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  "body": "\n## Proof\n\n1. Let $(X, d_X)$ and $(Y, d_Y)$ be [[metric spaces|063W5JMY85]] and let $f \\colon X \\to Y$ be an [[isometry|0878DN3WCS]].\n2. Let $x, y \\in X$ with $f(x) = f(y)$.\n3. A metric is $0$ on equal points, so $d_Y(f(x), f(y)) = 0$.\n4. Since $f$ is an isometry, $d_X(x, y) = d_Y(f(x), f(y)) = 0$.\n5. By the axiom of a metric, $x \\ne y$ would give $d_X(x, y) \u003e 0$. This contradicts step 4, so $x = y$.\n6. Thus $f(x) = f(y)$ implies $x = y$, and $f$ is [[injective|0PVG1CRAJN]].\n\n## Idea\n\nDistinct points have positive distance, and an isometry keeps distances.\nSo the images of distinct points also have positive distance, and they are distinct.\n",
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  "name": "Proof that isometries are injective",
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