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Math Help - functions of topologies

  1. #1
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    functions of topologies

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

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

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

    How do I go from this to showing what needs to be shown?
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    MHF Contributor Drexel28's Avatar
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    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 f,g:X\to Y be two continuous functions.

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

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

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