Proof Verification: C closed, convex, symmetric in Banach space X and $\cup_{n \in N \setminus 0} n.C= X$ then $B_\epsilon(0) \in C $.

Question

I have an outline of the proof of this which I've expanded (correctly or otherwise) below, I'd appreciate feedback on it. (I think that C has to be closed in order to assert that $\cup_{n \in N \setminus 0} n.C= X $, but I'm not sure about that)

$\cup_{n \in N \setminus 0} n.C= X $, and from the Baire category theorem, X is not the countable union of nowhere dense sets, so that at least one of the n.C must contain some open set. Since open sets are preserved under scaling, then C too must contain an open set, and therefore within it some open ball $B_\epsilon(x) \in C $.

Any vector $z \in B_\epsilon(-x) \subset X$ satisfies $||z – (-x)|| = ||-z - x|| < \epsilon$. So $-z \in B_\epsilon(x) \subset C$.

Since C is symmetric then $z = -(-z) \in C$ and therefore $B_\epsilon(-x) \subset C$.

If $w \in B_\epsilon(0)$ then $||w|| < \epsilon $

Let $z_1 = w + x$, so that $z_1 - x = w$, then $||z_1 – x|| < \epsilon$ and so $z_1 \in B_\epsilon(x)$. Similarly, $z_2 \in B_\epsilon(-x)$

But $w = z_1 + (z_2 – z_1)/2$ and so by the convexity of $C$, $w \in C$ and therefore $B_\epsilon(0) \subset C$.

Answer 1

In order to invoke Baire you have to assume the closedness of $C$. Otherwise your proof is fine.