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  "body": "\n## Statement\n\nLet $(X, d_X)$ and $(Y, d_Y)$ be [[metric spaces|063W5JMY85]] and let $f \\colon X \\to Y$ be a function.\nThen the following two conditions are equivalent:\n\n1. For each $a \\in X$ and each $\\varepsilon \u003e 0$ there is $\\delta \u003e 0$ such that $d_X(x, a) \u003c \\delta$ implies $d_Y(f(x), f(a)) \u003c \\varepsilon$ for all $x \\in X$.\n2. For each [[open set|01ZQQDCERP]] $V$ of $Y$, the [[preimage|0FGCFY1AZX]] $f^{-1}(V)$ is open in $X$.\n\n## Remarks\n\nCondition 2 is the definition of a [[continuous map|0SY9ECQ19Q]] between topological spaces. Thus, for metric spaces, the epsilon-delta definition and the topological definition agree.\nProof that 1 implies 2: let $a \\in f^{-1}(V)$. Choose $\\varepsilon$ with $B(f(a), \\varepsilon) \\subseteq V$, then $\\delta$ from condition 1. Then $B(a, \\delta) \\subseteq f^{-1}(V)$.\nImages of open sets need not be open: the constant map $\\mathbb{R} \\to \\mathbb{R}$, $x \\mapsto 0$, is continuous, but the image of $\\mathbb{R}$ is $\\{0\\}$.\n\n## Explanations\n\n### L1 human\n\nThe topological form of continuity uses only open sets and does not mention distances.\nThis form lets us define continuity in spaces that have no metric at all.\n",
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