other way around. Even the collection of sequences/nets that converge change when the topology changes; the fewer the open sets , the easier it is to eventually be in all the open sets (just like a larger collection of open sets allows you to find more covers that may not have finite subcovers). Maybe someone here can explain in more detail why one would choose, e.g., the strong operator topology over weaker topologies, and other reasons for preferring a smaller(larger) topology over a larger(smaller). Then you also have the issue of the initial and final topologies, (largest and smallest respectively, that make a collection of maps continuous), and why one would want to choose one over the other, or some in-between choice.

    Maybe a trivial example of how a topology shapes the conenctedness properties of the space is that, in the extreme case of the indiscrete topology (X,∅) , X is not just connected, but strongly-connected ( no elements can be separated by open sets), but (x, Indiscrete) is totally disconnected ( I mean, totally!) , i.e., for any two points x,y , there is a disconnection AUB with x in A and y in B; just take A={x} and B=X-{x} ; disjoint open sets whose union is X . It would be nice to see how other levels of connectedness would change with the topology; all I can think is that simple-connectedness assumes connectedness, so that when connectedness disappears, so does simple-connectedness (tho in a totally-disconnected space, curves are just single points, since continuous images of the unit interval I muse be connected.). It would be nice to see an example of a space that changes its fundamental group when the topology changes; I will try to come up with one myself.
     
    Last edited: Nov 28, 2011
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