Closure on topological vector space

Question

I came across a problem on topological vector spaces that seems to be simple yet it bothers me: Let $A,B$ be non empty subsets of the t.v.s $X$ with $\overline{A}$ compact. Then show that $$\overline{A+B}=\overline{A}+\overline{B}$$

Ok using that $f$ (let $f$ for addition $f:X\times X\to X$) is continuous on a t.v.s we get $$\overline{A}+\overline{B}=f(\overline{A}\times \overline{B})=f(\overline{A\times B})\subseteq \overline{f(A\times B)}=\overline{A+B}.$$

For the other inclusion can't i use that if $z\in \overline{A+B}$ then it belongs to every closed set containing $A+B$ . More specifically it belongs to the intersection of the family $$\overline{A}+ M$$ where $M$ closed, $B\subseteq M$. I guess something is wrong as i didn't use the compactness assumption... Any hints would be great.

Answer 1

Let $z\in \overline{A+B}$. Then, there is some net $\langle z_{\alpha}\rangle_{\alpha\in \mathcal{A}}\subset A+B$ such that $z_{\alpha}\to z$. We write $z_{\alpha} = x_{\alpha}+y_{\alpha}$ for $x_{\alpha}\in A$ and $y_{\alpha}\in B$. Then, as $\overline{A}$ is compact, $\langle x_{\alpha}\rangle_{\alpha\in \mathcal{A}}$ has a convergent subnet $\langle x_{\beta}^1\rangle_{\beta\in \mathcal{B}}$. We let $h : \mathcal{B}\to \mathcal{A}$ be a monotone final function such that $x_{\beta}^1 = x_{h(\beta)}$, and we define $y_{\beta}^1 = y_{h(\beta)}$ and $z_{\beta}^1 = z_{h(\beta)}$. We let $x_{\beta}^1\to x$. Then, $y_{\beta}^1\to y = z-x$ by the continuity of vector addition and scalar multiplication and the fact that $x_{\beta}^1+y_{\beta}^1 = z_{\beta}^1\to z$. Clearly, as $x\in \overline{A}$ and $y\in \overline{B}$, $z = x+y\in \overline{A}+\overline{B}$.

Your proof that $\overline{A}+\overline{B}\subseteq \overline{A+B}$ is correct. For consistency, we can also write it using the language of nets. For $z\in \overline{A}+\overline{B}$, $z = x+y$ for some $x\in \overline{A}$ and $y\in \overline{B}$. Then, $x_{\alpha}\to x$ and $y_{\beta}\to y$ for some nets $\langle x_{\alpha}\rangle_{\alpha\in \mathcal{A}}$ and $\langle y_{\beta}\rangle_{\beta\in \mathcal{B}}$. We define the directed set $\mathcal{A}\times \mathcal{B}$ with the product order. Then, consider the net $\langle x_{\alpha}+y_{\beta}\rangle_{(\alpha, \beta)\in \mathcal{A}\times \mathcal{B}}\subset A+B$. By the continuity of vector addition, we will have $x_{\alpha}+y_{\beta}\to x+y = z$ (as the nets $\langle x_{\alpha}\rangle_{(\alpha, \beta)\in \mathcal{A}\times \mathcal{B}}$ and $\langle y_{\beta}\rangle_{(\alpha, \beta)\in \mathcal{A}\times \mathcal{B}}$ converge to $x$ and $y$ respectively), so $z\in \overline{A+B}$.

Answer 2

(1). If $c\in Cl(A+B)$ there is a sequence $(c_n)_n$ in $A+B$ converging to $c.$ Let $c_n=a_n+b_n$ with $a_n\in A$ and $b_n\in B.$ Since $Cl(A)$ is compact and $(a_n)_n$ is a sequence in $Cl(A),$ there is a sub-sequence $(a_{n_j})_j$ converging to some $a\in Cl(A),$ so $(b_{n_j})_j =(c_{n_j}-a_{n_j})_j$ converges to $c-a.$ Obviously if $(b_{n_j})_j$ converges, then it converges to some to some $b\in Cl(B).$ So $c=a+b\in Cl(A)+Cl(B).$

(2). If $d\in Cl(A)+Cl(B)$ there exist $a\in Cl(A)$ and $b\in Cl(B)$ with $d=a+b. $ There are sequences $(a_n)_n, (b_n)_n$ in $A,B$ respectively, converging to $a,b$ respectively. The sequence $(a_n+b_n)_n$ in $A+B$ converges to $a+b= d.$ So $d\in Cl(A+B).$