Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 5 down vote favorite
1

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 $0\in U$.

share|improve this question
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 at 20:21
2  
@JyrkiLahtonen $2^{-n}<\alpha$ does not immediately imply $2^{-n}U\subseteq \alpha U$. – Hagen von Eitzen Feb 20 at 20:23
It is like @HagenvonEitzen said, and that's my problem in showing the statement. – Tomás Feb 20 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 at 21:44
Figured out how to fix that. See my answer for the gory details. – Jyrki Lahtonen Feb 20 at 22:29

2 Answers

up vote 1 down vote accepted

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.

share|improve this answer
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 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 at 7:32
up vote 5 down vote

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 $n$.

share|improve this answer
+1, but why is $0$ in the intersection? A set $U$ containing a single non-zero vector is bounded, no? – Jyrki Lahtonen Feb 20 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 at 22:48
@JyrkiLahtonenm you are right, I forgot to add the hypothesis that $0\in U$. I will add it – Tomás Feb 20 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 at 14:07

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.