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  "body": "\n## Example\n\nA *space-filling curve* is a [[continuous|0X3WGX95FE]] [[surjection|01RVCMSN5N]] from $[0, 1]$ onto the square $[0, 1]^2$.\nHere is one, due to Lebesgue.\nEach point $x$ of the [[Cantor set|0BXNDZFQME]] $C$ has a unique expansion $x = \\sum_{n \\ge 1} 2 a_n 3^{-n}$ with digits $a_n \\in \\set{0, 1}$.\nSet\n$$f(x) = \\left( \\sum_{n \\ge 1} a_{2n-1} 2^{-n}, \\ \\sum_{n \\ge 1} a_{2n} 2^{-n} \\right).$$\nOn each open interval removed in the construction of $C$, extend $f$ linearly between its values at the two ends.\n\nThe check has three steps.\nThe map $f$ is onto.\nTake binary expansions of both coordinates of a point of $[0, 1]^2$; interleaving their digits gives a point of $C$ that $f$ sends to it.\nThe map $f$ is continuous on $C$.\nTwo points of $C$ less than $3^{-2m}$ apart share their first $2m$ digits, so their images differ by at most $2^{-m}$ in each coordinate.\nThe two ends of a removed interval of length $3^{-k}$ share their first $k - 1$ digits.\nSo short removed intervals carry small linear pieces, and the extended map is continuous.\n\n## Remarks\n\nNo space-filling curve is injective.\nA continuous bijection $[0, 1] \\to [0, 1]^2$ would be a homeomorphism, since $[0, 1]$ is compact and the square is Hausdorff.\nThat is impossible: removing an inner point disconnects $[0, 1]$, but removing a point does not disconnect the square.\nThe first space-filling curve was found by Peano, and Hilbert gave another one.\n\n## Explanations\n\n### L1 human\n\nA space-filling curve is a pencil line drawn without lifting the pencil, in a finite time, that passes through every point of a filled square.\nIt comes as the limit of ever finer zigzag paths.\nThe curve must cross itself: it passes through some points more than once.\n",
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