[proofplan] The forward direction embeds $X$ as an open subspace of its one-point compactification $X^+$, which is compact Hausdorff by the [Characterisation of Hausdorff One-Point Compactification](/theorems/1060). The reverse direction uses the [Locally Compact Subspaces of LCH Spaces](/theorems/1059) theorem: an open subspace of a compact Hausdorff space is locally compact Hausdorff because it is locally closed (open sets are the intersection of an open set with the whole space, which is closed). [/proofplan] [step:Embed a locally compact Hausdorff space as an open subspace of a compact Hausdorff space] Assume $X$ is locally compact Hausdorff. If $X$ is compact, then $X$ is itself a compact Hausdorff space, and $X$ is open in $X$, so the statement holds with $Y = X$. If $X$ is not compact, form the one-point compactification $X^+ = X \cup \{\infty\}$. By the [Characterisation of Hausdorff One-Point Compactification](/theorems/1060), the space $X^+$ is compact Hausdorff (since $X$ is locally compact, Hausdorff, and not compact). The subspace $X \subset X^+$ is open in $X^+$: by definition of the one-point compactification topology, the complement $X^+ \setminus X = \{\infty\}$ is closed if and only if $\{\infty\}$ is not open, and a set containing $\infty$ is open if and only if its complement in $X$ is compact and closed. Since $X$ is not compact, $\{\infty\}$ is not open, hence $\{\infty\}$ is closed, and $X = X^+ \setminus \{\infty\}$ is open. Alternatively, every open set $U \subset X$ (in the original topology) is open in $X^+$ by definition, so $X$ is the union of all its own open sets and is therefore open in $X^+$. The inclusion map $\iota: X \hookrightarrow X^+$ is a homeomorphism onto the open subspace $X \subset X^+$. [guided] When $X$ is locally compact, Hausdorff, and not compact, the one-point compactification provides the required compact Hausdorff ambient space. The [Characterisation of Hausdorff One-Point Compactification](/theorems/1060) theorem guarantees that $X^+$ is compact Hausdorff under exactly these conditions. Why is $X$ open in $X^+$? In the one-point compactification topology, the open sets are: (i) all open sets of $X$, and (ii) sets of the form $\{\infty\} \cup (X \setminus K)$ where $K \subset X$ is compact and closed. The complement of $X$ in $X^+$ is $\{\infty\}$. For $\{\infty\}$ to be open, it would need to have the form $\{\infty\} \cup (X \setminus K)$ with $K = X$, requiring $X$ itself to be compact — which contradicts our assumption. Since $\{\infty\}$ is not open, its complement $X$ is not closed; but we need $X$ to be open, not closed. Indeed, $X$ is the union $\bigcup\{U : U \text{ open in } X\} = X$, which is open in $X^+$ since every open subset of $X$ is open in $X^+$ by construction. If $X$ is already compact, the argument is simpler: $X$ sits as an open subspace of itself. [/guided] [/step] [step:Show that an open subspace of a compact Hausdorff space is locally compact Hausdorff] Assume $X$ is homeomorphic to an open subspace $U$ of a compact Hausdorff space $Y$. We show $X$ (equivalently $U$) is locally compact Hausdorff. **Hausdorff.** Since $Y$ is Hausdorff and the Hausdorff property is inherited by subspaces (any two distinct points in $U$ can be separated by open sets in $Y$, and intersecting with $U$ gives open sets in $U$), the subspace $U$ is Hausdorff. **Locally compact.** A compact Hausdorff space is locally compact (for each $y \in Y$, take $K = Y$ itself as a compact neighbourhood). The subspace $U$ is open in $Y$, hence locally closed (since $U = U \cap Y$ is the intersection of the open set $U$ and the closed set $Y$). By the [Locally Compact Subspaces of LCH Spaces](/theorems/1059) theorem, $U$ is locally compact in the subspace topology. Alternatively, we give a direct argument. Fix $x \in U$. Since $Y$ is compact Hausdorff, it is regular (by [Compact Hausdorff Spaces Are Normal](/theorems/1027), compact Hausdorff spaces are normal, hence regular). By regularity, for the open set $U$ containing $x$, there exists an open set $V \subset Y$ with $x \in V \subset \overline{V} \subset U$. The closure $\overline{V}$ is taken in $Y$. Since $Y$ is compact and $\overline{V}$ is closed in $Y$, the set $\overline{V}$ is compact by [Compact Subspaces and Hausdorff Spaces](/theorems/307). Thus $V \cap U = V$ is an open neighbourhood of $x$ in $U$ contained in the compact set $\overline{V} \subset U$. The closure of $V$ in $U$ equals $\overline{V} \cap U = \overline{V}$ (since $\overline{V} \subset U$), which is compact. Since $U$ is locally compact and Hausdorff, so is $X$. [guided] The reverse direction must show that being an open subspace of a compact Hausdorff space forces local compactness. There are two ways to see this. The slick route appeals to the [Locally Compact Subspaces of LCH Spaces](/theorems/1059) theorem: any compact Hausdorff space is locally compact Hausdorff, and an open subspace is locally closed (write $U = U \cap Y$), so the theorem directly gives local compactness of $U$. The direct route is more instructive and does not require Theorem 1059. Fix $x \in U$. We need to find an open neighbourhood of $x$ in $U$ with compact closure in $U$. Since $Y$ is compact Hausdorff, it is normal by [Compact Hausdorff Spaces Are Normal](/theorems/1027), and normal spaces are regular. Regularity applied to the open set $U$ and the point $x \in U$ produces an open $V \subset Y$ with $x \in V \subset \overline{V} \subset U$. Why is $\overline{V}$ compact? Because $\overline{V}$ is a closed subset of the compact space $Y$, and closed subsets of compact spaces are compact by [Compact Subspaces and Hausdorff Spaces](/theorems/307). Why is $\overline{V}$ the closure in $U$? Because $\overline{V} \subset U$, the closure of $V$ in $U$ is $\overline{V}^Y \cap U = \overline{V}$. So $x$ has the open neighbourhood $V$ in $U$ with compact closure $\overline{V}$ in $U$. [/guided] [/step]