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I'm wondering about the following:

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](http://books.google.com/books?id=uf11K1wXEYUC&pg=PA19), and the results preceeding and following it.2011-09-18
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    @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!2011-09-18
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    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.2011-09-18

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