In a Topological Vector Space T0 implies T3½ (completely regular)? And other separation properties.

Question

I will describe my doubt.

I know that in a TVS T1 implies T2. Now since a TVS admits a uniformisable topology, we have that T2 implies the uniform structure is separating. Now a separating uniform structure can be given the structure of a gauge space (its topology can be generated by a set of pseudometrics). But now it is well known that completely regular spaces identify with gauge spaces. First of all: am I correct until now?

If so, the question boils down to ask if T0 implies T1 in a TVS. My answer would be positive, exploiting the translation invariance charachter of the topology. But I am not sure.

Are there any other interesting separation properties, of TVS? Which shed light on the implications of the linearity charachter of TVS topology? In particular can function like charachterization (Urysohn,Tietze...) be somehow enriched or exploited to study the vector space from its dual?

Answer 1

Any topological group (under $+$ in our case) induces a compatible uniform structure, so that we have a $T_{3\frac12}$ space as soon as we have a separated uniformity (so $T_0$ suffices to get that, and Hausdorff certainly does).

The scalar multiplication plays no role.

We don't get in general a $T_4$ space, as $\mathbb{R}^X$ for $X$ uncountable, in the product topology, is a locally convex TVS that is not $T_4$.

Answer 2

For your question the scalar multiplication of a TVS $X$ does not play a role. In fact, for any topological group $G$ property $T_0$ implies $T_{3\frac{1}{2}}$. See any book on topological groups.

Anyway, your arguments for $T_1 \Rightarrow T_{3\frac{1}{2}}$ are correct.