[proofplan] We prove the dense $G_\delta$ form: given dense open sets $G_1, G_2, \ldots$ and a nonempty open set $U$, we construct nested compact sets $K_1 \supset K_2 \supset \cdots$ with each $K_n$ contained in $G_n$ and having nonempty interior. The local compactness and Hausdorff hypotheses provide, at each stage, a compact neighbourhood inside the current open set. The [Cantor Intersection Theorem](/theorems/624) for compact sets in a Hausdorff space delivers a point in $U \cap \bigcap_{n=1}^\infty G_n$. The category form follows by the same complementation argument as in the metric space case. [/proofplan] [step:At each stage, use local compactness and Hausdorff to produce a compact neighbourhood inside $G_n$] Let $(G_n)_{n=1}^\infty$ be a sequence of dense open subsets of $X$, and let $U \subset X$ be a nonempty open set. We must show $U \cap \bigcap_{n=1}^\infty G_n \neq \varnothing$. We use the following property of locally compact Hausdorff spaces: for every point $x$ and every open neighbourhood $O$ of $x$, there exist an open set $V$ and a compact set $K$ with $x \in V \subset K \subset O$. This holds because local compactness provides a compact neighbourhood $C$ of $x$, the Hausdorff property ensures $C$ is closed ([Compact Subspaces and Hausdorff Spaces](/theorems/307), part 2), and intersecting with $O$ and taking the closure inside $C$ yields the desired compact neighbourhood contained in $O$. Since $G_1$ is dense and $U$ is nonempty and open, $U \cap G_1$ is nonempty and open. Choose $x_1 \in U \cap G_1$. By the property above applied to the open neighbourhood $U \cap G_1$ of $x_1$, there exist an open set $V_1$ and a compact set $K_1$ with \begin{align*} x_1 \in V_1 \subset K_1 \subset U \cap G_1, \quad K_1 \text{ compact}, \quad V_1 \text{ open}. \end{align*} Proceeding inductively: suppose we have an open set $V_n$ and a compact set $K_n$ with $V_n \subset K_n \subset G_n$ and $V_n$ nonempty. Since $G_{n+1}$ is dense and $V_n$ is nonempty and open, $V_n \cap G_{n+1}$ is nonempty and open. Choose $x_{n+1} \in V_n \cap G_{n+1}$. By the same property, applied to the open neighbourhood $V_n \cap G_{n+1}$ of $x_{n+1}$, there exist an open set $V_{n+1}$ and a compact set $K_{n+1}$ with \begin{align*} x_{n+1} \in V_{n+1} \subset K_{n+1} \subset V_n \cap G_{n+1}. \end{align*} Since $V_n \subset K_n$, we have $K_{n+1} \subset V_n \subset K_n$, giving the nesting $K_{n+1} \subset K_n$. Since $K_{n+1} \subset G_{n+1}$, the compact set $K_{n+1}$ is contained in the $(n+1)$-th dense open set. [guided] The core of the proof is this inductive construction. Unlike the complete metric space version, we do not have a metric, so we cannot build closed balls. Instead, we use local compactness to produce compact neighbourhoods. **Base case.** The set $G_1$ is dense and $U$ is nonempty and open, so $U \cap G_1 \neq \varnothing$ and this intersection is open. Choose $x_1 \in U \cap G_1$. By local compactness, $x_1$ has a compact neighbourhood — that is, there exists an open $V'$ and compact $C$ with $x_1 \in V' \subset C$. We want a compact neighbourhood entirely inside $U \cap G_1$. Here is where the Hausdorff hypothesis enters in an essential way. In a Hausdorff space, compact sets are closed ([Compact Subspaces and Hausdorff Spaces](/theorems/307), part 2). Consequently, the closure of any subset of a compact set is again compact (as a closed subset of a compact set, by part 1 of the same theorem). So we can refine: set $V_1 := V' \cap U \cap G_1$ (open, contains $x_1$), and $K_1 := \overline{V_1}$. Since $V_1 \subset V' \subset C$ and $C$ is compact and closed, $K_1 = \overline{V_1} \subset C$ is a closed subset of a compact set, hence compact. Since $U \cap G_1$ need not be closed, we cannot guarantee $K_1 \subset U \cap G_1$ from the closure alone. Instead, we use the stronger form of local compactness: in a locally compact Hausdorff space, every point has a local base of compact neighbourhoods. Concretely, by [Separation of Compact Sets in Hausdorff Spaces](/theorems/1026) (part 1), for the compact set $C$ and any point not in $U \cap G_1$, we can separate them by open sets, giving us a compact neighbourhood $K_1$ contained in $U \cap G_1$. Alternatively, we directly apply the fact that in a locally compact Hausdorff space, every neighbourhood of a point contains a compact neighbourhood: for $x_1 \in U \cap G_1$ (which is an open neighbourhood of $x_1$), there exist open $V_1$ and compact $K_1$ with $x_1 \in V_1 \subset K_1 \subset U \cap G_1$. **Inductive step.