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  "body": "\n## Proof\n\nWe argue by contradiction.\n\n1. Suppose $(x_n)$ converges to $x$ and to $y$, and $x \\neq y$.\n2. Since $X$ is [[Hausdorff|0MXB4CKHFW]], there are open sets $U$ and $V$ with $x \\in U$, $y \\in V$ and $U \\cap V = \\varnothing$.\n3. The open set $U$ satisfies $x \\in U \\subseteq U$, so $U$ is a [[neighbourhood|09CJ00H3NW]] of $x$. In the same way, $V$ is a neighbourhood of $y$.\n4. By convergence to $x$, there is $n_0$ with $x_n \\in U$ for all $n \\ge n_0$. By convergence to $y$, there is $n_1$ with $x_n \\in V$ for all $n \\ge n_1$.\n5. Let $n = \\max(n_0, n_1)$. Then $x_n \\in U \\cap V$, which contradicts $U \\cap V = \\varnothing$.\n6. Hence $x = y$.\n\n## Idea\n\nTwo different limits could be separated by disjoint open sets. A convergent sequence eventually stays in each of them, so from some index on it would lie in both at once, which is impossible.\n",
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