Special case of Baire's Theorem - Rudin ex 2.30 The question I have is regarding exercise 2.30 from Rudin's principles of mathematical analysis:

Prove the following result: If $\mathbb{R}^k = \cup_{n=1}^\infty F_n$ where each $F_n$ is a closed subset of $\mathbb{R}^k$, then at least one $F_n$ has a nonempty interior. Equivalent statement: If $G_n$ is a dense open subset of $\mathbb{R}^k$ for $n = 1,2,3,..,$ then $\cap_{n=1}^\infty G_n$ is not empty (in fact, it is dense in $\mathbb{R}^k$).

In the book, the statements are said to be equivalent with no further explanation but unfortunately I could not find a straightforward proof of this fact. The most obvious thing to do seemed to assume either result, and take complements, but for instance; given a sequence of closed sets as in the hypothesis, taking their complement does not appear guarantee me a sequence of dense open sets (if every closed set $F_n = \mathbb{R}^k$ for instance). Similarly, given a sequence of dense open subsets of $\mathbb{R}^k$, taking their complement does not appear to guarantee me a sequence of closed subsets whose union will be $\mathbb{R}^k$. If someone could explain how the statements are equivalent it would be a great help.
On the other hand, I think I do have a proof of the result using the statement of dense open subsets and would appreciate comments on the completeness of my argument. Let $G_n$ be a dense open subset of $\mathbb{R}^k$ for $n = 1,2,...$ and take an arbitrary point $x$ of $G_1$ to start.
$G_1$ is open, and so it has a compact neighbourhood $V_1$ contained in $G_1$. If this point is in $G_2$, we can look at $G_3$, otherwise it is a limit point of $G_2$, and so $V_1$ contains a point $y$ of $G_2$. By taking a neighbourhood small enough around $y$, we can obtain a compact neighbourhood $V_2$ such that $V_2 \subset V_1$. Proceeding in this manner we obtain a sequence of compact neighbourhoods $V_1 \supset V_2 \supset...$ and it is given in a theorem of the chapter that the intersection of such a sequence is nonempty. This intersection is a subset of $\cap_{n=1}^\infty G_n$ and proves the result.
Now, consider the intersection of $G_1$ with an arbitrary nonempty open set $O \subset \mathbb{R}^k$. If this intersection is empty, then there is an interior point of $O$, with a neighbourhood that has empty intersection with $G_1$. This point is not in $G_1$ and cannot be a limit point of $G_1$ contradicting $G_1$ being dense. Therefore the intersection is nonempty.
Using this; consider an arbitrary point $z$ in $\mathbb{R}^k$. If $z$ is not in $\cap_{n=1}^\infty G_n$, every neighbourhood of $z$, has nonempty intersection with $G_1$ and we can let an arbitrary point in that nonempty intersection be our starting point $x$ above. By choosing a sufficiently small $V_1$, we can argue that this neighbourhood of $z$ has nonempty intersection with $\cap_{n=1}^\infty G_n$. It follows that $z$ is either in $\cap_{n=1}^\infty G_n$ or $z$ is a limit point of $\cap_{n=1}^\infty G_n$, and so $\cap_{n=1}^\infty G_n$ is dense in $\mathbb{R}^k$.
 A: Your proof looks fine, though you could add more detail in the last paragraph about why a sufficiently small $V_1$ intersects $\bigcap G_n$.
On to your actual question, observe that there is a difference in the style of the two statements you're trying to say are equivalent. The first is "if $\bigcup F_n$ has some property then some $F_n$ has some other property," whereas the second is "if each $G_n$ has some property then $\bigcap G_n$ has some other property." These two are going in opposite directions so it's harder to see the equivalence, but this can be rectified with contraposition. Indeed, the contrapositive of the first statement is "if $F_n$ are closed and no $F_n$ has nonempty interior then $\bigcup F_n \neq \mathbb R^k$." Equivalently, this says that if $F_n$ are closed and nowhere dense then their union $\bigcup \mathbb F_n$ is not equal to $\mathbb R^k$. This is in the same form as the statement about open dense sets, and from here your idea of taking complements will work.
Indeed, suppose that this property of nowhere dense closed sets holds, and let $G_n$ be a collection of open dense sets. Then $F_n = \mathbb R^k - G_n$ is a collection of nowhere dense closed sets, so $\bigcup F_n \neq \mathbb R^k$. That is, $\mathbb R^k - (\bigcup F_n) \neq \emptyset$. The left hand side here is just $\bigcap G_n$, so we have shown that $\bigcap G_n \neq \emptyset$. The other direction is the same idea, which I'll leave to you to verify.
