Proof of Baire Category Theorem

The Proof Strategy: Shrinking Nested Balls

The Baire Category Theorem says that countably many open dense sets in a complete metric space have dense intersection. The previous tutorial used this principle to study residual sets and typical properties. Here we prove the theorem by a direct construction. The central idea is to choose closed balls that are nested, lie successively inside the required open sets, and have radii tending to zero. Completeness then supplies a point belonging to every ball.

We will prove the theorem for a nonempty complete metric space. The nonemptiness condition matters when stating the equivalent claim that nowhere-dense sets cannot cover the space: the empty space is complete and is covered by the empty set. The open-dense intersection formulation itself is also true for the empty space, since its only subset is empty and empty is dense in itself. We focus on the nonempty case, where the construction has a point at which to begin.

For a metric space \((X,d)\), write \(B(x,r)=\{y\in X:d(x,y)<r\}\) for the open ball and \(\overline{B}(x,r)=\{y\in X:d(x,y)\leq r\}\) for the closed ball. These are balls in the metric space \(X\), so they are understood relative to \(X\).

Proof. Since \(G_1\) is dense and \(U\) is nonempty and open, \(U\cap G_1\) is nonempty. It is open, so choose \(x_1\in U\cap G_1\) and \(a_1>0\) such that \(B(x_1,a_1)\subseteq U\cap G_1\). Choose \(r_1\) with \(0<r_1<\min(a_1,2^{-1})\). If \(y\in\overline{B}(x_1,r_1)\), then \(d(y,x_1)\leq r_1<a_1\), so \(y\in B(x_1,a_1)\). Therefore \(\overline{B}(x_1,r_1)\subseteq U\cap G_1\).

Suppose \(x_{n-1}\) and \(r_{n-1}\) have been chosen. The open ball \(B(x_{n-1},r_{n-1})\) is nonempty. Since \(G_n\) is dense, the intersection \(B(x_{n-1},r_{n-1})\cap G_n\) is nonempty; it is also open. Choose \(x_n\) in this intersection and choose \(a_n>0\) such that \(B(x_n,a_n)\subseteq B(x_{n-1},r_{n-1})\cap G_n\). Take \(r_n\) with \(0<r_n<\min(a_n,2^{-n})\). As before, the closed ball \(\overline{B}(x_n,r_n)\) lies inside \(B(x_n,a_n)\). Consequently, \(\overline{B}(x_n,r_n)\subseteq B(x_{n-1},r_{n-1})\cap G_n\). In particular, it is contained in the preceding closed ball, and it is contained in \(G_n\). This completes the recursive construction. The inequalities \(0<r_n<2^{-n}\) imply \(r_n\to0\).

It remains to use completeness. If \(m,k\geq n\), both \(x_m\) and \(x_k\) belong to \(\overline{B}(x_n,r_n)\), by nesting. The triangle inequality gives \(d(x_m,x_k)\leq d(x_m,x_n)+d(x_n,x_k)\leq2r_n\). Since \(r_n\to0\), the sequence \((x_n)\) is Cauchy. Completeness gives a point \(x\in X\) such that \(x_n\to x\). For any fixed \(n\), every \(x_m\) with \(m\geq n\) lies in the closed set \(\overline{B}(x_n,r_n)\). A closed set contains the limit of each convergent sequence in it, so \(x\in\overline{B}(x_n,r_n)\). This holds for every \(n\), proving the desired nested-ball construction. \(\square\)

The Baire Category Theorem

The lemma gives the density conclusion by allowing the first ball to be placed inside any chosen nonempty open set. That extra freedom is essential: to prove a set is dense, we must show that it meets every nonempty open set.

Proof. Let \(U\) be any nonempty open subset of \(X\). Apply the Nested-Ball Construction to \(U\) and the sequence \((G_n)\). It gives a point \(x\) in every closed ball \(\overline{B}(x_n,r_n)\). Since the first closed ball is contained in \(U\cap G_1\), we have \(x\in U\cap G_1\). For each \(n\geq2\), the \(n\)-th closed ball is contained in \(G_n\), so \(x\in G_n\) as well. Hence \(x\in U\cap\bigcap_{n=1}^{\infty}G_n\). Every nonempty open \(U\) meets the intersection, which is exactly the definition of density. \(\square\)

The proof isolates the role of each hypothesis. Density lets us choose the next center inside the current ball and the next open set. Openness gives room around that center for a smaller closed ball. The radii tending to zero make the centers Cauchy, while completeness guarantees that their limit exists in \(X\). Finally, closedness of the balls ensures that the limit remains in every one of them.

The Nowhere-Dense Formulation

Recall that a set \(A\subseteq X\) is nowhere dense when \(\overline{A}\) has empty interior. Its complement \(X\setminus\overline{A}\) is then open and dense. The Baire Category Theorem therefore has the following covering consequence.

Proof. Let \(A_1,A_2,\ldots\) be nowhere-dense subsets of \(X\), and put \(G_n=X\setminus\overline{A_n}\). Each \(G_n\) is open. It is dense because \(\overline{A_n}\) has empty interior: every nonempty open set must contain a point outside \(\overline{A_n}\). By the Baire Category Theorem, \(\bigcap_n G_n\) is dense. If a nonempty open set \(U\) were contained in \(\bigcup_n A_n\), density would give a point \(x\in U\cap\bigcap_n G_n\). But \(x\notin\overline{A_n}\), and thus \(x\notin A_n\), for every \(n\). This contradicts \(U\subseteq\bigcup_n A_n\). Taking \(U=X\) proves that the whole nonempty space cannot be such a union. \(\square\)

The conclusion concerns a nonempty open set, not just the whole space. A countable union of nowhere-dense sets cannot contain any nonempty open region in a complete metric space. This is the useful form when a problem proposes that a “small” countable union might contain a neighborhood.

Worked Applications of the Construction

Why Completeness Cannot Be Dropped

The nested-ball proof also shows why completeness is not a decorative assumption. In the incomplete space \(\mathbb{Q}\), with the metric inherited from \(\mathbb{R}\), each set \(G_n=\mathbb{Q}\setminus\{q_n\}\), where \((q_n)\) enumerates \(\mathbb{Q}\), is open and dense. Yet their intersection is empty, because every rational number occurs as some \(q_n\). The failure is exactly at the completeness step: the nested balls can have a limit in \(\mathbb{R}\) that is not in \(\mathbb{Q}\). There is no guarantee that the limiting point belongs to the incomplete space.

A second common pitfall is to replace “open and dense” by “dense.” Density alone does not make the recursive choice work: a dense set need not contain a neighborhood around the chosen center in which to place a closed ball. In the theorem, openness provides that neighborhood; density ensures it can be found at every stage. Together with completeness, these properties force a point to satisfy all countably many conditions.