Baire Space iff Open Sets are Non-Meager
Theorem
Let $T = \struct {S, \tau}$ be a topological space.
Then $T$ is a Baire space if and only if every non-empty open set of $T$ is non-meager in $T$.
Proof
We prove the contrapositive:
- $T$ is not a Baire space if and only if there exists a non-empty open set of $T$ which is meager in $T$.
We have by definition of Baire space:
- $T$ is a Baire space if and only if the interior of the union of any countable set of closed sets of $T$ which are nowhere dense is empty.
Therefore:
- $T$ is not a Baire space if and only if the interior of the union of some countable set of closed sets of $T$ which are nowhere dense is non-empty.
We have by definition of meager:
- $A$ is meager in $T$ if and only if it is a countable union of subsets of $S$ which are nowhere dense in $T$.
Here $I$ denotes a countable indexing set.
Sufficient Condition
Suppose $T$ is not a Baire space.
Then there is a countable set of closed sets of $T$ which are nowhere dense with the interior of its union non-empty.
Let $\FF = \set {F_i: i \in I}$ be such a set.
Then $\paren {\bigcup \FF}^\circ$ is a non-empty open set.
Consider the set $\FF' = \set {F_i \cap \paren {\bigcup \FF}^\circ: i \in I}$.
From Intersection is Subset:
- $\forall i \in I: F_i \cap \paren {\bigcup \FF}^\circ \subseteq F_i$
Since $F_i$ are nowhere dense, so is $F_i \cap \paren {\bigcup \FF}^\circ$.
We have:
\(\ds \bigcup \FF'\) | \(=\) | \(\ds \bigcup_{i \mathop \in I} \paren {F_i \cap \paren {\bigcup \FF}^\circ}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\bigcup \FF}^\circ \cap \bigcup_{i \mathop \in I} F_i\) | Intersection Distributes over Union of Family | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\bigcup \FF}^\circ \cap \bigcup \FF\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\bigcup \FF}^\circ\) | Set Interior is Largest Open Set; Intersection with Subset is Subset |
Therefore $\bigcup \FF'$ is meager in $T$.
Hence we have shown the existence of a non-empty open set of $T$ which is meager in $T$.
$\Box$
Necessary Condition
Suppose there exists a non-empty open set of $T$ which is meager in $T$.
Let $\ds A = \bigcup_{i \mathop \in I} A_i$ be such a set.
Then $A$ is a non-empty open set, and $A_i$ are nowhere dense for every $i \in I$.
Thus for every $i$, $\paren {A_i^-}^\circ = \O$.
By Closure of Topological Closure equals Closure:
- $\paren {\paren {A_i^-}^-}^\circ = \paren {A_i^-}^\circ = \O$.
By Set is Closed iff Equals Topological Closure, $A_i^-$ are closed sets of $T$ which are nowhere dense.
We also have:
\(\ds A\) | \(=\) | \(\ds A^\circ\) | Interior of Open Set | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\bigcup_{i \mathop \in I} A_i}^\circ\) | ||||||||||||
\(\ds \) | \(\subseteq\) | \(\ds \paren {\bigcup_{i \mathop \in I} A_i^-}^\circ\) | Set is Subset of its Topological Closure; Set Union Preserves Subsets; Interior of Subset |
Hence there is a countable set of closed sets of $T$ which are nowhere dense with the interior of its union non-empty.
Therefore $T$ is not a Baire space.
$\blacksquare$
Historical Note
This result was the original definition which René-Louis Baire gave for the Baire space.
The more modern approach is to define it directly in terms of interiors of countable unions of closed sets.