Every weak neighborhood of $0$ contains a finite co-dimensional subspace

36 Views Asked by At

I'm trying to prove a statement mentioned in this answer, i.e.,

In an infinite dimensional normed space $(E, |\cdot|)$, every weak neighborhood of $0$ contains a finite co-dimensional subspace.

Could you please have a check on my below attempt?


Let $X$ be a neighborhood of $0\in E$ in the weak topology $\sigma(E, E^*)$. By construction of $\sigma(E, E^*)$, there are $f_1, \ldots, f_n \in E^*$ and $\varepsilon_1, \ldots, \varepsilon_n >0$ such that $$ X \supseteq U:=\{x\in E \mid \forall k = 1, \ldots,n: |\langle f_k, x \rangle| < \varepsilon_k\}. $$

Consider the continuous linear map $$ T:E \to \mathbb R^n, x \mapsto \big ( \langle f_1, x \rangle, \ldots, \langle f_n, x \rangle \big ). $$

By first isomorphism theorem for rings, $E/ \ker T \cong R(T)$ where $R(T)$ is the range of $T$. On the other hand, $\dim R(T) \le n$ and $$ \ker T = \bigcap_{k=1}^n \ker f_k \subset U. $$

The claim then follows.