What is the probability of choosing $n$ linearly independent vectors in $\mathbb{R}$ over $\mathbb{Q}$? I recently made a question regarding the measure of a linearly independent set of vectors in $\mathbb{R}$ over $\mathbb{Q}$. As answered, such types of sets have measure zero. That question was made in the hope of knowing how frequently one would choose a linearly independent set of vectors. On the other hand, the answer I got does not seem to be reasonable, since finding a linearly independent set of vectors frequently occurs, when made randomly in finite-dimensional spaces, like that question answered here and the other one here. Now, in order to the question make more sense,

What is the probability of choosing $n$ linearly independent vectors in $\mathbb{R}$ over $\mathbb{Q}$? Or over other infinite dimensional space?

 A: If you restrict your choices to, say, the unit interval, and use the uniform distribution to choose $n$ real numbers then they will be independent over $\mathbb{Q}$ with probability $1$.
[This assertion needs a proof that I can't see immediately. Showing that any two particular irrational numbers are rationally independent is subtle, but the probability argument ought to be straightforward. My original answer below is to a slightly different question.]
The same is true for a set of $n$ vectors in $\mathbb{R}^n$
If you restrict your choices to, say, the unit sphere or the unit ball, and use the uniform distribution to choose  vectors in ℝ then they will be independent with probability 1.
One way to see that is to see that the set where the determinant is $0$ has $0$ volume.
The same argument will work for any nice enough  set on which you can define a reasonable distribution. You can't use the (nonexistent) uniform distribution on the whole space.
A: Claim. Let $\mathbb{P}$ be any probability distribution that is absolutely continuous w.r.t. Lebesgue measure (less fancily: "has a pdf").  Let $\{X_j\}_{j=1}^J$ be any sequence of iid draws from $\mathbb{P}$.  Then $\{X_j\}_j$ is linearly independent a.s.
Proof by induction:
If $J=1$, the claim is obvious.  Otherwise, suppose $\{X_j\}_{j=1}^{J-1}$ is linearly independent a.s. and condition on that event.
Since $\mathbb{Q}$ is countable, so is $\mathbb{Q}^{J-1}$.  The latter is isomorphic (as a $\mathbb{Q}$-vector space) to and thus in bijection with $\operatorname{span}_\mathbb{Q}{\{X_j\}_{j=1}^{J-1}}$.  But any countable set has Lebesgue measure $0$, so that the latter span has $\mathbb{P}$-measure $0$ as well.  In particular, $X_J\notin\operatorname{span}_\mathbb{Q}{\{X_j\}_{j=1}^{J-1}}$ a.s. too.  QED.
I do not know the answer in the case of an infinite-dimensional (topological) vector space of equal cardinality to the subspace.
