The Baire Category Theorem for Complete Metric Spaces

# The Baire Category Theorem for Complete Metric Spaces

Recall from The Cantor Intersection Theorem for Complete Metric Spaces page that if $X$ is a complete metric space, $(x_n)_{n=1}^{\infty}$ is a sequence in $X$, and $(r_n)_{n=1}^{\infty}$ is a sequence of positive real numbers such that $\displaystyle{\lim_{n \to \infty} r_n = 0}$ and $... \subseteq \overline{B}(x_{n+1}, r_{n+1}) \subseteq \overline{B}(x_n, r_n) \subseteq ... \subseteq \overline{B}(x_1, r_1)$ then there exists a point $x \in X$ such that:

(1)
\begin{align} \quad \bigcap_{n=1}^{\infty} \overline{B}(x_n, r_n) = \{ x \} \end{align}

We use this theorem to prove an important result in analysis known as the Baire category theorem.

 Theorem 1 (The Baire Category Theorem for Complete Metric Spaces): Let $X$ be a complete metric space. If $(D_n)_{n=1}^{\infty}$ is any countable collection of open dense sets in $X$ then $\displaystyle{\bigcap_{n=1}^{\infty} D_n}$ is dense in $X$.

Recall that if $X$ is a metric space then a subset $A \subseteq X$ is said to be dense in $X$ if for every open set $U \subseteq X$ we have that $A \cap U \neq \emptyset$. Equivalently, $A$ is dense in $X$ if for every open ball $B(x^*, r^*) \subseteq X$ we have that $A \cap B(x^*, r^*) \neq \emptyset$. We use this second characterization of the definition of a dense set in the proof of the Baire category theorem.

• Proof: Let $x^* \in X$, $r^* > 0$, and let $B(x^*, r^*)$ be the open ball centered at $x^*$ with radius $r^*$. We want to show that the intersection with $\displaystyle{\bigcap_{n=1}^{\infty} D_n}$ and this arbitrary open ball $B(x^*, r^*)$ is nonempty.
• Since $D_1$ is dense in $X$, we have that:
(2)
\begin{align} \quad D_1 \cap B(x^*, r^*) \neq \emptyset \end{align}
• So let $x_1 \in D_1 \cap B(x^*, r^*)$. Now since $D_1$ and $B(x^*, r^*)$ are both open sets, there exists an $r_1 > 0$ with $r_1 < 1$ such that:
(3)
\begin{align} \quad \overline{B}(x_1, r_1) \subseteq D_1 \cap B(x^*, r^*) \end{align}
• Now since $D_2$ is dense in $X$ , we have that:
(4)
\begin{align} \quad D_2 \cap B(x_1, r_1) \neq \emptyset \end{align}
• So let $x_2 \in D_2 \cap B(x_1, r_1)$. Now since $D_2$ and $B(x_1 r_1)$ are both open sets, there exists an $r_2 > 0$ with $r_2 < \frac{1}{2}$ such that:
(5)
\begin{align} \quad \overline{B}(x_2, r_2) \subseteq D_2 \cap B(x_1, r_1) \end{align}
• We continue in this process. Since $D_{n+1}$ is dense in $X$, we have that:
(6)
\begin{align} \quad D_{n+1} \cap B(x_n, r_n) \neq \emptyset \end{align}
• So let $x_{n+1} \in D_{n+1} \cap B(x_n, r_n)$. Now since $D_{n+1}$ and $B(x_n, r_n)$ are both open sets, there exists an $r_{n+1} > 0$ with $r_{n+1} < \frac{1}{n+1}$ such that:
(7)
\begin{align} \quad \overline{B}(x_{n+1}, r_{n+1}) \subseteq D_{n+1} \cap B(x_n, r_n) \end{align}
• So we obtain a sequence of points $(x_n)_{n=1}^{\infty}$ in $X$ with the properties that $(r_n)_{n=1}^{\infty}$ is a positive sequence of real numbers with $\displaystyle{\lim_{n \to \infty} r_n = 0}$ and such that for all $n \in \mathbb{N}$:
(8)
\begin{align} \quad \overline{B}(x_{n+1}, r_{n+1}) \subseteq D_{n+1} \cap B(x_n, r_n) \subseteq B(x_n, r_n) \end{align}
• So by the Cantor intersection theorem we have that:
(9)
\begin{align} \quad \bigcap_{n=1}^{\infty} \overline{B}(x_{n+1},r_{n+1}) \neq \emptyset \end{align}
• But by construction
(10)
\begin{align} \quad \emptyset \neq \bigcap_{n=1}^{\infty} \overline{B}(x_n, r_n) \subseteq \bigcap_{n=1}^{\infty} D_n \cap B(x_n, r_n) \subseteq \left ( \bigcap_{n=1}^{\infty} D_n \right ) \cap B(x^*, r^*) \end{align}
• So for every open ball $B(x^*, r^*)$ in $X$ we have that:
(11)
\begin{align} \quad \left ( \bigcap_{n=1}^{\infty} D_n \right ) \cap B(x^*, r^*) \neq \emptyset \end{align}
• Therefore $\displaystyle{\bigcap_{n=1}^{\infty} D_n}$ is dense in $X$.