Results 1 to 4 of 4
Like Tree1Thanks
  • 1 Post By Plato

Math Help - Countably Compact Sets

  1. #1
    Super Member Aryth's Avatar
    Joined
    Feb 2007
    From
    USA
    Posts
    652
    Thanks
    2
    Awards
    1

    Countably Compact Sets

    I'm working on this problem, and I'm not sure how to solve it.

    Let (X,\Omega) be a topological space and let A \subseteq X be countably compact. Prove that every subset of A that is closed relative to \Omega is also countably compact.

    I have a similar proof where this problem is stated without the 'countably' part. I'm unsure how to proceed with countably compact sets... Any help would be appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1

    Re: Countably Compact Sets

    Quote Originally Posted by Aryth View Post
    Let (X,\Omega) be a topological space and let A \subseteq X be countably compact. Prove that every subset of A that is closed relative to \Omega is also countably compact.
    As always this depends upon the definitions in use.
    If B\subset A that is closed relative to \Omega. Can you show that any countable open cover of B has a finite sub cover all relative to \Omega~?
    Thanks from Aryth
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Aryth's Avatar
    Joined
    Feb 2007
    From
    USA
    Posts
    652
    Thanks
    2
    Awards
    1

    Re: Countably Compact Sets

    Ah, so it is similar to the proof with compactness... I'll just change the proof a little, then. If you could verify that it is correct I would grateful.

    Suppose that A \subseteq X is countably compact and let B \subset A be closed. We will let \mathcal{F} = \{F_i : i \in \mathbb{N}\} be a countable cover of A. Since X \backslash B is open, it follows that \mathcal{F} along with X \backslash B is a countable open cover of A. We know that A is countably compact, so A can be covered by a finite number of sets from \mathcal{F}, F_1, F_2, \cdots , F_n, and possibly X \backslash B. Since B \subset A, F_1, F_2, \cdots, F_n covers B. Therefore, B is countably compact.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1

    Re: Countably Compact Sets

    Quote Originally Posted by Aryth View Post
    Ah, so it is similar to the proof with compactness... I'll just change the proof a little, then. If you could verify that it is correct I would grateful.

    Suppose that A \subseteq X is countably compact and let B \subset A be closed. We will let \mathcal{F} = \{F_i : i \in \mathbb{N}\} be a countable cover of A. Since X \backslash B is open, it follows that \mathcal{F} along with X \backslash B is a countable open cover of A. We know that A is countably compact, so A can be covered by a finite number of sets from \mathcal{F}, F_1, F_2, \cdots , F_n, and possibly X \backslash B. Since B \subset A, F_1, F_2, \cdots, F_n covers B. Therefore, B is countably compact.
    Good. That will work.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Intersection of two countably infinite sets
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: March 22nd 2011, 06:30 PM
  2. Union of Two countably infinite sets
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: March 22nd 2011, 06:25 PM
  3. countably many intersection of sets
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: March 26th 2010, 09:49 PM
  4. the intersection of a collection of compact sets is compact
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: February 28th 2010, 01:58 PM
  5. countably infinite / uncountable sets
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: February 9th 2010, 12:18 PM

Search Tags


/mathhelpforum @mathhelpforum