Equivalence of Definitions of Baire Space

From ProofWiki
Jump to navigation Jump to search

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.

$\blacksquare$