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.