The Locally Compact Hausdorff Baire Theorem

From Local Compactness to the Baire Property

A complete metric space has the Baire property because nested closed balls can be made smaller while meeting successive dense open sets. The same strategy works without a metric when the space is locally compact and Hausdorff: instead of small balls, we use open neighborhoods whose closures are compact. The Hausdorff condition ensures those compact closures are closed, and local compactness lets us keep each closure inside the open set we are trying to meet.

We use the following definitions. A space is locally compact here if every point has an open neighborhood whose closure is compact. It is Hausdorff if any two distinct points have disjoint open neighborhoods. A subset is relatively compact if its closure is compact. A space \(X\) is Baire if every countable intersection of dense open subsets of \(X\) is dense in \(X\). Equivalently, to prove that such an intersection is dense, it is enough to show that it meets every nonempty open set.

Theorem (Locally compact Hausdorff Baire theorem): Every locally compact Hausdorff space is Baire. More explicitly, if \((X,\mathcal{T})\) is locally compact and Hausdorff and \(G_n\) is dense and open in \(X\) for every \(n\in\mathbb{N}\), then \(\bigcap_{n\in\mathbb{N}}G_n\) is dense in \(X\).

The proof needs an open-neighborhood shrinking fact. We prove it first, including the role of both hypotheses.

Lemma (Relatively compact neighborhood inside an open set): Let \(X\) be locally compact and Hausdorff. If \(x\in U\) and \(U\) is open, then there is an open set \(V\) such that \(x\in V\), \(\overline{V}\) is compact, and \(\overline{V}\subseteq U\).

Proof. By local compactness, choose an open neighborhood \(W\) of \(x\) for which \(K=\overline{W}\) is compact. The set \(F=K\setminus(U\cap W)\) is closed in \(K\), so it is compact, and it does not contain \(x\). For each \(y\in F\), the Hausdorff condition gives disjoint open sets \(G_y,H_y\) with \(x\in G_y\) and \(y\in H_y\). The sets \(H_y\) cover the compact set \(F\), so finitely many, say \(H_{y_1},\ldots,H_{y_m}\), cover \(F\). Set \(G=\bigcap_{i=1}^{m}G_{y_i}\) and \(V=G\cap W\). Then \(V\) is open and contains \(x\). Also \(V\subseteq W\), so \(\overline{V}\subseteq K\). Since \(V\) misses each \(H_{y_i}\), its closure misses each \(H_{y_i}\): an open set disjoint from \(V\) is disjoint from \(\overline{V}\). Thus \(\overline{V}\) misses \(F\). Within \(K\), the complement of \(F\) is \(K\cap U\cap W\), so \(\overline{V}\subseteq U\). Finally, \(\overline{V}\) is closed in the compact set \(K\), hence compact. If \(F\) is empty, the same argument works with \(G=X\) and \(V=W\), since then \(K\subseteq U\cap W\). \(\square\)

The Nested-Neighborhood Proof

We will also use two elementary compactness facts. In a Hausdorff space every compact subset is closed: for a point outside a compact set, separate it from each point of that set by disjoint open neighborhoods, then use a finite subcover to obtain a neighborhood disjoint from the compact set. Also, a family of closed subsets of a compact space with the finite-intersection property has nonempty intersection. This follows directly from the open-cover definition of compactness: if the intersection were empty, their complements would cover the space; a finite subcover would say some finite intersection was empty.

Proof of the theorem. Let \(U\) be any nonempty open subset of \(X\). We will find a point of \(U\cap\bigcap_{n\in\mathbb{N}}G_n\). Since \(G_1\) is dense and open, \(U\cap G_1\) is a nonempty open set: density means every nonempty open set meets \(G_1\). Apply the lemma to choose a nonempty open \(U_1\) with \(\overline{U_1}\subseteq U\cap G_1\) and \(\overline{U_1}\) compact. Inductively, once \(U_n\) has been chosen, \(U_n\cap G_{n+1}\) is nonempty and open, so the lemma gives a nonempty open \(U_{n+1}\) with \(\overline{U_{n+1}}\subseteq U_n\cap G_{n+1}\) and compact closure. This yields nested nonempty compact sets \(C_n=\overline{U_n}\), with \(C_{n+1}\subseteq C_n\). Each \(C_n\) is closed in \(X\), because it is compact and \(X\) is Hausdorff. In the compact space \(C_1\), the sets \(C_n\) are closed and have the finite-intersection property: any finite intersection is the smallest-indexed set among them, and is nonempty. Therefore there is \(x\in\bigcap_{n\in\mathbb{N}}C_n\). Since \(C_1\subseteq U\), this point lies in \(U\). For every \(n\), \(C_n\subseteq G_n\), so \(x\in G_n\). We have found a point of \(U\cap\bigcap_{n\in\mathbb{N}}G_n\). As this holds for every nonempty open \(U\), the intersection is dense. If \(X\) is empty, density is vacuous and the conclusion holds as well. \(\square\)

The crucial point is not simply that each \(G_n\) is dense. At stage \(n\), we choose a neighborhood whose entire closure lies in \(G_n\), while also placing it inside the preceding neighborhood. Compactness then supplies one point that belongs to every stage, rather than a different point at each stage.

Worked Example: Nested Intervals in the Real Line

Let \(X=\mathbb{R}\) with its standard topology, and for \(n\in\mathbb{N}\) set \(G_n=\mathbb{R}\setminus\{1/(n+2)\}\). Each singleton is closed and has empty interior in \(\mathbb{R}\), so \(G_n\) is open and dense. Start with \(U=(0,1)\). Here is an explicit nested sequence of relatively compact open intervals that stays inside \(U\) and meets the required conditions:

U_n=\left(\frac12-\frac{1}{12\cdot 2^{n-1}},\frac12+\frac{1}{12\cdot 2^{n-1}}\right).

