Equivalence of Definitions of Baire Space

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Definition

Let $T = \left({S, \tau}\right)$ 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 $\displaystyle \exists x \in \left({\bigcup \mathcal U}\right)^\circ$.

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 $\forall U \in \mathcal U: U^\circ = \varnothing$.

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

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

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:

\(\displaystyle U^-\) \(=\) \(\displaystyle S\) $\quad$ $\quad$
\(\displaystyle \iff \ \ \) \(\displaystyle S \setminus U^-\) \(=\) \(\displaystyle \varnothing\) $\quad$ Relative Complement with Self is Empty Set $\quad$
\(\displaystyle \iff \ \ \) \(\displaystyle \left({S \setminus U}\right)^\circ\) \(=\) \(\displaystyle \varnothing\) $\quad$ Complement of Interior equals Closure of Complement $\quad$

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 $\displaystyle \bigcup_{U \in \mathcal U}\left({S \setminus U}\right)$.

We have that:

\(\displaystyle \left({\bigcup_{U \in \mathcal U}\left({S \setminus U}\right)}\right)^\circ\) \(=\) \(\displaystyle \varnothing\) $\quad$ as $T$ satisfies condition $(4)$ $\quad$
\(\displaystyle \iff \ \ \) \(\displaystyle \left({S \setminus \bigcap \mathcal U}\right)^\circ\) \(=\) \(\displaystyle \varnothing\) $\quad$ De Morgan's Laws: Difference with Intersection $\quad$
\(\displaystyle \iff \ \ \) \(\displaystyle S \setminus \left({\left({\bigcap \mathcal U}\right)^-}\right)\) \(=\) \(\displaystyle \varnothing\) $\quad$ Complement of Interior equals Closure of Complement $\quad$
\(\displaystyle \iff \ \ \) \(\displaystyle \left({\bigcap \mathcal U}\right)^-\) \(=\) \(\displaystyle S\) $\quad$ Relative Complement of Empty Set $\quad$

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:

$\forall V \in \mathcal V: V^\circ = \varnothing$


Then:

\(\displaystyle V^\circ\) \(=\) \(\displaystyle \varnothing\) $\quad$ $\quad$
\(\displaystyle \iff \ \ \) \(\displaystyle S \setminus V^\circ\) \(=\) \(\displaystyle S\) $\quad$ Relative Complement of Empty Set $\quad$
\(\displaystyle \iff \ \ \) \(\displaystyle \left({S \setminus V}\right)^-\) \(=\) \(\displaystyle S\) $\quad$ Complement of Interior equals Closure of Complement $\quad$

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


Now consider $\displaystyle \bigcap_{V \in \mathcal V}\left({S \setminus V}\right)$.

We have that:

\(\displaystyle \left({\bigcap_{V \in \mathcal V}\left({S \setminus V}\right)}\right)^-\) \(=\) \(\displaystyle S\) $\quad$ as $T$ satisfies condition $(1)$ $\quad$
\(\displaystyle \iff \ \ \) \(\displaystyle \left({S \setminus \bigcup \mathcal V}\right)^-\) \(=\) \(\displaystyle S\) $\quad$ De Morgan's Laws: Difference with Union $\quad$
\(\displaystyle \iff \ \ \) \(\displaystyle S \setminus \left({\left({\bigcup \mathcal V}\right)^\circ}\right)\) \(=\) \(\displaystyle S\) $\quad$ Complement of Interior equals Closure of Complement $\quad$
\(\displaystyle \iff \ \ \) \(\displaystyle \left({\bigcup \mathcal V}\right)^\circ\) \(=\) \(\displaystyle \varnothing\) $\quad$ Relative Complement with Self is Empty Set $\quad$


That is, $(1)$ holds.

$\Box$


All conditions have been shown to be equivalent.

$\blacksquare$