Two Ways to Recognize a Baire Space
Some topological spaces have a useful stability property: countably many open dense conditions can be imposed at once without losing density. The Baire property expresses this idea in terms of intersections, while an equivalent formulation says that no nonempty open region can be covered by a countable collection of nowhere-dense sets. The equivalence is useful because the two formulations turn different questions into manageable tests.
Throughout, \((X,\mathcal{T})\) is a topological space. A set \(D\subseteq X\) is dense if \(\overline{D}=X\); equivalently, every nonempty open subset of \(X\) intersects \(D\). The closure \(\overline{A}\) is the smallest closed set containing \(A\), and \(\operatorname{int}A\) is the largest open set contained in \(A\).
“Meager in \(X\)” is a statement about the ambient space: the nowhere-dense sets used in the cover are tested using closure and interior in \(X\). A set can be meager even if it is not itself nowhere dense, since it may be a countable union of nowhere-dense sets. The empty set is meager, and the definition of Baire space includes the empty space: in that space the only subset is empty, which is dense in \(X=\varnothing\).
Replacing Meager Covers by Closed Covers
In proofs it is often more convenient to work with closed nowhere-dense sets. Taking closures makes this possible without changing whether a set is meager.
Proof. Suppose \(M\) is meager. By definition, there are nowhere-dense sets \(N_n\subseteq X\), for \(n\in\mathbb{N}\), such that \(M\subseteq\bigcup_{n\in\mathbb{N}}N_n\). For each \(n\), the set \(\overline{N_n}\) is closed, and it is nowhere dense because \(\operatorname{int}\overline{N_n}=\varnothing\). Since \(N_n\subseteq\overline{N_n}\), we have \(M\subseteq\bigcup_{n\in\mathbb{N}}\overline{N_n}\), a cover by closed nowhere-dense sets. Conversely, every closed nowhere-dense set is nowhere dense by the definition, so containment in a countable union of such sets implies that \(M\) is meager. \(\square\)
This lemma lets us pass directly between a meager open set and a sequence of dense open sets: the complements of closed nowhere-dense sets are open and dense.
The Equivalence of the Two Criteria
(1) Every countable intersection of open dense subsets of \(X\) is dense in \(X\).
(2) No nonempty open subset of \(X\) is meager in \(X\).
Proof. First assume (1). Suppose, for a contradiction, that some nonempty open set \(U\subseteq X\) is meager. By the lemma, there are closed nowhere-dense sets \(F_n\) such that \(U\subseteq\bigcup_{n\in\mathbb{N}}F_n\). Put \(G_n=X\setminus F_n\). Each \(G_n\) is open because \(F_n\) is closed. It is also dense: if some nonempty open set \(V\) missed \(G_n\), then \(V\subseteq F_n\), so \(V\subseteq\operatorname{int}F_n\), contradicting that \(F_n\) is nowhere dense. Thus (1) says that \(\bigcap_{n\in\mathbb{N}}G_n\) is dense. A dense set must intersect the nonempty open set \(U\). But \(U\subseteq\bigcup_n F_n\), so no point of \(U\) belongs to every \(G_n=X\setminus F_n\). This contradiction proves (2).
Now assume (2), and let \((G_n)_{n\in\mathbb{N}}\) be any sequence of open dense subsets of \(X\). Suppose that \(G=\bigcap_{n\in\mathbb{N}}G_n\) is not dense. Then some nonempty open set \(U\) misses \(G\): this follows from the characterization of density by intersection with every nonempty open set. For each \(x\in U\), there is therefore some \(n\) with \(x\notin G_n\), and hence \(U\subseteq\bigcup_{n\in\mathbb{N}}(X\setminus G_n)\). Each \(X\setminus G_n\) is closed. It has empty interior, because a nonempty open subset contained in \(X\setminus G_n\) would fail to intersect the dense set \(G_n\). Thus each complement is nowhere dense, making \(U\) meager. This contradicts (2). Therefore \(G\) is dense, which proves (1). \(\square\)
The quantifiers in (1) matter: the sets being intersected must each be both open and dense. The theorem does not assert that an intersection of arbitrary open sets is dense. Its conclusion is also stronger than saying that the intersection is nonempty when \(X\) is nonempty: it must meet every nonempty open subset of \(X\).
Worked Examples
Worked Example: A Discrete Space
Let \(X\) be any set with the discrete topology, in which every subset is open. Determine whether \(X\) is Baire, including when \(X=\varnothing\).
For any \(A\subseteq X\), the set \(A\) is closed as well as open, so \(\overline{A}=A\) and \(\operatorname{int}\overline{A}=A\). It follows that \(A\) is nowhere dense exactly when \(A=\varnothing\). A countable union of nowhere-dense subsets is therefore empty, so the only meager subset of \(X\) is \(\varnothing\). No nonempty open subset is meager.