For \(n=1\), this gives \(U_1=(5/12,7/12)\), whose closure is \([5/12,7/12]\subset(0,1)\). For \(n=2\), it gives \(U_2=(11/24,13/24)\), whose closure lies inside \(U_1\). At every step the radius is halved, so \(\overline{U_{n+1}}\subset U_n\). Each removed point satisfies \(1/(n+2)\leq1/3=4/12<5/12\), while every \(U_n\) lies to the right of \(5/12\). Thus \(U_n\subseteq G_n\) for every \(n\). The closures are closed bounded intervals, hence compact by the Heine–Borel theorem.

Check: \(1/2\in U_n\) for every \(n\), since the distance from \(1/2\) to the center of \(U_n\) is \(0\), which is less than its positive radius. Also \(1/2\in U\) and \(1/2\in G_n\), because \(1/2\neq1/(n+2)\) for all \(n\geq1\). The construction exhibits a common point, as the theorem guarantees.

Worked Example: A Noncompact Locally Compact Space

Take \(X=(0,1)\) with its subspace topology from \(\mathbb{R}\). It is Hausdorff. It is locally compact: for \(x\in(0,1)\), let \(\delta=\min(x,1-x)/2>0\). The open neighborhood \((x-\delta,x+\delta)\) has closure \([x-\delta,x+\delta]\) in \(X\), a compact interval contained in \((0,1)\). The space \(X\) itself is not compact, since the open sets \((1/n,1)\), \(n\geq2\), cover \((0,1)\) but no finite subcollection covers points sufficiently close to \(0\).

For \(n\in\mathbb{N}\), let \(p_n=1/(n+2)\) and \(G_n=X\setminus\{p_n\}\). Each \(G_n\) is open because singletons are closed in a Hausdorff space. It is dense because every nonempty relatively open interval in \(X\) contains more than one point, and therefore cannot be contained in \(\{p_n\}\). The theorem implies that \(\bigcap_nG_n\) is dense in \(X\). For a direct check on the nonempty open interval \(U=(1/2,3/4)\), we have \(p_n\leq1/3<1/2\) for every \(n\), so \(U\subseteq G_n\) for every \(n\). In particular, \(U\) meets their intersection. The space need not be compact for the theorem to apply.

Worked Example: Why Local Compactness Cannot Be Dropped

Let \(X=\mathbb{Q}\) with the subspace topology from \(\mathbb{R}\), and enumerate its elements as \(q_1,q_2,\ldots\). For each \(n\), the set \(G_n=\mathbb{Q}\setminus\{q_n\}\) is open. It is dense because every nonempty relatively open set in \(\mathbb{Q}\) contains a rational other than \(q_n\). But \(\bigcap_nG_n=\varnothing\), since every rational is removed at its index. Thus \(\mathbb{Q}\) is not Baire, despite being Hausdorff.

This example does not contradict the theorem: \(\mathbb{Q}\) is not locally compact. If a point \(q\in\mathbb{Q}\) had an open neighborhood with compact closure \(K\) in \(\mathbb{Q}\), then \(K\) would contain \((q-\varepsilon,q+\varepsilon)\cap\mathbb{Q}\) for some \(\varepsilon>0\). The inclusion from \(\mathbb{Q}\) into \(\mathbb{R}\) is continuous, so \(K\), being compact in \(\mathbb{Q}\), would be compact in \(\mathbb{R}\) and therefore closed in \(\mathbb{R}\). Since \(K\) contains all rationals in that interval, its closedness forces it to contain the whole interval, including irrational numbers. That contradicts \(K\subseteq\mathbb{Q}\).

Check: the dense open sets above have empty intersection, so the Baire criterion fails exactly as claimed. The missing hypothesis is local compactness, not the Hausdorff condition.

A Common Pitfall

Common pitfall. It is not enough to choose one nonempty compact set at each stage; the sets must be nested. For example, \(C_n=[n,n+1]\subset\mathbb{R}\) is nonempty and compact for every \(n\), but \(\bigcap_{n\in\mathbb{N}}C_n=\varnothing\): already \(C_1=[1,2]\) and \(C_2=[2,3]\) intersect only at \(2\), while \(C_3=[3,4]\) does not contain \(2\). In the theorem’s proof, the inclusions \(\overline{U_{n+1}}\subseteq U_n\subseteq\overline{U_n}\) guarantee nestedness. That is what gives the finite-intersection property needed to invoke compactness.

The argument also explains why local compactness and the Hausdorff condition have distinct jobs. Local compactness gives a compact neighborhood to begin the shrinking construction. The Hausdorff condition lets us shrink inside the desired open set and ensures that the resulting compact closures are closed. With these tools, the complete-metric nested-ball argument becomes a compact-set argument that applies to spaces without any specified metric.

Check Your Understanding

For each question, identify where density, local compactness, or compactness is used.

In the neighborhood-shrinking lemma, why is the set \(F=K\setminus(U\cap W)\) compact?
Why does \(U_n\cap G_{n+1}\) have to be nonempty in the inductive step?
In the theorem’s proof, where is the Hausdorff condition used after the neighborhood-shrinking lemma?
Give a family of nonempty compact subsets of \(\mathbb{R}\) whose intersection is empty, and explain which feature of the theorem’s nested family it lacks.
For \(X=(0,1)\), verify that \(X\setminus\{1/(n+2)\}\) is dense and open for each \(n\), and state what the theorem says about their intersection.

The Locally Compact Hausdorff Baire Theorem

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