Topological vector spaces with a unique trivial nonempty convex set which is algebraically open

Question

Let $X$ be a topological vector space. We recall that a set $A\subseteq X$ is algebraically if, for each $x,y \in X$, the set $\{t \in \mathbf{R}: x+ty \in A\}$ is open.

Moreover, it is well known (and easy to prove) that every open set is algebraically open. Also, there are non-open sets which are algebraically open (see e.g. here).

In addition, if $A$ is convex, then the converse holds in $X=\mathbf{R}^n$. This is still false in an arbitrary topological vector space $X$ (see e.g. here).

Question. Does there exist a topological vector space $X$ such that the only convex algebraically open non-empty subset is $X$ itself?

Ps. Replacing "algebraically open" with the the stronger "open" the answer is positive, setting $X=L_p([0,1])$, for some $p \in (0,1)$.

Answer 1

Given a basis $v_i$, $i\in I$ of $X$ the set $$U=\left\{\sum_{i\in I}a_iv_i:\ |a_i|<1\text{ and only finitely many }a_i\neq0\right\}$$ is

1. Non-empty: $0\in U$.
2. Convex: If $x=\sum_{i\in A}a_iv_i,y=\sum_{i\in B}b_iv_i\in U$ we can assume $A=B$ and finite. Then $tx+(1-t)y=\sum_{i\in A}(ta_i+(1-t)b_i)v\in U$, since $(-1,1)$ is convex.
3. It is algebraically open: If $x,y$ are written as before, with $x\in U$ and $y$ arbitrary, then $x+ty=\sum_{i\in A}(a_i+tb_i)v_i$. Choosing $|t|<\frac{\min_{i\in A}\min(|a_i+1|,|a_i-1|)}{\max_{i\in A}|b_i|}$ should keep $x+ty\in U$.
4. $U\neq X$, when $X\neq \{0\}$. Since $2v_i\in X\setminus U$ for any element $v_i$ of the basis chosen.