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

[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.

 

[jgens]

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.

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.

 

[danzibr]

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.

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.

 

[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.

 

[Fredrik]

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

Use ## instead of $. PF LaTeX Guide.

 

[danzibr]

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.

Thanks, got it!

I actually did something a bit more abstract. Not much though. It goes like this:

Claim: Let $(X,\tau)$ be a locally bounded topological vector space.
If $U\subseteq X$ is such that there exists $0\neq x\in U$ satisfying
for all $n\in\mathbb{N}$ there exists $c_n\geq n$ for which $c_nx\in U$
then $U$ is not topologically bounded.

In other words, any topologically bounded set cannot contain
$cx$ for arbitrarily large $c$ where $x$ is a nonzero vector.

Proof: By way of contradiction suppose $U$ is topologically bounded
but there is such an $x$ as in the statement of the claim.
Let $N$ be a neighborhood of the origin. Since $U$ is topologically bounded
there exists a scalar $s$ so that $U\subseteq sN$, or $cx\in sN$
for arbitrarily large $c$, or $cx\in N$ for arbitrary large $c$.

Since $(X,\tau)$ is Hausdorff (recall the convention stating we deal with Hausdorff tvs's),
there exist disjoint neighborhoods of
$x$ and the origin. By the proposition stating every
tvs has a fundamental system of balanced, absorbing closed sets, the neighborhood of
the origin contains a neighborhood of the origin which is balanced, call it
$N_0$. By the preceding paragraph
$N_0$ contains $cx$ for arbitrarily large $c$. As $N_0$ is balanced
there holds $\frac{1}{c}N_0\subseteq N_0$ for $c\geq1$, hence $x\in N_0$,
but $N_0$ is contained in a set disjoint from a neighborhood of $x$ and
so cannot contain $x$, a contradiction.

Use ## instead of $. PF LaTeX Guide.

Thanks!