It is proved that the free topological vector space $\mathbb{V}([0,1])$ contains an isomorphic copy of the free topological vector space $\mathbb{V}([0,1]^{n})$ for every finite-dimensional cube $[0,1]^{n}$, thereby answering an open question in the literature. We show that this result cannot be extended from the closed unit interval $[0,1]$ to general metrisable spaces. Indeed, we prove that the free topological vector space $\mathbb{V}(X)$ does not even have a vector subspace isomorphic as a topological vector space to $\mathbb{V}(X\oplus X)$, where $X$ is a Cook continuum, which is a one-dimensional compact metric space. This is also shown to be the case for a rigid Bernstein set, which is a zero-dimensional subspace of the real line.

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.