Weakly open sets are unbounded

Question

I was studying the proof that every open set in the weak topology of an infinite dimensional space is unbounded and I came across the following argument. If $(f_1,...,f_n)$ are continuous linear operators defined on an infinite normed space E than the set $\cap_{i=1}^{n}kerf_i$ has a point $x\neq0$. The one who used this argument said that otherwise the map defined by $x \in E \mapsto(f_1(x),...,f_n(x)) \subset \mathbb{R}^n$ would be an injection. What I don't understand is why the proposed map can't be an injection, how does this set up a contradiction?

Answer 1

The proposed map can't be an injection because we can't inject an infinite-dimensional space into a finite-dimensional space. In particular, injective maps preserve linear independence, so mapping $n + 1$ linearly independent vectors from $E$ (which is possible to find, since $E$ is infinite-dimensional) will produce $n + 1$ vectors in $\Bbb{R}^n$, which cannot be linearly independent, contradicting injectivity.

Now, every weakly open set contains a basic weakly open set, i.e. one of the form: $$\mathcal{U} = \{x \in E : |f_i(x) - \alpha_i| < \varepsilon_i \text { for } i = 1, \ldots, n\}.$$ Let $K$ be the previously considered kernel. Note that, if $y \in K$, and $x \in \mathcal{U}$, then $x + y \in \mathcal{U}$. But $K$ is a non-trivial subspace, and hence is unbounded, and so we can make $x + y$ as large as we like, thus proving $\mathcal{U}$ (and hence the weakly open set) is unbounded too.