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  "body": "\n## Proof\n\nWe take a point outside $Y$, move it by a nearly closest point of $Y$, and normalise.\n\n1. Since $Y \\neq X$, choose $x_0 \\in X$ with $x_0 \\notin Y$. Let $r = \\inf_{y \\in Y} \\|x_0 - y\\|$. This is a real number $r \\ge 0$, since $Y$ contains $0$ and the norms are nonnegative.\n2. $r \u003e 0$. If $r = 0$, then for each $\\varepsilon \u003e 0$ there is $y \\in Y$ with $\\|x_0 - y\\| \u003c \\varepsilon$. So $x_0$ is in the closure of $Y$, by [[closure in a metric space|068P0P32DW]].\n3. As $Y$ is [[closed|0G5AV9Q4RR]], it equals its closure, by [[this result|0D2W7FS52T]]. So $x_0 \\in Y$, a contradiction.\n4. Since $0 \u003c \\alpha \u003c 1$ and $r \u003e 0$, we have $r / \\alpha \u003e r$. So $r / \\alpha$ is not a lower bound of the distances, and there is $y_0 \\in Y$ with $\\|x_0 - y_0\\| \u003c r / \\alpha$.\n5. Let $d = \\|x_0 - y_0\\|$. Then $d \\ge r \u003e 0$. Let $x = (x_0 - y_0) / d$. Then $\\|x\\| = 1$.\n6. Let $y \\in Y$. Then $x - y = (x_0 - y_1) / d$, where $y_1 = y_0 + d \\, y$.\n7. As $Y$ is a [[linear subspace|0EPTJA5QV4]], $y_1 \\in Y$. So $\\|x_0 - y_1\\| \\ge r$, by the definition of $r$.\n8. Hence $\\|x - y\\| = \\|x_0 - y_1\\| / d \\ge r / d \u003e r / (r / \\alpha) = \\alpha$.\n9. So $\\|x\\| = 1$ and $\\|x - y\\| \\ge \\alpha$ for all $y \\in Y$.\n\n## Idea\n\nA point outside a closed subspace has a positive distance $r$ from it. We pick a point of $Y$ almost as close as possible, within the factor $1/\\alpha$, and normalise the difference. The new vector has distance $r/d$ from $Y$, and this ratio is above $\\alpha$ because $d$ is almost $r$.\n",
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