Existence of balanced neighborhoods in a topological vector space

Let$\ X$ be a topological vector space. Then one could pick balanced neighborhoods$\ W$ and$\ U$ of$\ 0$ such that

$\ \overline{U} + \overline{U} \subset W$, where $\ U+U:=\{u_1 + u_2 | u_1,u_2 \in U \}$

I was faced this question while reading Rudin's "Functional Analysis". I'm able to prove this, but I think there should be a more elegant and easier way to do it.
Listed are the properties I used:

• using continuity of$\ +$ one can easily show that there exist $\ U, W$ as above such that $\ U+U \subset W$
• Use that every topological vector space has a balanced local base (local means here at$\ 0$.)
• If $\ \mathcal{B}$ is a local base (in the above sense) for a topological vector space$\ X$ then every member of$\ \mathcal{B}$ contains the closure of some member of$\ \mathcal{B}$.
• and the last property I used was:$\ \overline{U_1} + \overline{U_2} \subset \overline{U_1+U_2}$ where$\ U_1,U_2 \subset X$

As you can see I need a lot of theory / basic properties about topological vector spaces and I'm just wondering if there's not an easier way. Thx for suggestions.

cheers

math

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I don't think you can get it in a much cheaper way. Also, I would say that this is more a fact about topological groups, rather than about topological vector spaces, see e.g. Hewitt-Ross, Abstract Harmonic Analysis, Corollary 4.7, and the results preceeding and following it. – t.b. Sep 18 '11 at 11:25
@Theo Buehler: Thx for the link. You're right about the tag. Topological groups would be more general and TVS just an application of it. The reason for my choice suggests itself as I copied from Rudins chapter TVS. Again Thx for your commment! – math Sep 18 '11 at 11:47
Also observe that the result you ask about directly implies the first three bullet points at the end of your question, so some manipulations will have to enter the argument. – t.b. Sep 18 '11 at 12:07