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Suppose that $f$ is the continuous function from $X^2$ to $I=[0,1]$ and $K$ is the compact subset of $X$. Define the function $F$ from $X$ to $I^K$ by $F(x)(b)=f(x,b)$. The topology on the space $I^K$ is compact-open topology. Is $F$ continuous?

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Theorem 3.4.1 of Engelking’s General Topology says that for every pair of topological spaces $X$ and $Y$, the compact-open topology on $Y^X$ is proper. Just before Proposition 2.6.11 he defines the notion of a proper topology on $Y^X$ as follows:

A topology on $Y^X$ is proper if for every space $Z$ and any $f\in Y^{(Z\times X)}$, the function $F:Z\to Y^X$ defined by $\big(F(z)\big)(x)=f(z,x)$ is continuous.

Take his $X,Y,Z$ to be your $K,I,X$, respectively, and you have the desired result: your $F$ is continuous.

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It has the same result on the topology of uniform convergence? –  Paul Jan 4 '12 at 11:16
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@John: Not in general, no. However, if $X$ is compact, the uniform and compact open topologies on $C(X,Y)$ coincide. –  Brian M. Scott Jan 4 '12 at 12:52
    
Brian Sir, you are right but I have one problem with your answer. Actually in Engelking $Y^X$ means space of all continuous functions form X to Y. But here $Y^X$ does not mean that. –  Abcd J Jan 4 '12 at 15:42
    
$I^K$ denotes the function space here. –  Paul Jan 4 '12 at 23:44

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