Results 1 to 4 of 4

Math Help - Compactness II

  1. #1
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782

    Compactness II

    Given topologies T and T' on a set X with T \subset T', prove that if (X,T') is compact then so is (X,T).
    (X,T') compact gives:

    Let U be any open cover of (X,T'). Then there exists a finite subcover of U, say U_1=\{u_1, \ldots, u_n\} such that X \subset U_1 \subset \bigcup_{i=1}^n u_i.

    Each u_i is open (is the "finite subcover" always open?) so each u_i \in T'.

    From here I get stuck. If we consider cases, if every u_i \in T \subset T' then i'm done! However, if u_i \in T'-T then i'm not really sure what to do. I don't even think this is possible since u_i's are open so should be in T anyway.

    I'd really appreciate some help!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by Showcase_22 View Post
    (X,T') compact gives:

    Let U be any open cover of (X,T'). Then there exists a finite subcover of U, say U_1=\{u_1, \ldots, u_n\} such that X \subset U_1 \subset \bigcup_{i=1}^n u_i.

    Each u_i is open (is the "finite subcover" always open?) so each u_i \in T'.

    From here I get stuck. If we consider cases, if every u_i \in T \subset T' then i'm done! However, if u_i \in T'-T then i'm not really sure what to do. I don't even think this is possible since u_i's are open so should be in T anyway.

    I'd really appreciate some help!
    I think you're trying to do this the wrong way round. You are told that (X,T') is compact, and you want to show that (X,T) is compact.

    So let U be an open cover of (X,T). The members of U are T-open subsets of X. But T\subset T', so the members of U are T'-open. Therefore they form an open cover of (X,T'). So there is a finite subcover (and that's all that you need, to show that (X,T) is compact).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,413
    Thanks
    1853
    Quote Originally Posted by Showcase_22 View Post
    (X,T') compact gives:

    Let U be any open cover of (X,T'). Then there exists a finite subcover of U, say U_1=\{u_1, \ldots, u_n\} such that X \subset U_1 \subset \bigcup_{i=1}^n u_i.
    Are you sure this is what your notes say? The condition that X be compact is that there exist a finite subcover such that X\subset \bigcup_{i=1}^n u_i. There is no requirement that it be contained in one of the open sets so X\subset u_1 shouldn't be there.

    Each u_i is open (is the "finite subcover" always open?)
    You are misunderstanding- an "open cover" is a collection of sets in the topology, not a set in the topology itself so the word "open" does not apply to U itself but to the sets in U. Since U is an open cover, every set in it is open so every set in any sub-cover of U is also open.

    so each u_i \in T'.

    From here I get stuck. If we consider cases, if every u_i \in T \subset T' then i'm done! However, if u_i \in T'-T then i'm not really sure what to do. I don't even think this is possible since u_i's are open so should be in T anyway.

    I'd really appreciate some help!
    What exactly are you trying to prove? If you want to prove "if X is compact in T' then it is compact in T\subset T' you have it the wrong way around. Given that there is a finite subcover of X in T', it does NOT follow that there is a finite subcover in T. There might be some open set in that subcover in T' that is not in T.

    It is true that if X is compact in T' and T'\subset T then X is compact in T.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by Showcase_22 View Post
    (X,T') compact gives:

    Let U be any open cover of (X,T'). Then there exists a finite subcover of U, say U_1=\{u_1, \ldots, u_n\} such that X \subset U_1 \subset \bigcup_{i=1}^n u_i.

    Each u_i is open (is the "finite subcover" always open?) so each u_i \in T'.

    From here I get stuck. If we consider cases, if every u_i \in T \subset T' then i'm done! However, if u_i \in T'-T then i'm not really sure what to do. I don't even think this is possible since u_i's are open so should be in T anyway.

    I'd really appreciate some help!
    Define X,T')\to(X,T)" alt="\varphiX,T')\to(X,T)" /> by the identity map. Then clearly this is continuous. For, if E\subseteq (X,T) is open then E\in T\implies E=\varphi^{-1}(E)\in T' and thus \varphi^{-1}(E) is open. Thus, we see that (X,T) is the continuous image of a compact space, thus compact.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Compactness
    Posted in the Differential Geometry Forum
    Replies: 11
    Last Post: March 11th 2010, 12:17 PM
  2. Compactness
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: January 19th 2010, 02:12 AM
  3. compactness
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: October 26th 2009, 12:26 PM
  4. Topological Compactness and Compactness of a Set
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: September 22nd 2009, 01:16 AM
  5. Compactness
    Posted in the Calculus Forum
    Replies: 4
    Last Post: August 16th 2007, 07:00 AM

Search Tags


/mathhelpforum @mathhelpforum