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  "body": "\n## Counterexample\n\nLet $f \\colon \\mathbb{R} \\to \\mathbb{R}$ be given by $f(x) = 1$ for rational $x$ and $f(x) = 0$ for irrational $x$.\nThen $f$ is not [[continuous at|0EEE2CZB2A]] any point.\nIndeed, each real $a$ is the limit of a sequence of rationals and of a sequence of irrationals. Their images tend to $1$ and to $0$, so one of them does not tend to $f(a)$.\n\n## Remarks\n\nThe function $f$ is the indicator function of $\\mathbb{Q}$.\nEvery rational number is a period of $f$.\nThe function is [[not Riemann integrable|03Z4JC4YRJ]] on any interval with more than one point.\n\n## Explanations\n\n### L1 human\n\nRational and irrational numbers lie mixed together in every small interval.\nSo the function jumps between $0$ and $1$ near every point, and it has no limit anywhere.\n",
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