A Hausdorff abelian group with no character? Pontryagin Duality for locally compact abliean groups gives plenty of continuous (unitary) characters $\chi : A \to \mathbb{R} / \mathbb{Z}$, but if we do not assume local compactness, can anything be said? In particular, is the following true?

Every abelian Hausdorff topological group has a nontrivial continuous (unitary) character

I believe this to be false, but cannot find a concrete example.
 A: If a topological vector space $X$ is not locally convex, then it usually has not non-zero linear continuous functionals, and this means that there are no non-trivial continuous characters on $X$. For example, you can take the space of all measurable functions on $[0,1]$ with the metrics
$$
d(x,y)=\int_0^1\frac{|x(t)-y(t)|}{1+|x(t)-y(t)|}d t.
$$
A: The infinite-dimensional sphere $S^\infty$ (the evident colimit of finite-dimensional spheres) admits a topological group structure whose underlying abelian group is a torsion group (in fact a structure of $\mathbb{F}_2$-vector space). See this answer here: https://mathoverflow.net/a/43047/2926. 
If $\phi: S^\infty \to S^1$ is a continuous character, then $\phi(S^\infty)$ is evidently a connected torsion subgroup of $S^1$, which can only be trivial since the torsion subgroup $\mathbb{Q}/\mathbb{Z}$ of $S^1$ is totally disconnected. 
A: I'd think that a linearly ordered group extension of   $\mathbb R$,   especially any non-standard real field wouldn't have any non-trivial (Pontryagin) character.
