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  "body": "\n## Proof\n\n1. Let $(X, d)$ be a [[metric space|063W5JMY85]], let $x \\in X$ and let $r \u003e 0$. Let $B(x, r) = \\set{z \\in X : d(x, z) \u003c r}$ be the [[open ball|0DC3MPR7C9]].\n2. Let $y \\in B(x, r)$. Then $d(x, y) \u003c r$.\n3. Put $\\varepsilon = r - d(x, y)$. By step 2, $\\varepsilon \u003e 0$.\n4. Let $z \\in B(y, \\varepsilon)$, so $d(y, z) \u003c \\varepsilon$.\n5. By the [[triangle inequality|01M6CW9VM6]], $d(x, z) \\le d(x, y) + d(y, z) \u003c d(x, y) + \\varepsilon = r$.\n6. So $z \\in B(x, r)$. As $z$ was arbitrary, $B(y, \\varepsilon) \\subseteq B(x, r)$.\n7. Each point $y$ of $B(x, r)$ thus has a ball about it inside $B(x, r)$.\n8. By the [[description of the open sets of a metric space|0A26X79RTB]], $B(x, r)$ is open in $X$.\n\n## Idea\n\nA point $y$ inside the ball is at distance $d(x, y)$ from the centre. So it still has room $r - d(x, y)$ before it reaches the edge. The triangle inequality shows that the ball of that radius about $y$ stays inside.\n",
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