Results 1 to 7 of 7

Math Help - Question about ordered topology

  1. #1
    Junior Member
    Joined
    Feb 2010
    Posts
    37

    Question about ordered topology

    This is a Munkres's problem: Let Y be an ordered set with the ordered topology and f,g functions from X to Y, show that the set \{x:f(x)\leq g(x)\} is closed. I can show this, for intance, when Y=R(real numbers set) but I can't prove this if Y is just an ordered set because it may have consecutive elements. Any comments will be appreciated
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by facenian View Post
    This is a Munkres's problem: Let Y be an ordered set with the ordered topology and f,g functions from X to Y, show that the set \{x:f(x)\leq g(x)\} is closed. I can show this, for intance, when Y=R(real numbers set) but I can't prove this if Y is just an ordered set because it may have consecutive elements. Any comments will be appreciated
    Do you want a solution or advice? My advice would be to show that it's compliment is open. It's much more obvious.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Feb 2010
    Posts
    37
    yes that's what I did but found the problen I mentioned earlier
    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 facenian View Post
    yes that's what I did but found the problen I mentioned earlier
    Write your argument, I don't (although I haven't admittedly gone through the whole proof) see where you'd want to use that.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Feb 2010
    Posts
    37
    the proof would be \{x:f(x)>g(x)\}=\cup_{z\in Y}(\{x:f(x)>z\}\cap\{x:g(x)<z\})
    the problem is that if Y has consecutive elemets then the inclusion
    \{x:f(x)>g(x)\}\supset\cup_{z\in Y}(\{x:f(x)>z\}\cap\{x:g(x)<z\}) seems to be strict
    Last edited by facenian; April 6th 2010 at 04:57 PM. Reason: latex correction
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Nov 2008
    Posts
    394
    Quote Originally Posted by facenian View Post
    the proof would be \{x:f(x)>g(x)\}=\cup_{z\in Y}(\{x:f(x)>z\}\cap\{x:g(x)<z\})
    the problem is that if Y has consecutive elemets then the inclusion
    \{x:f(x)>g(x)\}\supset\cup_{z\in Y}(\{x:f(x)>z\}\cap\{x:g(x)<z\}) seems to be strict
    Let W=\{x:f(x)>g(x)\}. To show W is open in X, I shall show that for every x_0 \in W, there exists an open set containing x_0 \in W and is contained in W.

    As you said, there are two cases to consider. Let x_0 be an arbitrary point in W.

    1. There is no element between f(x_0) and g(x_0) such that f(x_0) > g(x_0). Then, f^{-1}(g(x_0), \infty) \cap g^{-1}(-\infty, f(x_0)) is an open set containing x_0 and is contained in W, since f and g are continuous, Y is an ordered set in the order topology, and an intersection of open sets is open.

    2. There exists element(s) between f(x_0) and g(x_0) such that f(x_0) > g(x_0). Let z be such an element f(x_0) > z > g(x_0). Then, f^{-1}(z, \infty) \cap g^{-1}(-\infty, z) is an open set containing x_0 and is contained in W.

    Thus W is open in X and X-W is closed.
    Last edited by aliceinwonderland; April 6th 2010 at 07:36 PM. Reason: Gramma fixing
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Feb 2010
    Posts
    37
    Thank you Guys! Your proof has been very helpfull aliceinwonderland
    Last edited by facenian; April 7th 2010 at 03:36 PM. Reason: grammar correcton
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Question about partially ordered sets
    Posted in the Discrete Math Forum
    Replies: 9
    Last Post: April 6th 2011, 12:54 PM
  2. ordered pairs question
    Posted in the Statistics Forum
    Replies: 2
    Last Post: January 18th 2011, 01:16 PM
  3. Ordered Pairs Question
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: September 20th 2010, 09:31 PM
  4. Question about ordered field...PLEASE HELP!!!
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: September 10th 2009, 02:48 AM
  5. Simple Question about Well-Ordered Sets
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: March 16th 2008, 12:01 PM

Search Tags


/mathhelpforum @mathhelpforum