5

Let $E$ be a Topological Vector Space and $U$ a bounded set of $E$ with $0\in U$, i.e. given any neighborhood $W$ of the origin, there exist $\alpha>0$ such that $\alpha U\subset W$. Is it true that $$\bigcap_{n=1}^\infty2^{-n}U=\{0\}$$

Note: I have added the hypothesis $0\in U$.

Tomás
  • 22,559
  • Aren't topological vector spaces Hausdorff? If $v\neq0$ there is a neighborhood of $W$ of the origin such that $v\notin W$. So $v\notin 2^{-n}U$ when $2^{-n}<\alpha$? – Jyrki Lahtonen Feb 20 '13 at 20:21
  • 2
    @JyrkiLahtonen $2^{-n}<\alpha$ does not immediately imply $2^{-n}U\subseteq \alpha U$. – Hagen von Eitzen Feb 20 '13 at 20:23
  • It is like @HagenvonEitzen said, and that's my problem in showing the statement. – Tomás Feb 20 '13 at 20:26
  • Yeah. For some strange reason I assumed $W$ to be convex in which case $\alpha U\subseteq W$ and $2^{-n}<\alpha$ do imply $2^{-n}U\subseteq W$. Never thought that $2^{-n}U$ should be contained in $\alpha U$. But for this to work we need a convex basis (or at least a basis at the origin such that each set contains the line segments from the origin to all of its points). – Jyrki Lahtonen Feb 20 '13 at 21:44
  • Figured out how to fix that. See my answer for the gory details. – Jyrki Lahtonen Feb 20 '13 at 22:29

2 Answers2

5

I assume $E$ is Hausdorff (not all authors require that). Then for any $v\ne 0$, the subspace topology on $v\mathbb R$ is just the standard topology. We observe that $U\cap v\mathbb R$ is bounded in the usual sense and hence $v\notin 2^{-n}U$ for suitable $n$.

Hagen von Eitzen
  • 374,180
  • +1, but why is $0$ in the intersection? A set $U$ containing a single non-zero vector is bounded, no? – Jyrki Lahtonen Feb 20 '13 at 21:46
  • @Lahtonen: Hagen’s argument is correct. In fact, it allows the possibility that $ \displaystyle \bigcap_{n \in \mathbb{N}} 2^{-n} \cdot U = \varnothing $. For example, consider the bounded subinterval $ U := (0,1) $ of $ \mathbb{R} $. Observe that $ \displaystyle \bigcap_{n \in \mathbb{N}} 2^{-n} \cdot U = \varnothing $. However, in general, if $ 0_{V} \in U $, then clearly $ \displaystyle \bigcap_{n \in \mathbb{N}} 2^{-n} \cdot U = { 0_{V} } $. – Haskell Curry Feb 20 '13 at 22:48
  • @JyrkiLahtonenm you are right, I forgot to add the hypothesis that $0\in U$. I will add it – Tomás Feb 20 '13 at 22:57
  • 1
    Sorry if this is obvious but can you explain why the subspace topology has to be the standard topology? I am not very familiar with tvs, beyond the definitions. – Julien Feb 21 '13 at 14:07
1

Nothing wrong with Hagen's solution, I just wanted to complete my suggestion to a full solution starting from first principles :-).

I assume that $E$ is Hausdorff. I argue that this implies that the intersection of those scaled versions of $U$ cannot contain a non-zero vector. Let $v\in E, v\neq0$. Then there exists an open neighborhood $V_1$ of the origin with the property that $v\notin V_1$. By continuity of the scalar multiplication there exists another neighborhood $V_2$ of the origin and an open interval $I_a=(-a,a),a>0,$ such that for all $x\in V_2, \alpha\in I_a$ we have $\alpha x\in V_1$. Let $V_3=(a/2)V_2$. Multiplication by $a/2$ is a homeomorphism, so $V_3$ is an open neighborhood of the origin. I claim that for all $t\in(0,1]$ we have $tV_3\subseteq V_1$. So let $t\in(0,1]$ and $x\in V_3$ be arbitrary. We have $x=(a/2)y$ for some $y\in V_2$. Then $tx=(ta/2) y\in V_1,$ because $(y,ta/2)\in V_2\times I_a$.

Now let $$ W=\bigcup_{t\in(0,1]} tV_3.$$ This is an open neighborhood of the origin as a union of open sets. We just checked that $W\subseteq V_1$, so $v\notin W$. Furthermore $W$ has the property that $tW\subseteq W$ for all $t\in(0,1]$. Finally we are in a position to attack the main claim. By boundedness of $U$, we have $\alpha U\subseteq W$ for some $\alpha>0$. If $2^{-n}<\alpha$, then $t=2^{-n}/\alpha\in(0,1]$, and therefore $$ 2^{-n}U=t\alpha U\subseteq tW\subseteq W. $$ Hence $v\notin 2^{-n}U$, and the claim follows.

Jyrki Lahtonen
  • 133,153
  • Interesting proof. You have proved that every topological vector space, has a local base of balanced neighborhoods and then you proved the statement for balanced neighborhoods. Thank you. – Tomás Feb 20 '13 at 23:01
  • @Tomás: Indeed. I didn't have my copy of Rudin at home, so I wasn't 100% certain that this is what balanced meant - and it's been a while since I really studied it anyway :-) – Jyrki Lahtonen Feb 21 '13 at 07:32