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  "body": "\n## Definition\n\nLet $X$ be a set, let $Y$ be a [[topological space|0CQHFPGJ45]], and let $Y^X$ be the set of all maps from $X$ to $Y$.\nFor $x \\in X$ and an [[open|02BH3MCBQB]] set $U \\subseteq Y$, put\n$$W(x, U) = \\set{f \\in Y^X : f(x) \\in U}.$$\nThe *topology of pointwise convergence* on $Y^X$ is the topology generated by the [[subbasis|09S8DV34Y0]] of all sets $W(x, U)$.\nOn a subset $F \\subseteq Y^X$, such as the set of continuous maps when $X$ is a space, it is the [[subspace topology|0WS9P12FQR]].\n\n## Remarks\n\nViewing $Y^X$ as the product $\\prod_{x \\in X} Y$, this topology is exactly the [[product topology|0GBG8B49AA]].\nA sequence $(f_n)$ converges to $f$ in it if and only if $f_n(x) \\to f(x)$ in $Y$ for every $x \\in X$, which is [[pointwise convergence|0GFGK40582]].\nThe same holds for nets.\n\nExample: on $\\R^{[0, 1]}$, let $f_n(x) = x^n$, and let $f(x) = 0$ for $x \u003c 1$ and $f(1) = 1$.\nFor each fixed $x$, the numbers $x^n$ tend to $f(x)$, so $f_n \\to f$ in this topology.\n\nNon-example: the discrete topology on $\\R^{[0, 1]}$ is not the topology of pointwise convergence.\nIn the discrete topology, the sequence $f_n$ above does not converge, because $\\set{f}$ is open and contains no $f_n$.\n\nFor an uncountable set $X$ and $Y = \\R$, the space $\\R^X$ with this topology is not first-countable, so it is not metrizable.\n",
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