$T_i(y_0)=0\:\:\:\forall i$ on the weak topology

Question

Let $E$ be a normed vector space over $\mathbb{K}$, of infinite dimension.

Let $x_0\in E$ and consider:

$$\mathscr{V}=V(x_0;T_1,T_2,...,T_n;\epsilon)$$ an open neighbourhood of $x_0$ on the weak topology $\sigma(E,E*)$ in $E$. Prove there exists $y_0\in E\setminus{0}$ such that $T_i(y_0)=0\:\:\:\forall i\in\{1,2,...,n\}$.

I tried to approach the question on the following way:

$n\in\mathbb{N}$ such that $x_0+\frac{1}{n}$ then:

$\exists N\in\mathbb{N},\text{for}|T_i(x_0)-T_i(x_0+\frac{1}{n})|<\epsilon\:\:\forall i$

$n\to\infty\implies|T_i(x_0)-T_i(x_0)|=0$ but this does not work, since I would end up with $x_0$ again.

It was then suggested to me to consider the following function:$\psi:E\to \mathbb{K}^n$ such that $\psi(x)=(T_1(x),...,T_n(x))$

However I am stuck and I do not see how to proceed.

Question:

How should I solve this problem?

Thanks in advance!

Answer 1

Consider the $\psi(x)$ you defined. It linearly maps $E$ into $\Bbb K^n$ so the $\dim(\ker(\psi))$ is infinite. So we can pick a non-zero vector $y_0$ in it.

(here we use $$\dim(\ker(A)) + \dim(\operatorname{Im}(A)) = \dim(E)$$

by a standard fact on linear maps and $\dim(\operatorname{Im}(\psi)) \le n$ and $\dim(E)\ge \aleph_0$. )