2

If $x\notin A+B$, then there is a neighborhood $V$ of $0$ such that $(x+V)\cap (A+B)=\emptyset$. But I am stuck in this step.

I have seen some solutions concerning about $\mathbb R^n$, but I don't know how to apply it to topological vector space which is more general.

Knt
  • 1,649
  • The argument for $\mathbb R^{n}$ works here too if you replace sequences by nets. – Kavi Rama Murthy Nov 18 '19 at 08:39
  • Sorry, I am not familiar with nets yet. – Knt Nov 18 '19 at 08:42
  • I am not sure if there is an elegant way to show this without using nets. Probably, you should start to get familiar with nets. If your topological vector space fulfills the first axiom of countability, then I guess you can work with sequences instead of nets. – Nathanael Skrepek Nov 18 '19 at 13:17

1 Answers1

3

Assume that $x\in\overline{A+B}$ but $x\notin A+B$, then $(-A+x)\cap B=\emptyset$. So $-A+x\subseteq B^{c}$ and note that $B^{c}$ is open and $-A+x$ is compact.

Assume at the moment that there is some neighborhood $V$ of $0$ such that $(-A+x)+V\subseteq B^{c}$, then $((-A+x)+V)\cap B=\emptyset$.

Note that $((-A+x)+V)\cap B=(x+V)\cap(A+B)$.

On the other hand, since $x\in\overline{A+B}$, we must have $(x+V)\cap(A+B)\ne\emptyset$, so we have a contradiction.

Now it is left to show the fact assumed.

We are to show that, for compact set $K$ and open set $G$ such that $K\subseteq G$, there is a neighborhood $V$ of $0$ such that $K+V\subseteq G$.

For every $x\in K$, there is a neighborhood $V_{x}$ of $0$ such that $x+V_{x}\subseteq G$. Since the addition is continuous, there is a neighborhood $W_{x}$ of $0$ such that $W_{x}+W_{x}\subseteq V_{x}$.

So $\{x+W_{x}\}_{x\in K}$ is an open cover for $K$ and hence there is some finite subcover $x_{i}+W_{i}$, $i=1,...,n$ for $K$. Let $V=\displaystyle\bigcap_{i=1}^{n}W_{i}$, then $V$ is also a neighborhood of $0$.

For $u\in K+V$, then $u=x+v$ for some $x\in K$ and $v\in V$, and we have some $i$ such that $x\in x_{i}+W_{i}$ and hence $u\in x_{i}+W_{i}+V\subseteq x_{i}+W_{i}+W_{i}\subseteq x_{i}+V_{x_{i}}\subseteq G$, this shows that $K+V\subseteq G$.

user284331
  • 55,591