# The weak topology on an infinite dimensional linear space is not first-countable

I thought I needed help proving the above statement, but during typing I found a proof. Since I had already written it all down I will post it anyway, maybe in the future someone can benefit from it.

One question remains though: I’m not sure whether/where I have used the property that the space is complete in the proof. It was stated explicitly that the space has to be a Banach space, so it’s likely that I’ve missed something. I would appreciate if somebody could point it out to me.

#### Solutions Collecting From Web of "The weak topology on an infinite dimensional linear space is not first-countable"

I’m going to prove the statement by contradiction.

The assumption that the weak topology $\tau_w$ on the $\infty$-dimensional linear space X is first-countable, is equivalent to saying that each point in $X$ has a countable neighborhood basis. In particular for $0_X \in X$ there exists a N.B. $(U_n)_n\subset\tau_w$.
We define the sequence $(A_n)_n\subset\tau_w$ by$$A_n:=\bigcap_{i=1}^n U_i$$
obtaining another neighborhood basis of $0_X$, but this time decreasing.

It’s easy to prove that in a normed linear space, each weakly open neighborhood of $0$ contains a non-trivial closed subspace. So for every $n$, $A_n$ contains a non-trivial subspace $Y_n$ of X, and we can pick a vector $y_n\in V_n\setminus\{0_x\}$.

We then consider $x_n:=n\frac{v_n}{\lvert\lvert v_n\rvert\rvert}\in V_n$. Because of $\lvert\lvert x_n\rvert\rvert=n$, $(x_n)_n$ is an unbounded sequence, so it is not weakly convergent (this is also easy to show, prove the contrapositive statement by using the canonical embedding and the uniform boundedness theorem).

On the other hand, for any $\varepsilon>0$ and $\phi \in X^*$ the set $$U_{\varepsilon,\phi}=\{x\in X \vert\ \lvert \phi(x) \rvert<\varepsilon\}$$ is a neighborhood of $0_X$, and because $(A_n)_n$ is a decreasing neighborhood basis, there exists an $N \in \mathbb{N}$ such that $$\forall n\ge N: U_{\varepsilon,\phi}\supseteq A_n\supseteq\ V_n$$ In particular $\lvert\phi(x_n)\rvert<\varepsilon,$ since $x_n\in V_n$. Because $\varepsilon$ was picked arbitrarily, we get $\lvert \phi(x)\rvert\rightarrow0$. We also considered an arbitrary $\phi \in X^*$, and hence $x\overset{w}\rightarrow 0$. This is a contradiction to what we showed in the above paragraph.

This proves the statement.