From Compactness to the Baire Property
The Baire property rules out a particular kind of countable smallness: countably many nowhere-dense sets cannot collectively contain an open region. Compactness is a global condition, while the locally compact Hausdorff Baire theorem connects a local compactness condition to this category conclusion. A compact Hausdorff space meets the local hypothesis directly: the whole space is a compact neighborhood of each of its points.
Recall that a subset \(A\) of a topological space \(X\) is nowhere dense if \(\operatorname{int}\overline{A}=\varnothing\). A space is Baire if every countable intersection of open dense subsets is dense. Here a subset is dense when its closure is the whole space. Equivalently, no nonempty open subset is contained in a countable union of nowhere-dense sets; we prove this equivalence below.
We use the Locally Compact Hausdorff Baire Theorem, proved in the earlier material on locally compact spaces: every locally compact Hausdorff space is Baire. The Hausdorff hypothesis is part of the theorem; compactness alone is not a substitute for it.
Proof. Let \((X,\mathcal{T})\) be compact and Hausdorff, and let \(x\in X\). In the definition of local compactness, take \(K=X\) and \(U=X\). The set \(U\) is open and contains \(x\), while \(K\) is compact by the assumed compactness of \(X\) and contains \(U\). Thus \(x\) has a compact neighborhood. This works for every \(x\in X\), so \(X\) is locally compact. If \(X=\varnothing\), the condition is vacuous, and the conclusion still holds. \(\square\)
Proof. Let \(X\) be compact and Hausdorff. The proposition shows that \(X\) is locally compact, so the Locally Compact Hausdorff Baire Theorem applies and gives that \(X\) is Baire. More explicitly, for any sequence \((G_n)_{n\in\mathbb{N}}\) of open dense subsets of \(X\), the intersection \(\bigcap_{n\in\mathbb{N}}G_n\) is dense in \(X\). \(\square\)
The theorem provides the implication; the compactness argument above explains why its local hypothesis is automatic. The conclusion is about countable intersections, not arbitrary intersections. It also does not say that the intersection is open or that it equals all of \(X\).
Two Equivalent Ways to Use the Conclusion
Proof. Suppose first that \(X\) is Baire, and let \((A_n)_{n\in\mathbb{N}}\) be nowhere-dense subsets. For each \(n\), set \(G_n=X\setminus\overline{A_n}\). Because \(\overline{A_n}\) is closed, \(G_n\) is open. Because \(\operatorname{int}\overline{A_n}=\varnothing\), every nonempty open set meets \(G_n\), so \(G_n\) is dense. The Baire property says that \(\bigcap_{n\in\mathbb{N}}G_n\) is dense. If a nonempty open set \(V\) were contained in \(\bigcup_{n\in\mathbb{N}}A_n\), it would be disjoint from every point of that intersection, contradicting its density. Hence no such \(V\) exists.
Conversely, suppose no nonempty open subset is contained in a countable union of nowhere-dense sets. Let \((G_n)_{n\in\mathbb{N}}\) be open dense subsets. The complement \(X\setminus G_n\) is closed. It has empty interior: a nonempty open subset of \(X\setminus G_n\) would be an open set missing the dense set \(G_n\). Thus each \(X\setminus G_n\) is nowhere dense. If \(\bigcap_{n\in\mathbb{N}}G_n\) were not dense, some nonempty open set \(V\) would miss it. Then \(V\subseteq\bigcup_{n\in\mathbb{N}}(X\setminus G_n)\), contrary to the assumed condition. So every countable intersection of open dense sets is dense, and \(X\) is Baire. \(\square\)
Worked Applications
Worked Example: A Compact Interval
Let \(a,b\in\mathbb{R}\) with \(a<b\), and give \([a,b]\) the subspace topology inherited from the standard topology on \(\mathbb{R}\). This interval is closed and bounded in \(\mathbb{R}\), so it is compact by the Heine–Borel theorem. It is Hausdorff: for distinct \(x,y\in[a,b]\), disjoint open intervals around \(x\) and \(y\) can be chosen in \(\mathbb{R}\), and their intersections with \([a,b]\) are disjoint relative neighborhoods. Therefore \([a,b]\) is Baire.
For a concrete consequence, take any countable family \((F_n)_{n\in\mathbb{N}}\) of nowhere-dense subsets of \([a,b]\), with nowhere denseness understood in the subspace topology. The Baire conclusion gives \(\bigcap_{n\in\mathbb{N}}([a,b]\setminus\overline{F_n})\) dense in \([a,b]\). In particular, it meets every nonempty relatively open subset of the interval. The equivalent criterion also says that no nonempty relatively open subset of \([a,b]\) can be covered by the sets \(F_n\).
If \(a=b\), the interval is the one-point space \(\{a\}\). It is compact and Hausdorff, hence Baire by the same theorem. In this case the only dense subset is \(\{a\}\), so every countable intersection of open dense subsets is \(\{a\}\). Thus the degenerate interval causes no exception.
