If $X = \bigcup_{n=1}^{\infty} F_n$ and each $F_n$ is a closed set, then $\bigcup_{n=1}^{\infty} F_n^{\circ}$ is dense in $X$. Show that for a metric space $X$, the following statements are equivalent:
a) X is a Baire Space.
b) Every nonempty open set in $X$ is a set of the second category.
c) If $X = \bigcup_{n=1}^{\infty} F_n$ and each $F_n$ is a closed set, then $\bigcup_{n=1}^{\infty} F_n^{\circ}$ is dense in $X$.
I am able to prove that a) implies b). But Unable to prove the other directions b) implies c) and c) implies a).
Help Needed. Thank You.
 A: $(a)\Longleftrightarrow(b)$ is quite straight forward.
I will show $(a)\Longleftrightarrow(c)$
$\underline{(a)\implies (c)}$
Clearly, $\overline{\bigcup_{n=1}^\infty F_n^\circ}\subseteq X$. For the reverse containment let, $$x\in X\setminus\overline{\bigcup_{n=1}^\infty F_n^\circ}\implies x\notin\overline{\bigcup_{n=1}^\infty F_n^\circ}$$ So there exists an open set $V$ of $X$ containig $x$ such that $V\cap{\bigcup_{n=1}^\infty F_n^\circ}=\emptyset$, that is $V\cap F_n^\circ=\emptyset$ for all $n$.
Now $V$ is a non empty open subset of the Baire space $X$ and hence is also Baire. Further, $V\subseteq \bigcup_{n=1}^\infty (F_n\cap V)$ each of which is closed in $V$ and have empty interior since $V\cap F_n^\circ=\emptyset$. This contrdicts the Baireness of $V$.
$\underline{(c)\implies(a)}$
Let $\{A_n\}$ be a countable collection of closed sets of $X$ with empty interior in $X$. To prove that $\left(\bigcup A_n\right)^\circ=\emptyset$.
Suppose $x\in\left(\bigcup A_n\right)^\circ\implies$ there is an open set $V$ of $X$ such that $x\in V\subseteq\left(\bigcup A_n\right)$.
Now we can write $X$ as $X=(\cup A_n)\bigcup(X\setminus V)$. By $(c)$ we have that $$X=\overline{(\cup A_n^\circ)\bigcup(X\setminus V)^\circ}=\overline{(X\setminus V)^\circ}\implies x\in\overline{(X\setminus V)^\circ}$$
By definition of closure we have that $V\cap(X\setminus V)^\circ\neq\emptyset$ which is clearly impossible.
Thus $\left(\bigcup A_n\right)^\circ=\emptyset$.
