How to prove in a topological vector space: cl(A) + cl(B) is a subset of cl(A+B), where cl denotes closure?

Question

I'm not sure where to really proceed. My process is as follows. Take any $x \in cl(A)+cl(B)$. Assume for a contradiction that $x \notin cl(A+B)$. Then there exists an open set $U$ such that $ x \in U$, and $U \cap (A+B) = \emptyset$. So for all $u \in U$ and $b \in B$ we have $u-b \notin A$. Since addition is continuous, there exists $U'$ open such that $u-b \in U'$ and $U' \cap A = \emptyset$. But then, we have that $X \setminus U$ is closed, and hence $u-b \notin cl(A)$. Am I on the right track, or is there a nicer solution that I've missed entirely?

Answer 1

You could write $\text{Cl}A+\text{Cl}B$ as the image of $\text{Cl}A\times \text{Cl}B=\text{Cl}(A\times B)$ under the addition map $+$. But $+$ is continuous, so $+(\text{Cl}(A\times B))\subseteq \text{Cl}(+(A\times B))=\text{Cl}(A+B)$.

You would still need to prove that $\text{Cl}A\times\text{Cl}B=\text{Cl}(A\times B)$, though. This isn't too difficult, and I think there are solutions to this on MSE. Note that this also holds for arbitrary products.

Answer 2

One could use nets as well (if you know them; they're pretty popular in functional analysis as a generalisation of sequences that work in all topological spaces): Take any $z \in \operatorname{cl}(A) + \operatorname{cl}(B)$, so $z= x + y$ with $x \in \operatorname{cl}(A)$, $y \in \operatorname{cl}(B)$, so we have a nets $a_i, b_i$, with a common index set $I$ wlog, such that all $a_i \in A$, all $b_i \in B$ and $a_i \rightarrow x$ and $b_i \rightarrow y$, so by continuity of addition (which is the crucial property we need) we get that the net $a_i+ b_i$ is in $A + B$ and converges to $x +y = z$, showing that $z \in \operatorname{cl}(A+B)$.