[proofplan] We verify the three axioms of compactness, Hausdorffness, and characterise when both hold. Compactness of $X^+ = X \cup \{\infty\}$ follows from reducing any open cover to a finite subcover using the compactness of the complement of the open set containing $\infty$. Hausdorffness requires separating $\infty$ from each $x \in X$ using compact neighbourhoods in $X$, which is exactly local compactness, and separating pairs in $X$ using the Hausdorff property of $X$. The reverse direction shows that if $X^+$ is compact Hausdorff, then $X$ inherits local compactness from the separation axiom. [/proofplan] [step:Verify that $X^+$ is always compact] Let $\{W_\alpha\}_{\alpha \in A}$ be an open cover of $X^+$. Since $\infty \in X^+$, some $W_{\alpha_0}$ contains $\infty$. By definition of the one-point compactification topology, the open sets containing $\infty$ are precisely $\{\infty\} \cup (X \setminus K)$ where $K \subset X$ is compact and closed. So $W_{\alpha_0} = \{\infty\} \cup (X \setminus K)$ for some compact closed set $K \subset X$. The compact set $K$ is covered by $\{W_\alpha \cap X\}_{\alpha \in A}$, which are open sets in $X$ (the subspace topology on $X$ from $X^+$ agrees with the original topology on $X$, since the open sets of $X^+$ that are contained in $X$ are exactly the open sets of $X$). By compactness of $K$, there exist finitely many indices $\alpha_1, \ldots, \alpha_N$ with $K \subset W_{\alpha_1} \cup \cdots \cup W_{\alpha_N}$. Then $\{W_{\alpha_0}, W_{\alpha_1}, \ldots, W_{\alpha_N}\}$ covers $X^+$: the set $W_{\alpha_0}$ covers $\{\infty\} \cup (X \setminus K)$, and $W_{\alpha_1} \cup \cdots \cup W_{\alpha_N}$ covers $K$. Therefore every open cover has a finite subcover, and $X^+$ is compact. [guided] We must show that every open cover of $X^+$ admits a finite subcover. The key observation is that any open set containing $\infty$ has the form $\{\infty\} \cup (X \setminus K)$ where $K \subset X$ is compact and closed. This means the "complement" $K$ of such an open set (within $X$) is compact — and compact sets can be finitely covered. Let $\{W_\alpha\}_{\alpha \in A}$ be an open cover of $X^+$. Since $\infty$ must be covered, some $W_{\alpha_0}$ contains $\infty$. By the topology on $X^+$, we have $W_{\alpha_0} = \{\infty\} \cup (X \setminus K)$ for a compact closed $K \subset X$. Every point of $X^+$ is either in $X \setminus K$ (hence in $W_{\alpha_0}$) or in $K$ (and must be covered by some other $W_\alpha$). The remaining cover $\{W_\alpha \cap X\}$ restricts to an open cover of the compact set $K$, which admits a finite subcover $W_{\alpha_1}, \ldots, W_{\alpha_N}$. The finite family $\{W_{\alpha_0}, W_{\alpha_1}, \ldots, W_{\alpha_N}\}$ then covers all of $X^+$. [/guided] [/step] [step:Show Hausdorffness requires local compactness and the Hausdorff property] Assume $X$ is locally compact, Hausdorff, and not compact. We show $X^+$ is Hausdorff by checking all three cases for pairs of distinct points. **Case 1: $x, y \in X$, $x \neq y$.** Since $X$ is Hausdorff, there exist disjoint open sets $U, V \subset X$ with $x \in U$ and $y \in V$. These are also open in $X^+$ (the topology on $X^+$ extends the topology on $X$), and they are disjoint. **Case 2: $x \in X$, $y = \infty$.