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  "body": "\n## Statement\n\nLet $X$ and $Y$ be [[topological spaces|0CNQXVBHCX]] and let $f \\colon X \\to Y$ be a function.\nThen $f$ is [[continuous|0SY9ECQ19Q]] if and only if for each [[closed set|0G5AV9Q4RR]] $F$ of $Y$, the [[preimage|0FGCFY1AZX]] $f^{-1}(F)$ is closed in $X$.\n\n## Remarks\n\nProof: preimages commute with complements, that is, $f^{-1}(Y \\setminus A) = X \\setminus f^{-1}(A)$ for each $A \\subseteq Y$. Then use that the [[complements of open sets are closed|0K67Q60Z8P]].\nFor example, for a continuous $f \\colon \\mathbb{R} \\to \\mathbb{R}$, the set of zeros $f^{-1}(\\{0\\})$ is closed.\n\n## Explanations\n\n### L1 human\n\nContinuity can be tested with closed sets as well as with open sets.\nThis form is useful to show that a set given by an equation, such as $\\{x : f(x) = 0\\}$, is closed.\n",
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