Worked Example: The Cantor Set
Let \(C\subseteq[0,1]\) be the middle-thirds Cantor set: start with \([0,1]\), remove the open middle third of each remaining interval at each stage, and take the intersection of the resulting sets \(C_n\). Each \(C_n\) is a finite union of closed intervals, hence is closed in \([0,1]\). Therefore \(C=\bigcap_{n\in\mathbb{N}}C_n\) is closed in \([0,1]\). Since \([0,1]\) is compact, its closed subset \(C\) is compact. With its subspace topology, \(C\) is Hausdorff because \([0,1]\) is Hausdorff. The compact Hausdorff theorem therefore says that \(C\) is Baire.
One can see the conclusion at work using particular dense open sets. Every point of \(C\) has a representation \(p=\sum_{j=1}^{\infty}2a_j/3^j\), where each \(a_j\) is \(0\) or \(1\). For any \(k\in\mathbb{N}\), changing just the \(k\)-th digit gives another point \(q\in C\) and changes the value by exactly \(2/3^k\). Since \(2/3^k\) tends to \(0\), every relative neighborhood of \(p\) contains a point of \(C\) different from \(p\). Thus \(C\) has no isolated points.
For each \(n\in\mathbb{N}\), let \(p_n=2/3^n\) and \(U_n=C\setminus\{p_n\}\). The points \(p_n\) belong to \(C\) and are distinct, since their ternary representations have a single digit \(2\), in different positions. A singleton is closed in a Hausdorff space, so \(U_n\) is open in \(C\). It is dense in \(C\): the removed singleton has empty interior because \(C\) has no isolated points. The Baire theorem now gives that \(\bigcap_{n\in\mathbb{N}}U_n=C\setminus\{p_n:n\in\mathbb{N}\}\) is dense in \(C\). This conclusion concerns density in the relative topology of \(C\).
Worked Example: The Circle
Give \(S^1=\{(x,y)\in\mathbb{R}^2:x^2+y^2=1\}\) the subspace topology from the standard topology on \(\mathbb{R}^2\). The circle is closed because it is the inverse image of the closed set \(\{1\}\) under the continuous function \((x,y)\mapsto x^2+y^2\), and it is bounded because every point on it has Euclidean norm \(1\). The Heine–Borel theorem makes it compact. The Euclidean topology is Hausdorff, and its subspace \(S^1\) is Hausdorff as well. Consequently, \(S^1\) is Baire.
For example, let \((F_n)_{n\in\mathbb{N}}\) be any sequence of nowhere-dense subsets of \(S^1\). It cannot cover the circle: if \(S^1=\bigcup_{n\in\mathbb{N}}F_n\), then the nonempty open set \(S^1\) itself would be contained in that countable union, contradicting the equivalent Baire criterion. More generally, no nonempty relatively open arc can be covered by such a family. The word “relatively” matters: openness here is openness in the topology on \(S^1\), not in the plane.
Why the Hausdorff Hypothesis Matters
Common pitfall: compactness by itself does not imply the Baire property. To check this, let \(X=\mathbb{N}\) have the finite-complement topology: its open sets are \(\varnothing\) and all subsets with finite complement. This space is compact. Indeed, from any open cover choose one nonempty member \(U\); only finitely many points lie outside \(U\), and finitely many additional members of the cover suffice to cover those points.
The space is not Hausdorff. Any two nonempty open sets have finite complements in the infinite set \(X\), so their intersection is nonempty; in particular, distinct points cannot have disjoint open neighborhoods. For each \(n\in\mathbb{N}\), set \(G_n=X\setminus\{n\}\). It is open by the definition of the topology. It is dense because every nonempty open set is infinite and therefore intersects \(G_n\). But \(\bigcap_{n\in\mathbb{N}}G_n=\varnothing\): each point \(m\in X\) is excluded from \(G_m\). The empty set is not dense in this nonempty space. Thus \(X\) is compact but not Baire, showing why compactness without the Hausdorff hypothesis cannot be used in the theorem.
For compact Hausdorff spaces, the useful chain is precise: compactness supplies a compact neighborhood at every point, Hausdorffness supplies the other hypothesis of the locally compact Hausdorff theorem, and that theorem supplies the Baire conclusion. The resulting dense-intersection statement is often the most convenient form; the nowhere-dense formulation is its equivalent way to rule out countable coverings of open regions.
Check Your Understanding
Use the hypotheses and conclusions in the statements above to answer each question.
- Which compact neighborhood can be used to show that a compact space is locally compact?
- Why does the compact Hausdorff Baire theorem apply to the Cantor set with its subspace topology?
- If \(G_n\) is open and dense for every \(n\), what does the Baire property say about \(\bigcap_{n\in\mathbb{N}}G_n\)? Does it say that this intersection is open?
- Explain why the sets \(X\setminus\{n\}\) in the finite-complement example are dense and open, although their intersection is empty.
- In the nowhere-dense criterion, is it permitted for the countable union to contain a nonempty open set? State exactly what the criterion rules out.