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.
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=== $(4) \iff (3)$ ===
==== $(4) \implies (3)$ ====
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 $\UU$ be a countable set of closed sets of $T$.
Let $\ds \bigcup \UU$ be their union.
Suppose $\exists U \in \UU$ such that $\ds \exists x \in \paren {\bigcup \UU}^\circ$.
That is, let $x$ be an interior point of $\ds \bigcup \UU$.
Then by hypothesis and the Rule of Transposition $\exists U \in \UU: x \in U^\circ$.
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 $\UU$ a countable set of closed sets of $T$.
Suppose that $\forall U \in \UU: U^\circ = \O$.
Then by hypothesis and the Rule of Transposition $\not \exists x \in \ds \bigcup \UU$.
That is, $\ds \bigcup \UU = \O$.
That is, $(4)$ holds.
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=== $(4) \iff (1)$ ===
==== $(4) \implies (1)$ ====
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 $\UU$ be a countable set of open sets of $T$ such that:
:$\forall U \in \UU: U^- = S$
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 $\ds \bigcup_{U \mathop \in \UU} \paren {S \setminus U}$.
We have that:
That is, by definition, $\bigcap \UU$ 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 $\VV$ be a countable set of closed sets of $T$ such that:
:$\forall V \in \VV: V^\circ = \O$
Then:
That is, by definition, $S \setminus V$ is an open set of $T$ which is everywhere dense.
Now consider $\ds \bigcap_{V \mathop \in \VV} \paren {S \setminus V}$.
We have that:
That is, $(4)$ holds.
All conditions have been shown to be equivalent.
Category:Baire Spaces