While I am studying FUNCTIONAL ANALYSIS by Walter Rudin, I found the following corollary.
Now, I wonder how the dual space(the set of all continuous linear functionals on $X$) of some pathetic topological vector spaces can't separate the points. Since a topological vector space (in this context) should be Hausdorff, if $X^*$ doesn't separate $x\neq y\in X,$ then $\forall f\in X^*$, there are disjoint open sets $G_x,\ G_y\subset X$ s.t. $x\in G_x,\ y\in G_y$ and $f(G_x)\cap f(G_y)\neq\emptyset$. (This is because if $f(x)\neq f(y)$, then we can find disjoint open sets $G_x, G_y$ s.t. $x\in G_x,\ y\in G_y$ and $f(G_x)\cap f(G_y)=\emptyset$.
At least finite dimensional topological vector space on $\mathbb{R}$ or $\mathbb{C}$ is homeomorphic to $\mathbb{R}^n$ or $\mathbb{C}^n$. So, to find an answer we should find an infinite dimensional case or change the scalar field to other fields such as the rational number field(?).
I thought of a situation where the cardinality of the topological vector space is too large that any continuous linear functional can't be onto. But still, I couldn't find the suitable example. For example, $l^p$ space is one of the non-locally convex spaces with larger cardinality than $\mathbb{R}$, but its dual space separates points on $l^p$.
Or in a pathetic topological vector space, the satisfying continuity is so difficult that only the zero functional is in $X^*$? (I don't know how this is possible for infinite-dimensional topological vector space though.)
The correct term is "pathological topological vector space" rather than "pathetic topological vector space".
In any case, some topological vector spaces that are not locally convex can have dual space $\{0\}$. A standard example is $L^p([0,1])$ when $0 < p < 1$. More generally, $L^p(X)^{*} = \{0\}$ when $0 < p < 1$ and $X$ is a nonatomic measure space. See Theorems 2.21 and 3.5 here.