Is U+V Open in a Topological Vector Space?

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SUMMARY

In a topological vector space X, the sum of two open sets U and V, denoted U+V, is proven to be open. This conclusion arises from the continuity of the addition operation in topological vector spaces. Specifically, if V is an open set, then for any vector u in X, the set u+V, defined as {u+v | v ∈ V}, is also open. This result confirms that the sum of open sets retains the property of openness in the context of topological vector spaces.

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I'm really stuck on this simple problem: Let X be a topological vector space and U, V are open sets in X. Prove that U+V is open.

It should be a direct consequence of the continuity of addition in topological vector spaces. But continuity states that the [tex]f^{-1}(V)[/tex] is open whenever V is open, but not the converse. It would work if I showed that adding a constant is a homeomorphism, but I don't think this is the way I should do it. Is there any more simple way, that I overlooked?
 
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In fact it is sufficient to assume the other one of the sets to be open. Suppose V is open. If you succeed in proving that for all vectors u the set

[tex] u+V = \{u+v\;|\;v\in V\}[/tex]

is open, then you are almost done.
 

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