Linear independency in cartesian product of $\mathbb{R}$ Let $\mathbb{R}^A $ be the vectorial space of functions from $A$ to $\mathbb{R}$. Suppose that I have $x_1, \ldots,x_n \in \mathbb{R}^A$ that are linearly independent. 
Can I always say that there exist $\alpha_1, \ldots, \alpha_n \in A$ such that $$(x_1(\alpha_1), \ldots, x_1(\alpha_n) )$$ $$\ldots$$ $$(x_n(\alpha_1), \ldots, x_1n(\alpha_n) )$$ are independent in $\mathbb{R}^n$? 
My intuition says that yes we can, but as I have no experience with that space, I don't know if there is some weird counterexample or something.
Thanks in advance.
 A: Yes, this is always possible. We can do this by considering the fact that linear dependence in $\mathbb R^A$ is the same as a linear dependence holding simultaneously at every coordinate. Then, we can use this to construct suitable $\alpha_i$, at each step working to eliminate one dimension of dependence among the restricted vectors.
A clean way to do this is to consider, for each $\alpha\in A$, the set of linear dependencies that arise if we project each $x_i$ to the coordinate $\alpha$. That is, we can define a subspace $V_{\alpha}\subseteq \mathbb R^n$ to be the space of tuples $(c_1,\ldots,c_n)$ such that $$c_1\cdot x_1(\alpha)+\ldots+c_n\cdot x_n(\alpha)=0.$$
Note that the linear dependencies among the vectors $x_1,\ldots,x_n$ are described exactly by the subspace $\bigcap_{\alpha\in A}V_{\alpha}$, as a linear dependence of the whole thing is just a dependence holding at every $\alpha$. More generally, for any subset $S\subseteq A$, the set of dependencies among $x_1,\ldots,x_n$ restricted to $S$ is just $\bigcap_{a\in S}V_a$.
We are given that $\bigcap_{\alpha\in A}V_{\alpha}=\{0\}$ by the condition of linear independence. In particular, for every $c\in\mathbb R^n\setminus\{0\}$, there exists some $\alpha$ such that $c\not\in V_{\alpha}$. Now, we can choose a sequence $\alpha_1,\ldots,\alpha_n$ based on the rule that we choose $\alpha_i$ such that there is some non-zero $x\in \bigcap_{j=1}^{i-1}S_{\alpha_j}$ which is not in $S_{\alpha_i}$. This ensures that $\bigcap_{j=1}^{i-1}S_{\alpha_j}$ has dimension exactly one greater than $\bigcap_{j=1}^{i}S_{\alpha_j}$. By induction, it implies that $\bigcap_{j=1}^{n}S_{\alpha_j}$ is zero dimensional; equivalently, this means that restriction to $\{\alpha_1,\ldots,\alpha_n\}$ preserves the linear independence of $\{x_1,\ldots,x_n\}$. 
