Baire's property iff first category has dense complement. Show that $(S, d)$ has Baire's property iff every set of first category has a dense complement. 

A set is of first category if it is a countable union of nowhere dense sets. First Category
Baire's Lemma: Let $(X, \rho)$ be a complete metric space and $\{U_n\}_{n=1}^{\infty}$ a sequence of open dense sets in $X$. Then the set $\bigcap_{n =1}^{\infty} U_n$ is also dense.
Note that A is nowhere dense iff $(\overline{A})^c$ is open and dense. Suppose $\forall A$, $A$ is first category. That is, $A$ is a countable union of nowhere dense sets. i.e., $A_i$ is nowhere dense. $\forall A, \; A = \bigcup_{n \in \mathbb{N}}A_i \implies (A)^c = \bigcap_{n \in \mathbb{N}}(A_{i})^{c}$. But we don't know if this is a countable dense set! We know that $(\overline{A}_i)^c$ would be, but not just $(A)^c$.

If $A$ is nowhere dense, what can we say about the compliment of $A$? 
Can anyone clear this up or provide advice?

More scratch work:
In my proposed proof, I said let all $A \subset S$ be of first category. i.e., each $A_i$ is nowhere dense, i.e, $(\overline{A}_i)^c$ is open and dense. So, $\forall A, (\overline{A}_i)^c = \bigcup (\overline{A}_i)^c \implies ((\overline{A}_i)^c)^c = \bigcap (((\overline{A}_i)^c)^c) = \bigcap (\overline{A}_i)$. Still though, what can I say about $\overline{A}_i$?

A nowhere dense set has empty interior. Therefore, the complement of a nowhere dense set is dense. Hence, I think my original solution is fine for one direction. If anyone else has comments to add, please do so. 
