Countable Union of Meager Sets is Meager

Theorem
Let $\struct {X, \tau}$ be a topological space.

Let $\family {A_n}_{n \in \N}$ be a countable set of meager subsets of $X$.

Let:
 * $\ds A = \bigcup_{n \mathop = 1}^\infty A_n$

Then $A$ is meager in $X$.

Proof
For each $n \in \N$, $A_n$ is meager and hence there exists a countable set $\family {A_{n, m} }_{m \in \N}$ of nowhere dense sets in $X$ such that:
 * $\ds A_n = \bigcup_{m \mathop = 1}^\infty A_{n, m}$

Hence:
 * $\ds A = \bigcup_{n \mathop = 1}^\infty \bigcup_{m \mathop = 1}^\infty A_{n, m} = \bigcup_{\tuple {n, m} \in \N^2} A_{n, m}$

From Cartesian Product of Countable Sets is Countable, $A$ is therefore the countable union of nowhere dense sets in $X$.

So $A$ is meager in $X$.