Can a topologically bounded set in a tvs contain a ray?

1. Jan 10, 2013

danzibr

Pretty much what the title says.

Suppose we have a topological vector space $(X,\tau)$ and $U\subseteq X$ is topologically bounded. Is it possible for there to be some $x\in X$ such that $cx\in U$ for arbitrarily large $c$? I'm thinking of a real vector space here.

If we try to prove this BWOC, suppose $U$ is topologically bounded but contains such an $x$. Right now I've gotten to the point where every neighborhood of the origin has to contain $cx$ for arbitrarily large $c$. This seems silly but... I see no contradiction.

How about if we add $(X,\tau)$ is topologically bounded? Or if that's not sufficient, what else should we add?

Sorry about the poor format. I don't see how to make the forum recognize my tex.

2. Jan 10, 2013

jgens

Assuming you take the Hausdorff separation axiom, that should give you a pretty immediate contradiction.

Edit: I want to expand upon my previous answer. Some people are crazy and do not require TVSs to be Hausdorff. In this case you can give your space the trivial topology and the continuity of addition and scalar multiplication is immediate. Then the whole space itself is bounded.

Last edited: Jan 10, 2013
3. Jan 10, 2013

danzibr

I see how taking $(X,\tau)$ to be be Hausdorff can result in some silly things, but I'm still not seeing any contradiction.

In this general setting we have 3 things to work with: our set $U$ being topologically bounded, $+$ being continuous and $\cdot$ being continuous. What would stop there being some nonzero $x\in X$ so that $cx$ is contained in every neighborhood of the origin for arbitrarily large $c$? It seems unreasonable but I just don't see what it contradicts.

I'm a total novice at this tvs stuff by the way, so I'm probably just missing something simple.

4. Jan 10, 2013

jgens

Suppose a ray were bounded. Then every neighborhood of the identity necessarily contains this ray. Now fix a non-zero point on the ray and use the Hausdorff separation axiom to get a neighborhood of zero not containing this point. But this is a contradiction.

5. Jan 10, 2013

Fredrik

Staff Emeritus
Thanks!