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  "body": "\n## Proof\n\nWe show that the inverse map is continuous by testing closed sets.\n\n1. Since $f$ is bijective, it has an [[inverse function|0W766JAPSN]] $g = f^{-1} \\colon Y \\to X$. By the definition of a [[homeomorphism|05GZ11PWRY]], it remains to show that $g$ is continuous.\n2. By [[continuity via closed sets|0B8GPXFSP7]], it is enough to show that $g^{-1}(F)$ is closed in $Y$ for each closed $F \\subseteq X$.\n3. Let $F \\subseteq X$ be closed. The [[preimage|0FGCFY1AZX]] is $g^{-1}(F) = \\set{y \\in Y : g(y) \\in F}$.\n4. For $y \\in Y$, we have $g(y) \\in F$ if and only if $y = f(x)$ for some $x \\in F$, because $g(y)$ is the unique $x$ with $f(x) = y$. So $g^{-1}(F) = f(F)$.\n5. The [[subspace topology|0D5YHQGDJY]] on $X$ as a subset of itself is the topology of $X$. So $X$ is a [[compact set|00R0X1WGPX]] in $X$.\n6. Since [[closed subsets of compact sets are compact|066HW6D9A7]], $F$ is compact.\n7. The [[continuous image of a compact set is compact|0K9AAJ8G51]], so $f(F)$ is compact in $Y$.\n8. Since $Y$ is Hausdorff, [[compact subsets of a Hausdorff space are closed|0MYFJVMN7M]]. So $f(F)$ is closed in $Y$.\n9. By steps 4 and 8, $g^{-1}(F)$ is closed for each closed $F$. So $g$ is continuous, and $f$ is a homeomorphism.\n\n## Idea\n\nThe inverse of $f$ is continuous exactly when $f$ sends closed sets to closed sets. In a compact space, closed sets are compact, and compactness survives the continuous map $f$. In a Hausdorff space, compact sets are closed again.\n",
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