Published online by Cambridge University Press: 07 August 2017
For a Tychonoff space $X$, let
$\mathbb{V}(X)$ be the free topological vector space over
$X$,
$A(X)$ the free abelian topological group over
$X$ and
$\mathbb{I}$ the unit interval with its usual topology. It is proved here that if
$X$ is a subspace of
$\mathbb{I}$, then the following are equivalent:
$\mathbb{V}(X)$ can be embedded in
$\mathbb{V}(\mathbb{I})$ as a topological vector subspace;
$A(X)$ can be embedded in
$A(\mathbb{I})$ as a topological subgroup;
$X$ is locally compact.