# Show that no topological vector space is bounded.

I am studying the concept of topological vector spaces in Grubb's Distributions and Operators.

A vector space $X$ (over $\mathbb{L} = \mathbb{R}$ or $\mathbb{C}$) is called a topological vector space if $X$ is equipped with a topology $\tau$ such that

1. $\{x\} \subset X$ is closed for all $x \in X$.

2. $X \times \mathbb{L} \to X, (x, \lambda) \mapsto \lambda x$ is continuous.

3. $X \times X \to X, (x,y) \mapsto x+y$ is continuous.

Here, $\mathbb{L}$ ($= \mathbb{R}$ or $\mathbb{C}$) has the usual topology, and both $X \times \mathbb{L}$ and $X \times X$ are considered to have the product topology.

I am familiar with the fact that imposing properties 1 through 3 ensures that $\tau$ is a Hausdorff topology.

Next, we say that $C \subseteq X$ is balanced if $x \in C$ and $|\lambda| \le 1$ implies $\lambda x \in C$. Furthermore, $C \subseteq X$ is bounded if for each open set $U$ containing $0$, there is a $t > 0$ so that $C \subset tU$. I do know that an equivalent way of defining $C$ bounded is that for all $U \ni 0$ there is a $t >0$ so that $C \subset sU$ all $s \ge t$.

I am also familiar with the fact that, for each open neighborhood $U$ of $0$, there exists balanced open $V \subset U$ such that $0 \in V$ (one uses this fact to show that the two definitions of a bounded set are equivalent).

I would like to show that every nontrivial vector subspace of $X$ is not bounded.

This is a problem given in Appendix B of Grubb, and I've had some trouble solving it. I'm hoping that I can solve it with the facts I've stated above, but perhaps I need something else.

Certainly, it suffices to take $x \in X \setminus \{0\}$ and show that $S= \operatorname{span}\{x\}$ is not bounded. Intuitively, this makes since because elements of the form $\lambda x$ should get large (in some sense) as $|\lambda| \to \infty$. But I am not able to make this precise yet. I need to find some specific open neighborhood $\tilde{U}$ of 0, and show there is no $t > 0$ such that $S \subseteq t\tilde{U}$. But it seems that $\tilde{U}$ might be tricky to identify.

Hints or solutions are greatly appreciated.

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We don't have the condition $(\lambda_1, (\lambda_2, x)) = (\lambda_1 \lambda_2, x)$? –  nayrb Jun 7 '14 at 17:32
@nayrb - We do indeed have that condition. Although, I think this property holds because it is part of the definition of a vector space. So that property would hold even if $X$ didn't have any topological structure. –  jtms88 Jun 7 '14 at 17:35
gotcha. I see now the point of the additional conditions... –  nayrb Jun 7 '14 at 17:42

Let $X$ be a topological vector space and $Y$ a non-trivial subspace. Take any $y\in Y\setminus\left\{0\right\}$. Then $X\setminus\left\{y\right\}$ is a neighbourhood of $0$, but for any $t\in L$, the space $Y$ is not contained in $t(X\setminus\left\{y\right\})=X\setminus\left\{ty\right\}$.

So, we've found a neighbourhood $U(=X\setminus\left\{y\right\})$ of $0$ such that there does not exist $t\in L$ with $Y\subseteq tU$, and that means that $Y$ is not bounded.

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Must a subspace be open? –  nayrb Jun 7 '14 at 17:31
I think this solution checks out, as the only open set that is being utilized is $X \setminus \{ty\}$. And this set is certainly open since we are assuming one point sets are closed. –  jtms88 Jun 7 '14 at 17:46
Yes, my bad. Realized all this. Good answer! +1 –  nayrb Jun 7 '14 at 17:46

Let me give you a "high level" solution: It is well known (maybe this is also a statement in your book), that any finite dimensional (Hausdorff) topological vector space is automatically equipped with the "usual" topology (I.e. induced by an arbitrary norm), see e.g. here (http://terrytao.wordpress.com/2011/05/24/locally-compact-topological-vector-spaces/)

Using this it is then easy to see that $S$ is not bounded in the relative topology and hence also not bounded as a subset of $X$.

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