📂Topology

Proof of the Baire Category Theorem

Definition

A topological space $X$ is called a Baire space if $\displaystyle \bigcap_{n=1}^{\infty} O_{n}$ is dense for every sequence $\left\{ O_{n} \right\}_{n=1}^{\infty}$ of dense open sets.

Baire Category Theorem¹

Every complete metric space is a Baire space.

Proof

Claim: For every open set $U \subset X$, we have $\displaystyle U \cap \left( \bigcap_{n=1}^{\infty} O_{n} \right) \ne \emptyset$.

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Part 1.

Since $X$ is a metric space, an open set can be written as follows for some $x^{ \ast } \in X$ and $r^{ \ast } > 0$. $$ U = B (x^{ \ast },r^{ \ast }) $$

Since $O_{1}$ is a dense set in $X$, for arbitrary $x_{0} \in X$ and $r_{0} > 0$, there exists $$ x_{1} \in \left( B(x_{0},r_{0}) \cap O_{1} \right) $$ Since $B(x_{0},r_{0})$ and $O_{1}$ are open sets, there exist $x_{1} \in O_{1}$ and $0< r_{1} < 1$ satisfying $$ B[x_{1} , r_{1} ] \subset \left( B(x_{0}, r_{0}) \cap O_{1} \right) $$

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Part 2.

Likewise, since $O_{2}$ is a dense set in $X$, there exists $$ x_{2} \in \left( B(x_{1},r_{1}) \cap O_{2} \right) $$ Since $B(x_{1},r_{1})$ and $O_{2}$ are open sets, there exist $x_{2} \in O_{2}$ and $\displaystyle 0< r_{2} < {{1} \over {2}}$ satisfying $$ B[x_{2} , r_{2} ] \subset \left( B(x_{1}, r_{1}) \cap O_{2} \right) $$

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Part 3.

Repeating this, we can keep choosing $x_{n} \in O_{n}$ and $\displaystyle 0 < r_{n} < {{1} \over {n}}$ satisfying $$ B[x_{n}, r_{n}] \subset \left( B(x_{n-1}, r_{n-1}) \cap O_{n} \right) $$ Since $X$ is a complete space, the Cauchy sequence $\left\{ x_{n} \right\}$ converges to some $x \in B[x_{n},r_{n}] $. For all $n \in \mathbb{N}$, $$ x \in B[x_{n},r_{n}] \subset B(x_{0},r_{0}) $$ and $x \in O_{n}$, so the following holds. $$ B(x_{0},r_{0}) \cap \left( \bigcap_{n=1}^{\infty} O_{n} \right) \ne \emptyset $$

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Explanation

The Baire category theorem is useful for determining the cardinality of a set or as a lemma in fields such as functional analysis.