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  "body": "\n## Definition\n\nLet $X$ be a set, let $(Y, d)$ be a [[metric space|0VKD1TP7F3]], and let $Y^X$ be the set of all maps from $X$ to $Y$.\nThe uniform metric on $Y^X$ is\n$$\\bar d(f, g) = \\sup_{x \\in X} \\min\\bigl(d(f(x), g(x)), 1\\bigr).$$\nThe *topology of uniform convergence* on $Y^X$ is the [[metric topology|0YVKNB7V5F]] of $\\bar d$.\nOn a subset of $Y^X$, it is the [[subspace topology|0WS9P12FQR]].\n\n## Remarks\n\nThe cut-off at $1$ makes $\\bar d$ finite even for unbounded maps; it does not change which sequences converge.\nA sequence $(f_n)$ converges to $f$ in this topology if and only if $f_n \\to f$ [[uniformly|08VZ57PT26]] on $X$.\nOn bounded maps, the sup metric $\\sup_x d(f(x), g(x))$ gives the same topology.\nIf $(Y, d)$ is complete, then $(Y^X, \\bar d)$ is complete.\nIf $X$ is a topological space, the set $C(X, Y)$ of continuous maps is closed in $Y^X$, since a uniform limit of continuous maps is continuous.\n\nExample: on $\\R^{[0, 1]}$, let $f_n(x) = x / n$.\nThen $\\bar d(f_n, 0) \\le 1/n$, so $f_n \\to 0$ in this topology.\n\nNon-example: let $f_n(x) = x^n$ on $X = [0, 1)$ and $Y = \\R$.\nThen $f_n \\to 0$ pointwise, but $\\bar d(f_n, 0) = 1$ for every $n$, since $x^n$ takes values close to $1$ near $x = 1$.\nSo $(f_n)$ converges in the topology of pointwise convergence, but not in the topology of uniform convergence.\n",
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