# If $X$ is infinite dimensional, all open sets in the $\sigma(X,X^{\ast})$ topology are unbounded.

As in the title, if $X$ is infinite dimensional, all open sets in the $\sigma(X,X^{\ast})$ topology are unbounded. The $\sigma(X,X^{\ast})$ topology is the weakest topology that makes linear functionals on $X^\ast$ continuous. How does one show this? How does having an infinite basis relate to open sets being unbounded? I can't see this, please help and thanks in advance!

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Hint: The basic open neighborhoods of $0$ are of the form $\bigcap_{i=1}^n \{x : \lvert \varphi_i(x)\rvert \lt \varepsilon_i\}$, so such a neighborhood contains the subspace $\bigcap_{i=1}^n \ker{\varphi_i}$. – commenter Oct 29 '12 at 3:02

It's enough to show it for basic non-empty open sets which contain $0$ (for the others, do a translation). These ones are of the form $$V_{N,\delta,f_1,\dots,f_N}=\bigcap_{j=1}^N\{x\in X, |f_j(x)|<\delta\},$$ where $N$ is an integer, $f_j\in X^*$ and $\delta>0$, $1\leq j\leq N$. Then $$\bigcap_{j=1}^N\ker f_j\subset V_{N,\delta,f_1,\dots,f_N}.$$ As $X$ is infinite dimensional, $\bigcap_{j=1}^N\ker f_j$ is not reduced to $0$ (otherwise the map $x\in X\mapsto (f_1(x),\dots,f_N(x))\in\Bbb R^n$ would be injective). So it contains a non-zero vector $x_0$, and $\lambda x_0$ for all scalar $\lambda$, proving that $V_{N,\delta,f_1,\dots,f_N}$ is not bounded.
Presumably $X$ is assumed to be a normed space and to be bounded means "bounded in norm". In topological vector spaces one calls a set $A$ bounded if it is completely absorbed by every zero neighborhood $U$, that is: there is $t \gt 0$ such that $A \subset tU$. Then it is not very interesting to prove that weakly open subsets are in general not bounded in this sense. More interesting is the fact that the uniform boundedness principle shows that one can deduce that "norm bounded" and "bounded in the weak topology" are one and the same notion for a normed space. – commenter Oct 29 '12 at 11:01
@MattN. :-)${}{}$ – commenter Oct 29 '12 at 11:29