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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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.