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  "body": "\n## Proof\n\nWe prove each implication. Each $T_N$ is nonempty, so its [[diameter|0692CGQF00]] is defined, with a value in $[0, \\infty]$.\n\n1. Let $(X, d)$ be a [[metric space|063W5JMY85]], let $(x_n)_{n \\ge 1}$ be a sequence in $X$, and let $T_N = \\set{x_n : n \\ge N}$.\n2. Assume condition 1. Let $\\varepsilon \u003e 0$ be real.\n3. By condition 1 applied to $\\varepsilon / 2$, there is an integer $N$ with $d(x_m, x_n) \u003c \\varepsilon / 2$ for all $m, n \\ge N$.\n4. Let $M \\ge N$. Two points of $T_M$ have the form $x_m$ and $x_n$ with $m, n \\ge M \\ge N$. So their distance is less than $\\varepsilon / 2$ by step 3.\n5. So $\\varepsilon / 2$ is an upper bound of all distances in $T_M$. The [[supremum|0BTMQ04556]] is the least upper bound, so $\\operatorname{diam}(T_M) \\le \\varepsilon / 2 \u003c \\varepsilon$.\n6. Step 5 holds for every $M \\ge N$. Since $\\varepsilon$ was arbitrary, $\\operatorname{diam}(T_N) \\to 0$ as $N \\to \\infty$. So condition 2 holds.\n7. Now assume condition 2. Let $\\varepsilon \u003e 0$ be real.\n8. By condition 2, there is an integer $N$ with $\\operatorname{diam}(T_N) \u003c \\varepsilon$.\n9. Let $m, n \\ge N$. Then $x_m$ and $x_n$ lie in $T_N$.\n10. The diameter is an upper bound of the distances in $T_N$. So $d(x_m, x_n) \\le \\operatorname{diam}(T_N) \u003c \\varepsilon$.\n11. Thus condition 1 holds. By steps 2 to 6 and 7 to 10, the two conditions are equivalent.\n\n## Idea\n\nThe diameter of a tail is the largest distance between two of its terms, up to a supremum.\nSo \"all late terms are within $\\varepsilon$ of each other\" says the same as \"some tail has diameter at most $\\varepsilon$\".\n",
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