Is the unit ball bounded for all metrizable topological vector spaces?

Question

If not, what would a counterexample be?

Answer 1

Assuming the "unit ball" to mean $\{y: d(y,0) < 1\}$, where $d$ is any metric compatible with the topology, then the answer is no. For example, take $\mathbb R$ with the metric $d(x,y) = \dfrac{|x-y|}{1+|x-y|}$. Then the unit ball consists of all of $\mathbb R$, and this is not bounded (in the topological vector space sense).

Answer 2

if it's a unit ball, it contains all points within a distance of $1$ from an origin $z$. A not bounded set would include points $x,y$ arbitrary far away from each other, but using the triangle inequality this cannot be true, since: $$d(x,y) \le d(x,z) + d(z,y) < 2$$

Am I missing something?