# 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:

- $F_i \cap \paren {\bigcup \FF}^\circ \subseteq F_i$ for any $i \in I$

Since $F_i$ are nowhere dense, so is $F_i \cap \paren {\bigcup \FF}^\circ$.

We have:

\(\displaystyle \bigcup \FF'\) | \(=\) | \(\displaystyle \bigcup_{i \mathop \in I} \paren {F_i \cap \paren {\bigcup \FF}^\circ}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {\bigcup \FF}^\circ \cap \bigcup_{i \mathop \in I} F_i\) | Intersection Distributes over Union | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {\bigcup \FF}^\circ \cap \bigcup \FF\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {\bigcup \FF}^\circ\) | 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 $\displaystyle A = \bigcup_{i \in I} A_i$ 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:

\(\displaystyle A\) | \(=\) | \(\displaystyle A^\circ\) | Interior of Open Set | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {\bigcup_{i \in I} A_i}^\circ\) | |||||||||||

\(\displaystyle \) | \(\subseteq\) | \(\displaystyle \paren {\bigcup_{i \in I} A_i^-}^\circ\) | 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.