# 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?

## 2 Answers

(I have withdrawn my earlier answer based on Tsang's justified criticism)

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
• Good point. Give me a few hours (not too many downvotes please) and if I cannot repair it I will delete my answer. – Justpassingby Nov 27 '15 at 9:01
• Ah yes, that's what I was missing, wish Rudin would quote which result he was using more often. Thanks anyways! – Chi Cheuk Tsang Nov 27 '15 at 11:34

As mentioned in the answer above, it is quite elementary to prove that the existence a a metric inducing the topology of a topological vector space (t.v.s.) is equivalent to the existence of a translation invariant one. However, if you then want to speak about F-spaces, i.e. complete metrizable t.v.s., the equivalence of having a complete metric or a complete translation invariant metric is not so easy. This had been a question by Banach which was solved only about 30 years later by Victor Klee.