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 November 5th, 2010, 08:25 AM #1 Newbie   Joined: Jan 2009 Posts: 20 Thanks: 0 continuous map Let f : (X x Y) --> R be continuous where Y is compact. Show that the map g : X --> R defined as g(x) = sup{f(x, y), taken over y, is continuous. Any help would be appreciated.
 November 5th, 2010, 09:48 AM #2 Newbie   Joined: Oct 2010 Posts: 17 Thanks: 0 Re: continuous map $g$ is well-defined since $Y$ is compact. $g^{-1}((a,\infty))=\{x\in X : \exists y\in Y(f(x,y)>a\}=\cup_{y\in Y}\{x\in X : f(x,y)>a\}=$, which is open. If $x_0\in g^{-1}((-\infty, b)),$then $f(x_0,y) For every $z\in Y$ there is an open neighborhood $U_z$ of $x_0$ in $X$, and an open neighborhood $V_z$ of $z$ such that $f(x,y) whenever $(x,y)\in U_z\times V_z.$ Since $Y$ is compact, there are $z_1, \dots, z_n$ such that $Y=\cup_{i=1}^nV_{z_i}.$ Let $U=\cap_{i=1}^nU_{z_i}$ Then $U$ is open, and $U\times V_{z_i}\subseteq U_{z_i}\times V_{z_i},$ so $f on $U\times V_{z_i}.$ Furthermore, $\cup_{i=1}^nU\times V_{z_i}=U\times V,$ implying that $U\subseteq g^{-1}((-\infty, b)).$ Maybe there´s a shorter argument, but I don´t see it.
 November 5th, 2010, 01:46 PM #3 Newbie   Joined: Jan 2009 Posts: 20 Thanks: 0 Re: continuous map Perhaps I should mention that X, Y are locally compact and Hausdorff. Regardless, thanks for the help; your proof looks correct.

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