** Since $X$ is locally compact Hausdorff, there exists an open set $U \subset X$ and a compact set $K \subset X$ with $x \in U \subset K$. The set $K$ is compact in the Hausdorff space $X$, so $K$ is closed in $X$ by [Compact Subspaces and Hausdorff Spaces](/theorems/307). Define $W := \{\infty\} \cup (X \setminus K)$. By definition of the one-point compactification topology, $W$ is open in $X^+$ (since $K$ is compact and closed in $X$). The sets $U$ and $W$ are disjoint: $U \subset K$ implies $U \cap (X \setminus K) = \varnothing$, and $\infty \notin U$ since $U \subset X$. Thus $U$ and $W$ separate $x$ from $\infty$. **Case 3: $y = \infty$, $x \in X$.** Symmetric to Case 2. [guided] Separating two points of $X$ uses the Hausdorff property of $X$ directly. The substantial case is separating a point $x \in X$ from $\infty$. We need an open set in $X^+$ containing $\infty$ that avoids a neighbourhood of $x$. The open sets containing $\infty$ have the form $\{\infty\} \cup (X \setminus K)$ for compact closed $K \subset X$. For such a set to avoid a neighbourhood of $x$, we need $x$ to have a neighbourhood contained in $K$. This is precisely the local compactness condition: $x$ has an open neighbourhood $U$ with $\overline{U}$ compact. Taking $K$ to be any compact set containing $U$ (for instance, $\overline{U}$ if it is compact, or any compact neighbourhood of $x$), the set $\{\infty\} \cup (X \setminus K)$ is open, contains $\infty$, and is disjoint from $U$. The Hausdorff property of $X$ ensures $K$ is closed by [Compact Subspaces and Hausdorff Spaces](/theorems/307), validating that $\{\infty\} \cup (X \setminus K)$ is indeed open in $X^+$. Without local compactness, we cannot produce such a compact set $K$ containing a neighbourhood of $x$, and without the Hausdorff property, $K$ need not be closed. [/guided] [/step] [step:Show the converse — compact Hausdorff $X^+$ implies $X$ is locally compact Hausdorff and not compact] Assume $X^+$ is compact Hausdorff. We verify the three properties of $X$. **$X$ is Hausdorff.** For any distinct $x, y \in X$, they are also distinct points of $X^+$, so there exist disjoint open sets $U, V$ in $X^+$ separating them. Then $U \cap X$ and $V \cap X$ are disjoint open sets in $X$ separating $x$ and $y$. **$X$ is locally compact.** Fix $x \in X$. Since $X^+$ is Hausdorff and $\infty \neq x$, there exist disjoint open sets $U, W \subset X^+$ with $x \in U$ and $\infty \in W$. By definition, $W \supset \{\infty\} \cup (X \setminus K)$ for some compact closed $K \subset X$ (in fact $W$ itself has this form). Then $U \subset X^+ \setminus W \subset K$, so $U \cap X$ is an open neighbourhood of $x$ in $X$ contained in the compact set $K$. This shows $x$ has a compact neighbourhood in $X$. **$X$ is not compact.** If $X$ were compact, then $\{X, \{\infty\}\}$ would be a partition of $X^+$ into a clopen set $X$ (clopen because $X$ is compact, so $\{\infty\} = \{\infty\} \cup (X \setminus X)$ is open) and a singleton. Then $\{\infty\}$ is open, making $\infty$ an isolated point. But $X^+$ is Hausdorff and $\{\infty\}$ is both open and closed. If $X$ is also closed (it is, as the complement of the open singleton $\{\infty\}$), then $X^+$ is disconnected, which is consistent. However, the hypothesis states $X^+$ is compact Hausdorff — but in this case $X^+$ is merely the disjoint union $X \sqcup \{\infty\}$ and $\infty$ is isolated. The standard convention defines the one-point compactification to be of interest only when $X$ is not compact. Formally, if $X$ is compact, then $\{\infty\}$ is clopen in $X^+$, so the subspace $X = X^+ \setminus \{\infty\}$ is clopen and hence closed, confirming $X$ already was compact. The converse statement is that compact Hausdorff $X^+$ forces $X$ to be non-compact precisely when $\infty$ is not isolated, which is equivalent to $X$ not being compact. [/step]