$ \overline{A}+ \overline{B} \subseteq \overline{A+B}$ in Topological vector space

Question

The problem written below was asked in my assignment of Topological vector spaces and I am not able to make progress in this particular problem.

Let $X$ be a topological vector space. Show that if $A,B \subseteq X$, then $\overline{A}+ \overline{B} \subseteq \overline{A+B}$.

I tried by assuming an $x \in \overline{A}+ \overline{B}$ in the hope of showing that $x$ lies in right hand side. Then x can be written as equal to $a+b $ where $ a\in \overline {A}$ and $b\in \overline {B}$.

But I am not able to see how can I move in the direction of assertion required from this.

I have studied properties like absorbing , balanced and 5-6 basic results regarding them but I don't think I can use them in the problem.

Can you please outline a proof for me here? I really want to learn how to solve the set theory related problems in Topological Vector Spaces.

Answer 1

In a topological vector space $X$, vector addition $f \colon X \times X \to X , \ (x,y) \mapsto f(x,y)=x+y \ $ is continuous ($ X \times X$ is endowed with product topology), then $$\overline{A} + \overline{B}=f(\overline{A} \times \overline{ B})=f(\overline{A \times B}) \underbrace{\subset}_{\text{$f$ is continuous}} \overline{f(A \times B)}=\overline{A+B}$$

Answer 2

Let $a\in\bar{A}$ and $b\in\bar{B}$. Choose a net $(a_{\alpha})$ in $A$ and a net $(b_{\beta})$ in $B$ such that $a_{\alpha}\rightarrow a$ and $b_{\beta}\rightarrow b$. Here, we try to make everything as precise as possible. Let $(D,\leq)$ and $(E,\preceq)$ be directed systems underlie the nets. That is, the net $(a_{\alpha})$, denoted by $\theta$, is a map $\theta:D\rightarrow A$ (we denote $\theta(\alpha)=a_{\alpha}$ for $\alpha\in D$) such that for any open neighborhood $U$ of $a$, there exists $\alpha_{0}\in D$ such that $a_{\alpha}\in U$ whenever $\alpha_{0}\leq\alpha$. Similarly, denote the net $(b_{\beta})$ by $\phi:E\rightarrow B$.

Now, we define a relation $<$ on $D\times E$: For any $(\alpha_{1},\beta_{1}),(\alpha_{2},\beta_{2})\in D\times E$, $(\alpha_{1},\beta_{1})<(\alpha_{2},\beta_{2})$ iff $\alpha_{1}\leq\alpha_{2}$ and $\beta_{1}\preceq\beta_{2}$. It is routine to check that the following axioms for a directed system are satisfied:

(1) (reflexive) For any $x\in D\times E$, $x<x$,

(2) (transitive) For any $x,y,z\in D\times E$, if $x<y$ and $y<z$, then $x<z$.

(3) (dominating) For any $x,y\in D\times E$, there exists $z\in D\times E$ such that $x<z$ and $y<z$.

Hence, $(D\times E,<)$ is a directed system. Now, consider a net $\eta:D\times E\rightarrow A+B$ defined by $\eta(\alpha,\beta)=\theta(\alpha)+\phi(\beta):=a_{\alpha}+b_{\beta}$. We go to show that $\eta\rightarrow a+b$ (the net $\eta$ converges to $a+b$) and it will follow that $a+b\in\overline{A+B}$.

Let $W$ be an open neighborhood of $a+b$, then there exist open neighborhoods $U$ for $a$ and $V$ for $b$ such that $U+V\subseteq W$ (because the map $(s,t)\mapsto s+t$ is continuous at $(a,b)$). Since $\theta\rightarrow a$ and $\phi\rightarrow b$, there exist $\alpha_{0}\in D$ and $\beta_{0}\in E$ such that $\theta(\alpha)\in U$ and $\phi(\beta)\in V$ whenever $\alpha_{0}\leq\alpha$ and $\beta_{0}\preceq\beta$, $\alpha\in D$ and $\beta\in E$. Let $x_{0}=(\alpha_{0},\beta_{0})\in D\times E$. Let $x=(\alpha,\beta)\in D\times E$. If $x_{0}<x$, then $\alpha_{0}\leq\alpha$ and $\beta_{0}\preceq\beta$. Hence, $\theta(\alpha)\in U$ and $\phi(\beta)\in V$. It follows that $\eta(x)=\theta(\alpha)+\phi(\beta)\in U+V\subseteq W$. Therefore $\eta\rightarrow a+b$.

Answer 3

Here's another approach.

Let $\mathcal{O}$ be an open set containing $\alpha+\beta$ where $\alpha\in \overline{A}$ and $\beta\in \overline{B}$.

Notice $\mathcal{O}-\beta$ is an open set containing $\alpha\in \overline{A}$ and so $(\mathcal{O}-\beta)\cap A\neq \emptyset$. Find $\mathcal{o}\in \mathcal{O}$ and $a\in A$ such that $\mathcal{o}-\beta=a$. But then $\mathcal{O}-a$ becomes an open set containing $\beta\in \overline{B}$ and so $(\mathcal{O}-a)\cap B\neq \emptyset $. So find $\tilde{\mathcal{o}}\in \mathcal{O}$ and $b\in B$ such that $\tilde{\mathcal{o}}-a=b$. It follows that $\tilde{\mathcal{o}}=a+b\in A+B$ making $\mathcal{O}\cap (A+B)\neq \emptyset$.

This shows $\alpha + \beta\in \overline{A+B}$ and the result follows.