Does weak topology preserve topological vector space structure?

Question

Consider a topological vector space $X$ (not necessarily normed or metrisable or anything). Its topological dual is a well defined set of linear maps. With this in place, the weak topology is defined as the coarsest topology on $X$ such that all of the dual still consists only of continuous linear forms. It it possible that the weak topology be so coarse that $X$ equipped with its weak topology is no longer a topological vector space ?

Answer 1

The weak topology that you speak about is defined by the seminorms (indexed by $f \in X^*$)

$$p_f : X \to [0, \infty), \ p_f (x) = |f(x)| .$$

In other words, a net $(x_i)_{i \in I}$ converges in this topology to $x$ if and only if

$$\lim \limits _{i \in I} |f (x_i) -f(x)| = 0 \; \forall f \in X^* .$$

This is the same topology as the one that you mention in your question, but describing it in terms of seminorms makes working with it much easier.

This topology is necessarily compatible with the algebraic structure of $X$, so the answer to your question is negative.