The finite intersection of absorbing set is absorbing

Question

I got stuck proving the finite intersection of absorbing set is absorbing. Can anyone help?

Answer 1

What's the picture here? Take any point and connect it to $0$. Then a set is absorbing provided a small neighborhood of $0$ on that ray intersects the set, for every point in the vector space. So this means that any two absorbing sets will always intersect a little bit on every ray, which means that their intersection will always be absorbing.

Let's make this intuition precise. Suppose $S$ and $T$ are two absorbing sets in $V$, which is an $\Bbb{R}$-vector space. Take $x\in V$. Then we know there are $r_S$ (resp. $r_T$) such that for all $a\in\Bbb{R}$ with $|a|>r_S$ (resp. $a>r_T$) we have $x\in aS$ (resp. $x\in aT$).

This means that $x/a\in S$ for all $a > r_S$ and $x/a\in T$ for all $a > r_T$. Thus $x/a\in S\cap T$ for all $a > \max\{r_S,r_T\}$. Therefore, if we let $r = \max\{r_S,r_T\}$ then we will have for all $a > r$, that $x\in a(S\cap T)$. Note that $r_S,r_T$, and thus $r$ will depend on $x$, but that's okay -- this is a challenge-response definition, where for every $x$ we have to be able to provide an $r$ such that the criterion is satisfied. But we've done so, which means we've finished the proof.