Prove that in a Banach space every nonempty open set is of the second category

Question

I am new in learning such topics. What I got is:

Let S be a nonempty open set in a Banach space X.\

Then $\exists E_i $ countable closed nowhere dense sets, s.t. $S = \cup E_i$.

Then $V_i = E_i^C$ will be dense set and $\cap V_i$ is dense (Baire Theorem) since $V_i \subset X$.

Then I lost from here.

Also, I am a little confused here. $\Bbb R$ is a Banach space. $S=\{\frac{1}{n} : n \in \Bbb N \}$ is nonempty, but $S$ is a nowhere dense set. How can it be category II?

Answer 1

Let $S \subset X$ be a nonempty open subset of $X$. Then $S$ contains a closed ball $B = \overline{B} \subset S$. Then $B$ is a closed subset of a complete metric space, and hence complete. Therefore $B$ is of the second category.

Therefore $S$ can't be of the first category, for if it were, so would be $B$.

Answer 2

If $S$ is a nonempty open set, let $x$ be an element of $S$. Then $S-x=\{y-x:y\in S\}$ is an open set containing $0$, and $\bigcup\limits_{n=1}^\infty n(S-x)=X$. Translation and scaling preserve meagerness, and a countable union of meager sets is meager, so this implies that $S$ is of second category.

---

In your example you are lacking "open".