** Suppose we have an open set $V_n$ (nonempty) and a compact set $K_n$ with $V_n \subset K_n \subset G_n$. Since $G_{n+1}$ is dense and $V_n$ is nonempty and open, $V_n \cap G_{n+1} \neq \varnothing$. This intersection is open. Pick $x_{n+1} \in V_n \cap G_{n+1}$. By local compactness, there exist open $V_{n+1}$ and compact $K_{n+1}$ with \begin{align*} x_{n+1} \in V_{n+1} \subset K_{n+1} \subset V_n \cap G_{n+1}. \end{align*} The inclusion $K_{n+1} \subset V_n \subset K_n$ gives the nesting $K_{n+1} \subset K_n$, and $K_{n+1} \subset G_{n+1}$ gives containment in the $n+1$-th dense open set. Why does this require $X$ to be Hausdorff, not merely locally compact? Without the Hausdorff property, compact sets need not be closed, and a compact neighbourhood of a point need not be containable inside a given open set. The Hausdorff property ensures that we can always shrink a compact neighbourhood to fit inside any prescribed open neighbourhood. [/guided] [/step] [step:Apply the Cantor Intersection Theorem to the nested compact sets] The construction produces a decreasing sequence of nonempty compact sets: \begin{align*} K_1 \supset K_2 \supset K_3 \supset \cdots \end{align*} Each $K_n$ is nonempty (it contains $V_n$, which contains $x_n$). Since $X$ is Hausdorff, the [Cantor Intersection Theorem](/theorems/624) (compact version) applies: the intersection of a decreasing sequence of nonempty compact subsets of a Hausdorff space is nonempty. Therefore \begin{align*} \bigcap_{n=1}^\infty K_n \neq \varnothing. \end{align*} Let $x \in \bigcap_{n=1}^\infty K_n$. For every $n \ge 1$, $x \in K_n \subset G_n$, so $x \in \bigcap_{n=1}^\infty G_n$. Moreover, $x \in K_1 \subset U$, so $x \in U \cap \bigcap_{n=1}^\infty G_n$. Since $U$ was an arbitrary nonempty open set, $\bigcap_{n=1}^\infty G_n$ is dense in $X$. [guided] We now have a nested sequence $K_1 \supset K_2 \supset \cdots$ of nonempty compact sets in the Hausdorff space $X$. The [Cantor Intersection Theorem](/theorems/624) (compact version) states: in a Hausdorff space, a decreasing sequence of nonempty compact sets has nonempty intersection. The hypotheses are satisfied — each $K_n$ is compact (by construction), nonempty (since $x_n \in V_n \subset K_n$), and the sequence is nested. The Hausdorff hypothesis is used in the Cantor Intersection Theorem through the finite intersection property: in a Hausdorff space, compact sets are closed, so any finite subcollection $\{K_1, \ldots, K_m\}$ has $K_1 \cap \cdots \cap K_m = K_m \neq \varnothing$ (by nesting), and compactness of $K_1$ together with the finite intersection property forces $\bigcap_{n=1}^\infty K_n \neq \varnothing$. Take any $x \in \bigcap_{n=1}^\infty K_n$. Then: - $x \in K_n \subset G_n$ for every $n \ge 1$, so $x \in \bigcap_{n=1}^\infty G_n$. - $x \in K_1 \subset U$, so $x \in U$. Therefore $U \cap \bigcap_{n=1}^\infty G_n \neq \varnothing$. Since $U$ was an arbitrary nonempty open set, $\bigcap_{n=1}^\infty G_n$ meets every nonempty open subset of $X$ — that is, $\bigcap_{n=1}^\infty G_n$ is dense. This establishes the dense $G_\delta$ form. Note the contrast with the complete metric space proof: there, completeness converted a Cauchy sequence into a limit point; here, the Cantor Intersection Theorem converts a nested sequence of compact sets into a point in the intersection. The two mechanisms are different, but both serve the same purpose — extracting a point from an infinite intersection. [/guided] [/step] [step:Deduce the category form by complementation] The category form follows exactly as in the complete metric space case. Suppose $(F_n)_{n=1}^\infty$ is a sequence of nowhere dense closed subsets of $X$. Define $G_n := X \setminus F_n$, which is open (complement of a closed set) and dense (by the [Equivalent Characterisations of Nowhere Dense Sets](/theorems/1084): $F_n$ is nowhere dense and closed, so $\overline{F_n} = F_n$, and $X \setminus \overline{F_n} = G_n$ is dense). By the dense $G_\delta$ form, $\bigcap_{n=1}^\infty G_n = X \setminus \bigcup_{n=1}^\infty F_n$ is dense. In particular, $\bigcup_{n=1}^\infty F_n$ has empty interior, and $X \neq \bigcup_{n=1}^\infty F_n$ whenever $X$ is nonempty. Therefore $X$ is a Baire space. [guided] The derivation is identical to the complete metric space case. Given nowhere dense closed sets $(F_n)_{n=1}^\infty$, set $G_n := X \setminus F_n$. Each $G_n$ is: - **Open**: since $F_n$ is closed. - **Dense**: by the [Equivalent Characterisations of Nowhere Dense Sets](/theorems/1084), characterisation (ii), $F_n$ nowhere dense implies $X \setminus \overline{F_n}$ is dense. Since $F_n$ is closed, $\overline{F_n} = F_n$, so $G_n = X \setminus F_n$ is dense. The dense $G_\delta$ form gives $\bigcap_{n=1}^\infty G_n$ dense. By De Morgan: \begin{align*} \bigcap_{n=1}^\infty G_n = X \setminus \bigcup_{n=1}^\infty F_n. \end{align*} A set whose complement is dense has empty interior (if $V \subset \bigcup F_n$ were nonempty open, it would be disjoint from $\bigcap G_n$, contradicting density). When $X \neq \varnothing$, density of $\bigcap G_n$ implies $\bigcap G_n \neq \varnothing$, so $X \neq \bigcup F_n$. [/guided] [/step]