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  "body": "\n## Proof\n\nWe argue by contradiction.\n\n1. Let $(X, d)$ be a [[metric space|063W5JMY85]] and let $F$ be a [[closed subset|0G5AV9Q4RR]] of $X$.\n2. Let $(x_n)$ be a sequence of points of $F$ that [[converges|06JYPJ4WHH]] to $x \\in X$.\n3. Suppose that $x \\notin F$. Then $x \\in X \\setminus F$, and $X \\setminus F$ is open because $F$ is closed.\n4. By the [[description of the open sets of a metric space|0A26X79RTB]], there is $\\varepsilon \u003e 0$ with $B(x, \\varepsilon) \\subseteq X \\setminus F$. Here $B(x, \\varepsilon)$ is the [[open ball|0DC3MPR7C9]].\n5. Since $x_n \\to x$, there is an integer $N$ with $d(x_n, x) \u003c \\varepsilon$ for all $n \\ge N$.\n6. By the symmetry of $d$, $d(x, x_N) \u003c \\varepsilon$, so $x_N \\in B(x, \\varepsilon) \\subseteq X \\setminus F$.\n7. This contradicts $x_N \\in F$. Hence $x \\in F$.\n\n## Idea\n\nIf the limit were outside $F$, it would have a ball around it that misses $F$. The terms of the sequence eventually enter every ball around the limit. They would then leave $F$, which they never do.\n",
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