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  "body": "\n## Proof\n\nWe prove each direction separately. Both metric spaces carry their metric topologies.\n\n1. Let $(X, d_X)$ and $(Y, d_Y)$ be [[metric spaces|063W5JMY85]], let $f \\colon X \\to Y$ and let $a \\in X$.\n2. Write $B_X(a, \\delta)$ and $B_Y(f(a), \\varepsilon)$ for the [[open balls|0DC3MPR7C9]] of $X$ and $Y$. By the symmetry of a metric, $x \\in B_X(a, \\delta)$ if and only if $d_X(x, a) \u003c \\delta$.\n3. Recall that a set $N$ is a [[neighbourhood|09CJ00H3NW]] of a point $p$ when some open set $U$ satisfies $p \\in U \\subseteq N$.\n4. First suppose that $f$ is [[continuous at|07N7W5C29X]] $a$. Let $\\varepsilon \u003e 0$ and put $V = B_Y(f(a), \\varepsilon)$.\n5. By [[open balls are open|0B2305GNAZ]], $V$ is open. Also $f(a) \\in V$, as $d_Y(f(a), f(a)) = 0 \u003c \\varepsilon$. So $V$ is a neighbourhood of $f(a)$.\n6. By continuity at $a$, the [[preimage|0FGCFY1AZX]] $f^{-1}(V)$ is a neighbourhood of $a$. So there is an open set $U$ with $a \\in U \\subseteq f^{-1}(V)$.\n7. By the [[description of the open sets of a metric space|0A26X79RTB]], there is $\\delta \u003e 0$ with $B_X(a, \\delta) \\subseteq U$.\n8. Let $x \\in X$ with $d_X(x, a) \u003c \\delta$. By step 2, $x \\in B_X(a, \\delta) \\subseteq f^{-1}(V)$, so $f(x) \\in V$.\n9. Thus $d_Y(f(a), f(x)) \u003c \\varepsilon$, and by symmetry $d_Y(f(x), f(a)) \u003c \\varepsilon$. This is the epsilon-delta condition.\n10. Conversely, suppose the epsilon-delta condition holds at $a$. Let $V$ be a neighbourhood of $f(a)$. Choose an open set $W$ with $f(a) \\in W \\subseteq V$.\n11. By the description of open sets, there is $\\varepsilon \u003e 0$ with $B_Y(f(a), \\varepsilon) \\subseteq W$. Take $\\delta \u003e 0$ as given by the condition for this $\\varepsilon$.\n12. Let $x \\in B_X(a, \\delta)$. By step 2, $d_X(x, a) \u003c \\delta$, so $d_Y(f(x), f(a)) \u003c \\varepsilon$.\n13. By symmetry, $f(x) \\in B_Y(f(a), \\varepsilon) \\subseteq V$. Hence $B_X(a, \\delta) \\subseteq f^{-1}(V)$.\n14. The ball $B_X(a, \\delta)$ is open, because open balls are open, and it contains $a$. So $f^{-1}(V)$ is a neighbourhood of $a$.\n15. As $V$ was arbitrary, $f$ is continuous at $a$.\n\n## Idea\n\nIn a metric space, the open balls about a point are small enough to stand in for all its neighbourhoods. Every neighbourhood contains a ball, and every ball is a neighbourhood. So the condition on neighbourhoods becomes a condition on balls, which is the epsilon-delta condition.\n",
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