For a direct check of the other criterion, an open dense subset \(G\) must equal \(X\): its closure is \(\overline{G}=G\), so density \(\overline{G}=X\) forces \(G=X\). Every countable intersection of open dense sets is thus \(X\), which is dense in \(X\). If \(X=\varnothing\), the only possible intersection is again \(\varnothing=X\), and the density statement still holds.
Check: both criteria give the same conclusion: every discrete space is Baire.
Worked Example: The Sierpinski Space
Let \(X=\{a,b\}\) with topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\). Find its nowhere-dense sets and verify the Baire property.
The closed sets, obtained by taking complements of the open sets, are \(\varnothing,\{b\},X\). The closure of \(\{a\}\) is \(X\), since \(X\) is the only closed set containing \(a\); its closure has interior \(X\), so \(\{a\}\) is not nowhere dense. The closure of \(\{b\}\) is \(\{b\}\). The only open subset contained in \(\{b\}\) is \(\varnothing\), so \(\operatorname{int}\{b\}=\varnothing\), and \(\{b\}\) is nowhere dense. The empty set is nowhere dense, while \(X\) is not. Thus the nowhere-dense subsets are exactly \(\varnothing\) and \(\{b\}\).
Every countable union of nowhere-dense subsets is contained in \(\{b\}\). The nonempty open subsets are \(\{a\}\) and \(X\), and neither is contained in \(\{b\}\). Therefore no nonempty open set is meager. To check the intersection criterion, the open dense sets are \(\{a\}\) and \(X\): \(\{a\}\) is dense because its closure is \(X\). Any countable sequence chosen from these two sets has intersection either \(X\), if all terms are \(X\), or \(\{a\}\), if at least one term is \(\{a\}\). Both intersections are dense.
Check: the space is Baire, even though it is not Hausdorff: the open neighborhoods of \(b\) include only \(X\), so \(a\) and \(b\) cannot have disjoint open neighborhoods.
Worked Example: A Countably Infinite Cofinite Space
Let \(X=\mathbb{N}=\{1,2,\ldots\}\) with the finite-complement topology: the open sets are \(\varnothing\) and all subsets whose complements in \(X\) are finite. Show that \(X\) is not Baire in two ways.
First consider a singleton \(\{n\}\). Its complement is open, so \(\{n\}\) is closed. It is not open: its complement in the infinite set \(X\) is infinite, so \(\{n\}\) does not have finite complement. Thus its interior is empty, and because it is closed, its closure is itself. Hence \(\{n\}\) is nowhere dense. Since \(X=\bigcup_{n\in\mathbb{N}}\{n\}\), the whole space \(X\) is meager. It is nonempty and open, so criterion (2) fails.
For a direct check, define \(G_n=X\setminus\{n\}\) for every \(n\in\mathbb{N}\). Each \(G_n\) is open because its complement is finite. It is dense: if \(V\) is any nonempty open set, then \(V\) has finite complement in \(X\), and \(V\cap G_n\) has finite complement too. Since \(X\) is infinite, \(V\cap G_n\ne\varnothing\). However, a point \(k\in X\) is absent from \(G_k\), so \(\bigcap_{n\in\mathbb{N}}G_n=\varnothing\). The empty set is not dense in the nonempty space \(X\). Thus criterion (1) fails as well.
Check: the same countable collection demonstrates both failures: its members are dense and open, but their intersection is empty, while the singleton sets show that \(X\) itself is meager.
A Common Pitfall
One tempting but incorrect argument is that a countable intersection of dense open sets must be dense because each set in the intersection is dense. Density guarantees that each individual set meets every nonempty open set; it does not, by itself, guarantee that infinitely many such requirements can be met simultaneously. In the cofinite example, every \(G_n=X\setminus\{n\}\) is dense and open, but the calculation gives \(\bigcap_n G_n=\varnothing\), which is not dense in \(X\).
The theorem identifies precisely what distinguishes Baire spaces: in them, this simultaneous intersection is dense, equivalently no nonempty open region is meager. Outside the Baire setting, the two explicit failures in the cofinite example show why the conclusion cannot be assumed without checking the space.
Check Your Understanding
For each question, specify the ambient topology when deciding whether a set is open, dense, nowhere dense, or meager.
- In the standard topology on \(\mathbb{R}\), explain why \(\mathbb{Z}\) is nowhere dense.
- Use the theorem to show that if a nonempty open set is meager, then there is a sequence of dense open sets whose intersection misses that open set.
- In the countably infinite cofinite space, verify directly that \(G_n=X\setminus\{n\}\) is dense by testing its intersection with an arbitrary nonempty open set.
- Can a meager set contain a nonempty open subset of a Baire space? Justify your answer using the equivalent criteria.
- In the Sierpinski space above, list all the open dense sets and determine the intersection of a sequence containing \(\{a\}\) at least once.