Generating vector spaces according to some sets that shall serve as a basis Give a family of elements $\left(v_{1},\ldots,v_{n}\right)$ (where $v_{1},\ldots,v_{n}$  are just some sets), how many
vector spaces are there, such that this family is a basis for that
vector space ?
For example, if $n=1$ and $\left(v_{1}\right)=(1)$, there are at least two vector spaces, namely $\mathbb{Q}$ and  $\mathbb{R}$ over themselves (as fields), such that this is a basis.
Is there even a reasonable way to answer this ?
 A: You are talking about vector spaces over different fields: $\mathbb{Q}$ is a $\mathbb{Q}$-vector space for which $\{1\}$ is a basis, and $\mathbb{R}$ is an $\mathbb{R}$-vector space for which $\{1\}$ is a basis, but $\mathbb{R}$ is a $\mathbb{Q}$-vector space for which $\{1\}$ is clearly not a basis. Part of the structure of a vector space is the field of scalars, so $\mathbb{R}$-as-an-$\mathbb{R}$-vector space and $\mathbb{R}$-as-a-$\mathbb{Q}$-vector-space are different mathematical objects.
So, if you want to know how many $K$-vector spaces there are having a set $S$ as a basis, where $K$ can be any field, there are "at least as many" as there are fields (so, certainly infinitely many, and in fact there should be "as many" as there are sets, because any set can be chosen to be a transcendence basis over some given field; speaking precisely, they form a proper class).
Even restricting yourself to only a specific field $K$, it is still the case that given a set $S$, there are as many $K$-vector spaces with basis $S$ as there are sets, because  (for example) you can always replace the zero element of your vector space with something arbitrary. For example, the $\mathbb{Q}$-vector space $\mathbb{Q}$ has $\{1\}$ as a basis, but so also does the vector space
$$\mathbb{Q}\cup\{\star\}\setminus\{0\}$$
(where $\star$ is the zero element of the vector space), and $\star$ can be anything.
But if you're only interested in the number of different $K$-vector spaces with basis $S$ for a given field $K$ up to isomorphism, then there is only one, because any two $K$-vector spaces with equinumerous bases are isomorphic.
