# Uniquely Promoting a TVS over the Rationals to a TVS over the Reals

I have a Topological Vector Space (TVS) ($$I, \oplus , \otimes$$) over $$\mathbb Q$$ and I want to uniquely extend its scalar multiplication to $$\mathbb R$$, so that it is promoted to a TVS over $$\mathbb R$$.

The question is, what additional conditions does $$I$$ need to fulfill for the promotion to be possible in a unique way?

In the specific problem I am attempting to solve, topologically $$I$$ is given to be isomorphic to $$\mathbb R$$. This of course makes it metrizable (and possessing a bunch of other nice properties for that matter) but it is not a metric space, so is it enough? In fact, its promotion is an intermediary step for endowing $$I$$ with a metric other than the Euclidean (which it could easily inherit through its homeomorphism with $$\mathbb R$$).

So my primary interest is in finding out whether homeomorphism to $$\mathbb R$$ is enough and in what way. Of course if general conditions under which the promotion is possible can be given, without utilizing the homeomorphism to $$\mathbb R$$, it would be even better.

I am aware of the fact that, being a TVS, makes $$I$$ a Uniform Space. I am also aware that scalar multiplication mapping

$$\otimes_v : \mathbb Q \rightarrow I$$

$$\rho \rightarrow \rho \otimes v$$

is a Uniformly Continuous mapping for every $$v \in I$$. I expected that this would be enough to allow the unique (uniformly) continuous extension of $$\otimes_v$$ to the closure of $$\mathbb Q$$ which is $$\mathbb R$$ for every $$v \in I$$, which is adequate for my purposes.

Unfortunately, in my search, all I have managed to come across is the possibility of (uniquely and continuously) extending a Uniformly Continuous mapping from a subset to its closure, between metric spaces.

So, I guess, another way to pose the question would be:

Are there more general conditions which allow for such an extension?

In particular, is metrizability (or any other topological property which $$I$$ inherits from $$\mathbb R$$) enough for the extension?

• Why I is homeomorphic to R? – William Elliot Apr 15 at 11:51
• It is a given from the particular problem I am attempting to solve. My first priority it to find out whether under the assumption of homeomorphism to R, the promotion is doable. Of course I am also interested in the most general conditions under which it is possible. I am editing the question to reflect this. – Crispost Apr 15 at 12:04
• It is true that if $A \subseteq X$ for $(X, \mathcal U)$ a uniform space, and if $(Y, \mathcal V)$ is a complete $T_0$ uniform space, then every uniformly continuous map on $A$ into $Y$ extends uniquely to a uniformly continuous map on $\overline A$. See e.g. Kelley (or any book covering uniform spaces I suppose). – LinearOperator32 Apr 16 at 6:49