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  "body": "\n## Proof\n\nWe prove each implication. The second one is proved by contraposition.\n\n1. Let $(X, d_X)$ and $(Y, d_Y)$ be [[metric spaces|063W5JMY85]] and let $f \\colon X \\to Y$ be a function.\n2. Assume condition 1. Let $(x_n)$ and $(y_n)$ be sequences in $X$ with $d_X(x_n, y_n) \\to 0$. Let $\\varepsilon \u003e 0$ be real.\n3. By condition 1, there is $\\delta \u003e 0$ with $d_Y(f(x), f(y)) \u003c \\varepsilon$ whenever $d_X(x, y) \u003c \\delta$.\n4. Since $d_X(x_n, y_n) \\to 0$ in $\\R$, there is an integer $N$ with $d_X(x_n, y_n) \u003c \\delta$ for all $n \\ge N$.\n5. By steps 3 and 4, $d_Y(f(x_n), f(y_n)) \u003c \\varepsilon$ for all $n \\ge N$.\n6. Since $\\varepsilon$ was arbitrary, $d_Y(f(x_n), f(y_n)) \\to 0$ in the sense of [[convergence|06JYPJ4WHH]] in $\\R$. So condition 2 holds.\n7. Now assume that condition 1 fails. Then there is a real $\\varepsilon \u003e 0$ with the following property.\n8. For each real $\\delta \u003e 0$ there are $x, y \\in X$ with $d_X(x, y) \u003c \\delta$ and $d_Y(f(x), f(y)) \\ge \\varepsilon$.\n9. For each integer $n \\ge 1$, apply step 8 with $\\delta = 1/n$. This gives points $x_n, y_n \\in X$ with $d_X(x_n, y_n) \u003c 1/n$ and $d_Y(f(x_n), f(y_n)) \\ge \\varepsilon$.\n10. Let $\\eta \u003e 0$. Since [[the reals are Archimedean|0A3GWZ86AY]], there is an integer $N \u003e 1/\\eta$. For $n \\ge N$ we get $d_X(x_n, y_n) \u003c 1/n \\le 1/N \u003c \\eta$. So $d_X(x_n, y_n) \\to 0$.\n11. But $d_Y(f(x_n), f(y_n)) \\ge \\varepsilon$ for every $n$, so this sequence does not tend to $0$.\n12. By steps 10 and 11, condition 2 fails. So condition 2 implies condition 1.\n13. By steps 2 to 6 and step 12, the two conditions are equivalent.\n\n## Idea\n\nIf one $\\delta$ works everywhere, then pairs of points that get close are eventually within $\\delta$, and their images get close.\nIf no $\\delta$ works, take $\\delta = 1/n$ for each $n$ and collect the bad pairs: they form two sequences that come together while their images stay apart.\n",
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