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  "body": "\n## Statement\n\nLet $X$ be a normed space over $\\mathbb{R}$ or $\\mathbb{C}$, and let $Y \\neq X$ be a [[closed|0G5AV9Q4RR]] linear subspace of $X$.\nLet $\\alpha$ be a real number with $0 \u003c \\alpha \u003c 1$.\nThen there is $x \\in X$ with $\\|x\\| = 1$ and $\\|x - y\\| \\ge \\alpha$ for all $y \\in Y$.\n\n## Remarks\n\nProof idea: take $x_0 \\notin Y$ and let $r = \\inf_{y \\in Y} \\|x_0 - y\\| \u003e 0$, which is positive because $Y$ is closed. Choose $y_0 \\in Y$ with $\\|x_0 - y_0\\| \\le r / \\alpha$, and let $x = (x_0 - y_0) / \\|x_0 - y_0\\|$.\nIf $Y$ is finite-dimensional, one can take $\\alpha = 1$.\nThe lemma is a substitute for orthogonal vectors in normed spaces without an inner product.\nIts main application is that a [[compact closed unit ball forces finite dimension|0VZVV8VME5]].\n\n## Explanations\n\n### L1 human\n\nFor each closed proper subspace, there is a unit vector that is almost as far from the subspace as possible.\nIn an inner product space, a unit vector orthogonal to the subspace works with distance $1$.\n",
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