Results 1 to 2 of 2

Math Help - Closed subset of sequentially compact set is sequentially compact.

  1. #1
    Member
    Joined
    Mar 2009
    Posts
    168

    Closed subset of sequentially compact set is sequentially compact.

    Let B be a subset of Y, and let B be sequentially compact. Let D be a closed subset of B. Then D is sequentially compact.

    Can anyone help me out and give me a general direction to go with proving this?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Apr 2009
    From
    México
    Posts
    721

    Re: Closed subset of sequentially compact set is sequentially compact.

    Let (x_n)\subset D be an infinite sequence, since B is sequentially compact there is a subsequence (I'll call it the same) and x\in B such that x_n \to x, but D is closed so...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. sequentially compact? Show bounded?
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: April 11th 2010, 10:11 PM
  2. Sequentially Compact and composition
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 16th 2008, 07:54 PM
  3. [SOLVED] Sequentially Compact
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: September 25th 2008, 04:32 AM
  4. [SOLVED] Closet Point/Sequentially compact subspace
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: August 12th 2008, 12:13 PM
  5. Compact and sequentially compact
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 2nd 2008, 01:57 AM

Search Tags


/mathhelpforum @mathhelpforum