Baire’s_Theorem

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Definition

The set of real numbers $\mathbb{R}$ cannot be written as the countable union of nowhere-dense sets.

Equivalently, if $\{O_n : n \in \mathbb{N}\}$ is a countable collection of open, dense sets, then the intersection $\bigcap_{n=1}^{\infty} O_n$ is not empty. In fact, it is dense in $\mathbb{R}$.

Proof

Let $\{O_n\}$ be a countable collection of open, dense sets. We show $\bigcap O_n$ is dense.

Let $(a,b)$ be any open interval. Since $O_1$ is dense, $O_1 \cap (a,b) \neq \emptyset$. Choose a closed interval $I_1 \subseteq O_1 \cap (a,b)$.

Since $O_2$ is dense and open, $O_2 \cap \text{int}(I_1) \neq \emptyset$. Choose $I_2 \subseteq O_2 \cap I_1$.

Continuing, we get nested closed intervals $I_1 \supseteq I_2 \supseteq \cdots$ with $I_n \subseteq O_n$.

By the Nested_Interval_Property, $\bigcap I_n \neq \emptyset$, and any point in this intersection lies in $\bigcap O_n \cap (a,b)$.