Is U+V Open in a Topological Vector Space?

[r4nd0m]

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 $$f^{-1}(V)$$ 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?

 

[jostpuur]

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

$$u+V = \{u+v\;|\;v\in V\}$$

is open, then you are almost done.

 

[tacman]

Hi there,

I can understand why you might be stuck on this problem. It can be tricky to apply general topological concepts to specific situations. However, there is a simpler way to prove that U+V is open in this case.

First, we know that the addition operation in a topological vector space is continuous. This means that for any open set W in X, the preimage of W under addition, denoted as U+V, is also open.

Now, let's consider the set U+V. We can express it as the union of all possible sums of elements from U and V. In other words, for any x in U+V, there exists u in U and v in V such that x = u+v. This means that x is a sum of two open sets, and since open sets are closed under addition, x must also be in an open set.

Therefore, we can conclude that U+V is a union of open sets, and hence it is open. This proves that the addition operation preserves openness, and hence U+V is open in X.

I hope this helps in understanding the problem better. Keep practicing and you'll get the hang of it!