Equivalence of Definitions of Baire Space

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

$T$ is a Baire space :


 * $(1): \quad$ 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 the interior of its closure is empty.

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

$(4) \implies (3)$
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.

$(3) \implies (4)$
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.

$(4) \implies (1)$
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:

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:

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

That is, $(1)$ holds.

$(1) \implies (4)$
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:

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:

That is, $(4)$ holds.

All conditions have been shown to be equivalent.