Countable Union of Meager Sets is Meager

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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 $X$.


There is believed to be a mistake here, possibly a typo.
In particular: Should this not invoke Countable Union of Countable Sets is Countable I see no cartesian products?
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So $A$ is meager in $X$.

$\blacksquare$


Axiom of Countable Choice

This theorem depends on the Axiom of Countable Choice, by way of Countable Union of Countable Sets is Countable.

Although not as strong as the Axiom of Choice, the Axiom of Countable Choice is similarly independent of the Zermelo-Fraenkel axioms.

As such, mathematicians are generally convinced of its truth and believe that it should be generally accepted.


Sources

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