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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?

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    $\begingroup$ Why I is homeomorphic to R? $\endgroup$ – William Elliot Apr 15 at 11:51
  • $\begingroup$ 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. $\endgroup$ – Crispost Apr 15 at 12:04
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    $\begingroup$ 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). $\endgroup$ – LinearOperator32 Apr 16 at 6:49

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