Meagre sets have empty interior-A false statement.

Question

Suppose $A$ is a meagre set i.e. a set of first category.Then $A=\large\cup_{n\in \mathbb N} $$A_n$,where each $A_n$ is nowhere dense.Now $A^o=(\large \cup_n A_n)^o\subset\large \cup_n A_n^o=\phi$ as $A_n^o\subset (\bar A_n)^o=\phi$.Thus we are done.Is this proof correct? Then we can also prove Baire Category that complete metric spaces are not meagre.By showing that if $X$ is meagre,then $X^o=\phi$,which is a contradiction,but where do we require completeness,am I making mistake?Is this not true $(\large \cup_n A_n)^o\subset\large \cup_n A_n^o$.

Answer 1

If $A_1$ is the set of all rational numbers, $A_2$ is the set of all irrational numbers, and $A_n$ is the empty set for $n>2$ the interrior of $\cup_n A_n $ is the entire real line whereas $A_n^{0} $ is empty for each $n$.