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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.
