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  "body": "\n## Statement\n\nLet $X$ and $Y$ be [[topological spaces|0CNQXVBHCX]], let $f \\colon X \\to Y$ be a [[continuous map|0SY9ECQ19Q]] and let $K \\subseteq X$ be [[compact|00R0X1WGPX]].\nThen the [[image|0Q3F20XB0A]] $f(K)$ is compact in $Y$.\n\n## Remarks\n\nProof: let $(V_i)$ be open sets of $Y$ that cover $f(K)$. The sets $f^{-1}(V_i)$ are open and cover $K$. A finite subfamily covers $K$, and the matching $V_i$ cover $f(K)$.\nA consequence is that a [[continuous real function on a compact set attains its bounds|0V4DET59NW]].\nCompactness is a [[topological property|02SMVGC690]].\n\n## Explanations\n\n### L1 human\n\nContinuous maps keep compactness.\nFor example, the image of $[0, 1]$ under a continuous map into the plane is a compact curve, so it is closed and bounded.\n",
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