Let $E$ be a vector space with a non countable number of elements. Define $$\mathscr{F}=\{U \subset E:0 \in U \hbox{ and } \#E\setminus U<\infty\}.$$ It's easy to see that $\mathscr{F}$ defines a filter on $E$.
I have already proven that:
- $0 \in U$ for all $U \in \mathscr{F}$;
- There exists $U \in \mathscr{F}$ such that $V+V \not\subset U$ for all $V \in \mathscr{F}$;
- For every $U \in \mathscr{F}$ and for all $\lambda \in \mathbb{C}$, $\lambda \neq 0$ we have $\lambda U \in \mathscr{F}$
My question: I would like to know if every $U \in \mathscr{F}$ is absorbing and if every $U \in \mathscr{F}$ contains some $V \in \mathscr{F}$ which is balanced.
I tried to work with the set $U=E\setminus\{p\}$ where $p \neq 0$, but I was not able to give a counter example.
Let $E$ be a vector space over $\Bbb C$. Let $\Bbb T=\{z\in \mathbb{C} :|z|=1\}$ be the unit circle.
If $V\subset E$ is a balanced set and $p\not\in V$, then $\Bbb T\cdot\{p\}\cap V=\varnothing$. Since $p\ne 0$, a set $\Bbb T\cdot\{p\}$ is infinite, so a set $E\setminus\{p\}\in\mathcal F$ contains no balanced set of $\mathcal F$.
Any set $U\in\mathcal F$ is absorbing. Indeed, for any element $x\in E$ a set $\Lambda_x=\{\lambda\in\Bbb C:\lambda x\not\in U\}$ is finite and does not contains zero. It follows that $x\in\alpha U$ for each $\alpha\in \mathbb{C}$ with $|\alpha|>\max \{|\lambda|^{-1}:\lambda\in\Lambda_x\}$. Indeed, if $x\not\in \alpha U$ then $\alpha^{-1}x\not\in U$, so $\alpha^{-1}\in \Lambda_x$. Thus $|\alpha^{-1}|\ge \min \{|\lambda|:\lambda\in\Lambda_x\}$, that is $|\alpha|\le \max \{|\lambda|^{-1}:\lambda\in\Lambda_x\}$, a contradiction.