# Can a metrizable TVS be induced by a non-translation invariant metric?

Is it possible to have a topological vector space $(X, \tau)$ with its topology induced by a metric $d$ which is not translation invariant?

I'm asking this because in Rudin's 'Functional Analysis' Theorem 1.28, he automatically assumed the metric of a metrizable TVS is translation invariant (he defined a metrizable TVS to be one which topology can be induced by a metric, no requirement on the metric being translation invariant or not). It seems that Rudin is usually careful about his assumptions, so I wonder if I'm missing something?

Metrizable topological spaces always satisfy the first axiom of countability (take the open balls with radius $1/n$).
In theorem 1.24 Rudin proves that if $X$ is a TVS with a countable local base then there is an invariant metric that is compatible with the topology. The proof involves a construction that is rather more elaborate than the one I tried in my earlier answer.
• Is $d'$ a metric? Seems that $d'(x,y)=d'(y,x)$ and $d'(x,y)+d'(y,z) \geq d'(x,z)$ may not hold... – Chi Cheuk Tsang Nov 27 '15 at 8:51