We use cookies to distinguish you from other users and to provide you with a better experience on our websites. Close this message to accept cookies or find out how to manage your cookie settings.
We use cookies to distinguish you from other users and to provide you with a better experience on our websites. Close this message to accept cookies or find out how to manage your cookie settings.
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.
