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  "body": "\n## Proof\n\nWe show that the complement of $K$ is open, by separating each outside point from $K$.\n\n1. Let $X$ be a [[Hausdorff space|0MXB4CKHFW]] and let $K$ be a [[compact|00R0X1WGPX]] subset of $X$.\n2. If $K = \\emptyset$, then $X \\setminus K = X$ is open, and $K$ is [[closed|0G5AV9Q4RR]]. So assume $K \\ne \\emptyset$.\n3. Fix $y \\in X \\setminus K$. For each $x \\in K$ we have $x \\ne y$. By the Hausdorff property, choose [[open sets|01ZQQDCERP]] $U_x \\ni x$ and $V_x \\ni y$ with $U_x \\cap V_x = \\emptyset$.\n4. The sets $U_x$ with $x \\in K$ are open and cover $K$. By [[compactness via open covers|0P6PMJB4MA]], there are $x_1, \\ldots, x_n \\in K$ with $K \\subseteq U_{x_1} \\cup \\cdots \\cup U_{x_n}$, and $n \\ge 1$.\n5. Let $V_y = V_{x_1} \\cap \\cdots \\cap V_{x_n}$. A [[topology|0CNQXVBHCX]] is closed under finite intersections, so $V_y$ is open. Also $y \\in V_y$.\n6. Let $z \\in K$. By step 4, $z \\in U_{x_j}$ for some $j$. Since $U_{x_j} \\cap V_{x_j} = \\emptyset$ and $V_y \\subseteq V_{x_j}$, we get $z \\notin V_y$.\n7. So $V_y \\subseteq X \\setminus K$ for each $y \\in X \\setminus K$.\n8. Hence $X \\setminus K = \\bigcup_{y \\in X \\setminus K} V_y$, a union of open sets. It is open by the axioms of a topology.\n9. Thus $K$ is closed in $X$.\n\n## Idea\n\nEach point outside $K$ can be separated from each point of $K$ by disjoint open sets.\nCompactness lets us use only finitely many of these separations, and a finite intersection of open sets is still open. This gives an open neighbourhood of the outside point that misses $K$.\n",
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