Register to reply

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

by danzibr
Tags: bounded, topologically
Share this thread:
danzibr
#1
Jan10-13, 10:54 AM
P: 9
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.
Phys.Org News Partner Science news on Phys.org
Experts defend operational earthquake forecasting, counter critiques
EU urged to convert TV frequencies to mobile broadband
Sierra Nevada freshwater runoff could drop 26 percent by 2100
jgens
#2
Jan10-13, 11:18 AM
P: 1,622
Quote Quote by danzibr View Post
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
#3
Jan10-13, 12:06 PM
P: 9
Quote Quote by jgens View Post
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
#4
Jan10-13, 12:10 PM
P: 1,622
Can a topologically bounded set in a tvs contain a ray?

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
#5
Jan10-13, 04:14 PM
Emeritus
Sci Advisor
PF Gold
Fredrik's Avatar
P: 9,420
Quote Quote by danzibr View Post
Sorry about the poor format. I don't see how to make the forum recognize my tex.
Use ## instead of $. PF LaTeX Guide.
danzibr
#6
Jan12-13, 09:32 AM
P: 9
Quote Quote by jgens View Post
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.
Quote Quote by Fredrik View Post
Use ## instead of $. PF LaTeX Guide.
Thanks!


Register to reply

Related Discussions
Proof of analytic function being bounded inside the region if bounded on the boundary Calculus & Beyond Homework 6
Topologically complete space in the product topology Calculus & Beyond Homework 12
Are these two shapes topologically same Differential Geometry 10
Topologically Conjugate Calculus & Beyond Homework 0
Topologically seperated Linear & Abstract Algebra 0