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  "body": "\n## Counterexample\n\nUse base $13$ with the thirteen digits $0, 1, \\ldots, 9, \\mathtt{+}, \\mathtt{-}, \\mathtt{.}$, in this order.\nFor $x \\in \\R$, take the base-$13$ expansion of the fractional part of $x$ that does not end in an infinite string of the last digit.\nSuppose that from some position on, this expansion has the form\n$$\\mathtt{\\pm}\\, a_1 a_2 \\ldots a_k \\,\\mathtt{.}\\, b_1 b_2 b_3 \\ldots$$\nThis tail consists of a sign symbol and finitely many digits $a_i \\in \\set{0, \\ldots, 9}$.\nThen comes the point symbol, and after it only digits $b_j \\in \\set{0, \\ldots, 9}$.\nThen set $f(x) = \\pm\\, a_1 a_2 \\ldots a_k . b_1 b_2 b_3 \\ldots$, read as an ordinary decimal number.\nOtherwise set $f(x) = 0$.\n\nEach open interval $J$ contains all numbers with a fixed integer part whose expansion begins with a suitable finite block of digits.\nAfter that block, any tail can be appended.\nSo for each real $y$, some $x \\in J$ has a tail that encodes $y$, and $f(J) = \\R$.\n\n## Remarks\n\nThe function $f$ maps every interval onto $\\R$, so it is a [[Darboux function|7BJEH1GCTE]]: it has the intermediate value property.\nBut $f$ is [[continuous|0P27W9RW0G]] at no point, since its values on every interval are unbounded.\nSo the converse of the [[intermediate value theorem|00KAFBR32E]] fails badly.\nAlmost every real number has a base-$13$ expansion with infinitely many point symbols, so $f = 0$ [[almost everywhere|RARTMA0ZEC]].\n\n## Explanations\n\n### L1 human\n\nConway's function hides a decimal number in the end of the base-$13$ digits of $x$.\nEvery tiny interval contains numbers with every possible ending, so the function takes every real value on every tiny interval.\nIt never skips a value, yet it is wildly discontinuous everywhere.\nSo \"never skips values\" is much weaker than \"continuous\".\n",
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