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  "body": "\n## Definition\n\nLet $X$ and $Y$ be [[topological spaces|0CQHFPGJ45]], and let $C(X, Y)$ be the set of [[continuous maps|0VGFHTM75Y]] from $X$ to $Y$.\nFor a [[compact|0R6EWFKRCF]] subset $K \\subseteq X$ and an [[open|02BH3MCBQB]] subset $U \\subseteq Y$, put\n$$V(K, U) = \\set{f \\in C(X, Y) : f(K) \\subseteq U}.$$\nThe *compact-open topology* on $C(X, Y)$ is the topology generated by the [[subbasis|09S8DV34Y0]] of all sets $V(K, U)$.\n\n## Remarks\n\nIf $(Y, d)$ is a metric space, the compact-open topology is the topology of uniform convergence on compact subsets of $X$.\nIn particular, a sequence $(f_n)$ converges to $f$ if and only if $\\sup_{x \\in K} d(f_n(x), f(x)) \\to 0$ for every compact $K \\subseteq X$.\nIf $X$ is locally compact Hausdorff, the evaluation map $C(X, Y) \\times X \\to Y$, $(f, x) \\mapsto f(x)$, is continuous.\n\nExample: for $X = [0, 1]$ and $Y = \\R$, the space $X$ is compact, so convergence in $C([0, 1], \\R)$ is uniform convergence on $[0, 1]$.\n\nNon-example: the topology of pointwise convergence on $C([0, 1], \\R)$ is not the compact-open topology.\nLet $f_n$ be the piecewise linear map with $f_n(0) = 0$, $f_n(1/n) = 1$ and $f_n(x) = 0$ for $x \\ge 2/n$.\nThen $f_n(x) \\to 0$ for each $x$, but $f_n \\notin V([0, 1], (-1, 1))$ for every $n$, so $f_n$ does not converge to $0$ in the compact-open topology.\n",
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