Results 1 to 2 of 2

Thread: functions of topologies

  1. #1
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    10

    functions of topologies

    Let X be a topological space and let Y be a simply ordered set in the order topology. Let $\displaystyle f,g:X\to Y$ be two continuous functions.

    Show that the set $\displaystyle \{x\in X:f(x)\leq g(x)\}$ is closed in X.

    Since f is continuous, for every open set $\displaystyle V\subset Y$, $\displaystyle f^{-1}(V)=\{x\in X: f(x)\in V\}$ is an open subset of X. Similar, for every open set $\displaystyle U\subset Y$, $\displaystyle g^{-1}(U)=\{x\in X: g(x)\in U\}$.

    How do I go from this to showing what needs to be shown?
    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
    22

    Re: functions of topologies

    Quote Originally Posted by dwsmith View Post
    Let X be a topological space and let Y be a simply ordered set in the order topology. Let $\displaystyle f,g:X\to Y$ be two continuous functions.

    Show that the set $\displaystyle \{x\in X:f(x)\leq g(x)\}$ is closed in X.

    Since f is continuous, for every open set $\displaystyle V\subset Y$, $\displaystyle f^{-1}(V)=\{x\in X: f(x)\in V\}$ is an open subset of X. Similar, for every open set $\displaystyle U\subset Y$, $\displaystyle g^{-1}(U)=\{x\in X: g(x)\in U\}$.

    How do I go from this to showing what needs to be shown?
    Let $\displaystyle \Sigma=\left\{(x,y)\in Y^2:x\leqslant y\right\}$ then our set is the preimage of $\displaystyle \Sigma$ under the map the continuous map $\displaystyle X\to Y^2$ given by $\displaystyle x\to (f(x),g(x))$ and so it suffices to prove that $\displaystyle \Sigma$ is closed in $\displaystyle Y^2$. To see this we merely note that $\displaystyle Y^2-\Sigma=\left\{(x,y)\in Y^2:x>y\right\}$ and let $\displaystyle (x,y)\in Y^2-\Sigma$. We have two choices, if there exists $\displaystyle x<z<y$ then set $\displaystyle U=(z,\infty)\times (-\infty,z)$. If there is no point $\displaystyle z$ such that $\displaystyle x<z<y$ then set $\displaystyle U=(y,\infty)\times(-\infty,x)$. Either way this is an open set in $\displaystyle Y^2$ with the order topology which contains $\displaystyle (x,y)$ and is disjoint from $\displaystyle \Sigma$. It thus follows that $\displaystyle Y^2-\Sigma$ is open and so $\displaystyle \Sigma$ is closed, so that the conclusion follows from previous comment.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. continuous functions between Topologies
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: Oct 19th 2011, 07:59 AM
  2. Subspace topologies
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Sep 24th 2011, 09:39 PM
  3. Easier way to look at topologies?
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Oct 10th 2010, 04:18 AM
  4. IVT with different topologies
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Apr 10th 2009, 10:27 PM
  5. Comparing topologies
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: Jan 22nd 2009, 09:48 PM

/mathhelpforum @mathhelpforum