Equivalence of Definitions of Baire Space

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.

$\blacksquare$