# Equivalence of Definitions of Baire Space

## Contents

## Definition

Let $T = \struct {S, \tau}$ be a topological space.

The following definitions of the concept of **Baire Space** in the context of **Topology** are equivalent:

$T$ is a Baire space if and only if:

- $(1): \quad$ The intersection of any countable set of open sets of $T$ which are everywhere dense is everywhere dense.

- $(2): \quad$ The interior of the union of any countable set of closed sets of $T$ which are nowhere dense is empty.

- $(3): \quad$ Whenever the union of any countable set of closed sets of $T$ has an interior point, then one of those closed sets must have an interior point.

- $(4): \quad$ The union of any countable set of closed sets of $T$ whose interiors are empty also has an interior which is empty.

## Proof

First, let:

- $H^\circ$ denote the interior of any $H \subseteq S$
- $H^-$ denote the closure of any $H \subseteq S$.

### $(2) \iff (4)$

We have that a Closed Set Equals its Closure.

By definition, a subset $H$ is nowhere dense if and only if the interior of its closure is empty.

Hence we see that $(2)$ and $(4)$ are saying the same thing using different words.

$\Box$

### $(4) \iff (3)$

#### $(4) \implies (3)$

Let $T$ be a topological space such that:

- The union of any countable set of closed sets of $T$ whose interiors are empty also has an empty interior.

That is, let $(4)$ hold.

Let $\mathcal U$ be a countable set of closed sets of $T$.

Let $\displaystyle \bigcup \mathcal U$ be their union.

Suppose $\exists U \in \mathcal U$ such that $\displaystyle \exists x \in \paren {\bigcup \mathcal U}^\circ$.

That is, let $x$ be an interior point of $\displaystyle \bigcup \mathcal U$.

Then by hypothesis and the Rule of Transposition $\exists U \in \mathcal U: x \in U^\circ$.

That is, $x$ is an interior point of $U$.

That is, $(3)$ holds.

$\Box$

#### $(3) \implies (4)$

Let $T$ be a topological space such that:

- Whenever the union of any countable set of closed sets of $T$ has an interior point, then one of those closed sets must have an interior point.

That is, let $(3)$ hold.

Let $\mathcal U$ a countable set of closed sets of $T$.

Suppose that $\forall U \in \mathcal U: U^\circ = \O$.

Then by hypothesis and the Rule of Transposition $\not \exists x \in \displaystyle \bigcup \mathcal U$.

That is, $\displaystyle \bigcup \mathcal U = \O$.

That is, $(4)$ holds.

$\Box$

### $(4) \iff (1)$

#### $(4) \implies (1)$

Let $T$ be a topological space such that:

- The union of any countable set of closed sets of $T$ whose interiors are empty also has an empty interior.

That is, let $(4)$ hold.

Let $\mathcal U$ be a countable set of open sets of $T$ such that:

- $\forall U \in \mathcal U: U^- = S$

That is, all of $U$ are everywhere dense.

We have that:

\(\displaystyle U^-\) | \(=\) | \(\displaystyle S\) | |||||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle S \setminus U^-\) | \(=\) | \(\displaystyle \O\) | Relative Complement with Self is Empty Set | |||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle \paren {S \setminus U}^\circ\) | \(=\) | \(\displaystyle \O\) | Complement of Interior equals Closure of Complement |

That is, by definition, $S \setminus U$ is nowhere dense.

By definition of closed set we have that $S \setminus U$ is closed.

Now consider $\displaystyle \bigcup_{U \mathop \in \mathcal U} \paren {S \setminus U}$.

We have that:

\(\displaystyle \paren {\bigcup_{U \mathop \in \mathcal U} \paren {S \setminus U} }^\circ\) | \(=\) | \(\displaystyle \O\) | as $T$ satisfies condition $(4)$ | ||||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle \paren {S \setminus \bigcap \mathcal U}^\circ\) | \(=\) | \(\displaystyle \O\) | De Morgan's Laws: Difference with Intersection | |||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle S \setminus \paren {\paren {\bigcap \mathcal U}^-}\) | \(=\) | \(\displaystyle \O\) | Complement of Interior equals Closure of Complement | |||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle \paren {\bigcap \mathcal U}^-\) | \(=\) | \(\displaystyle S\) | Relative Complement of Empty Set |

That is, by definition, $\bigcap \mathcal U$ is everywhere dense.

That is, $(1)$ holds.

$\Box$

#### $(1) \implies (4)$

Let $T$ be a topological space such that:

- The intersection of any countable set of open sets of $T$ which are everywhere dense is everywhere dense.

That is, let $(1)$ hold.

Let $\mathcal V$ be a countable set of closed sets of $T$ such that:

- $\forall V \in \mathcal V: V^\circ = \O$

Then:

\(\displaystyle V^\circ\) | \(=\) | \(\displaystyle \O\) | |||||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle S \setminus V^\circ\) | \(=\) | \(\displaystyle S\) | Relative Complement of Empty Set | |||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle \paren {S \setminus V}^-\) | \(=\) | \(\displaystyle S\) | Complement of Interior equals Closure of Complement |

That is, by definition, $S \setminus V$ is an open set of $T$ which is everywhere dense.

Now consider $\displaystyle \bigcap_{V \mathop \in \mathcal V} \paren {S \setminus V}$.

We have that:

\(\displaystyle \paren {\bigcap_{V \mathop \in \mathcal V} \paren {S \setminus V} }^-\) | \(=\) | \(\displaystyle S\) | as $T$ satisfies condition $(1)$ | ||||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle \paren {S \setminus \bigcup \mathcal V}^-\) | \(=\) | \(\displaystyle S\) | De Morgan's Laws: Difference with Union | |||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle S \setminus \paren {\paren {\bigcup \mathcal V}^\circ}\) | \(=\) | \(\displaystyle S\) | Complement of Interior equals Closure of Complement | |||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle \paren {\bigcup \mathcal V}^\circ\) | \(=\) | \(\displaystyle \O\) | Relative Complement with Self is Empty Set |

That is, $(4)$ holds.

$\Box$

All conditions have been shown to be equivalent.

$\blacksquare$