Sets of first and second category
Definition (Nowhere dense set)
Given a metric space \((X, d)\), \(A \subseteq X\) is nowhere dense if \((\overline{A})^{c}\) is dense in \(X\). Equivalently, \((\overline{A})^{\circ} = \emptyset\).
Note (Characterisation via the interior of the closure)
\(A\) is nowhere dense if and only if for every \(x \in X\) and \(r > 0\),
\[(\overline{A})^{c} \cap B(x, r) \neq \emptyset.\]Definition (First and second category (meagre))
- A set \(A \subseteq X\) is of first category (or meagre) if it is a countable union of nowhere dense sets.
- A set is of second category if it is not of first category.
The Baire theorems
Theorem (Baire (countable intersection of dense open sets))
Let \((X, d)\) be complete. If \(\{G_{n}\}_{n=1}^{\infty}\) is a sequence of open dense sets in \(X\), then \(\bigcap_{n=1}^{\infty} G_{n}\) is dense in \(X\).
Proof: Let \(A \subseteq X\) be a non-empty open set. Since \(G_{1}\) is dense, there exists \(x_{1} \in A \cap G_{1}\), and since \(A \cap G_{1}\) is open, there exists \(r_{1} > 0\) with \(B(x_{1}, r_{1}) \subseteq A \cap G_{1}\). Take \(B_{1} = B(x_{1}, r_{1}/2)\), so \(\overline{B}_{1} \subseteq B(x_{1}, r_{1}) \subseteq A \cap G_{1}\).
Applying the same argument to \(B_{1}\) and \(G_{2}\), there exist \(x_{2} \in G_{2}\) and \(0 < r_{2} < r_{1}/2\) such that \(B(x_{2}, r_{2}) \subseteq B_{1} \cap G_{2} \subseteq A \cap G_{1} \cap G_{2}\) and, defining \(B_{2} = B(x_{2}, r_{2}/2)\), \(\overline{B}_{2} \subseteq B_{1} \cap G_{2}\).
Iterating, we obtain a sequence \(\{x_{n}\}\) with \(x_{n} \in G_{n}\) and \(r_{n} < r_{n-1}/2 \leq r_{1}/2^{n-1}\) such that
\[B(x_{n}, r_{n}) \subseteq B_{n-1} \cap G_{n} \subseteq A \cap \bigcap_{i=1}^{n} G_{i}, \quad \overline{B}_{n} \subseteq B_{n-1} \cap G_{n}.\]If \(n \geq m\) then \(B_{n} \subseteq B_{m}\) and \(x_{n}, x_{m} \in B_{m}\), so \(d(x_{n}, x_{m}) < 2 r_{m} < r_{1}/2^{m-2}\). Thus \(\{x_{n}\}\) is Cauchy. By completeness there exists \(x = \lim_{n} x_{n}\), and since \(\{x_{n}\}_{n=m}^{\infty} \subseteq B_{m}\), \(x \in \overline{B}_{m} \subseteq A \cap \bigcap_{i=1}^{m} G_{i}\) for every \(m \geq 1\), so \(x \in A \cap \bigcap_{i=1}^{\infty} G_{i}\). We conclude that \(\bigcap_{i=1}^{\infty} G_{i}\) is dense.
Theorem (Baire categories)
Let \((X, d)\) be a complete metric space. If \(G \subseteq X\) is a non-empty open set, then \(G\) is of second category.
Proof: Suppose, for a contradiction, that \(G\) is of first category. There exist nowhere dense sets \(A_{n}\) with \(G = \bigcup_{n=1}^{\infty} A_{n}\). Let \(G_{n} = (\overline{A}_{n})^{c}\); each \(G_{n}\) is open and dense (by the definition of nowhere dense). By Baire’s theorem,
\[\bigcap_{n=1}^{\infty} G_{n} \;=\; \bigcap_{n=1}^{\infty} (\overline{A}_{n})^{c} \;=\; \Big(\bigcup_{n=1}^{\infty} \overline{A}_{n}\Big)^{c}\]is dense in \(X\). On the other hand, since \(A_{n} \subseteq \overline{A}_{n}\) for every \(n\),
\[G \;=\; \bigcup_{n=1}^{\infty} A_{n} \;\subseteq\; \bigcup_{n=1}^{\infty} \overline{A}_{n},\]whence
\[G \cap \Big(\bigcup_{n=1}^{\infty} \overline{A}_{n}\Big)^{c} \;=\; G \cap \bigcap_{n=1}^{\infty} G_{n} \;=\; \emptyset.\]But \(G\) is a non-empty open set, so \(G\) must meet every dense set. This contradicts the density of \(\bigcap_{n=1}^{\infty} G_{n}\), concluding the result.