Linearly independent family of sequences of rationals with a cardinal equal to the continuum I'm coming back to this question. Is it possible to have "an explicit" linearly independent family of sequences of rationals with a cardinal equal to the continuum?
PS: sorry for the duplicate on the previous question.
 A: Choose a bijection $\alpha:\mathbb{N}\to\mathbb{Q}$, and for each $x\in\mathbb{R}$ let 
$$a(x)_i=\begin{cases}0&\mbox{ if $\alpha(i)<x$}\\1&\mbox{ if $\alpha(i)\geq x$}\end{cases}.$$
Then the set of sequences $\{a(x):x\in\mathbb{R}\}$ is linearly independent.
A: Take a family $A_\alpha$ of continuum non finite ``almost disjoint'' subsets of $\textbf{N}$.  That is the sets $A_\alpha\cap A_\beta$ are  finite 
for any $\alpha\ne \beta$, and each $A_\alpha$ is infinite.
Then the characteristic functions  $x(\alpha)$ of the set $A_\alpha$ is a 
sequence of $0$ and $1$, so rational numbers.  These $x(\alpha)$ are independent.
In fact any relation 
$$x(\alpha)=q_1 x(\beta_1)+\dots + q_n x(\beta_n)$$
(with $\alpha$, $\beta_1$, $\dots$, $\beta_n$ distincts)
is false on any non null coordinate of the $x(\alpha)$ that is not common with any 
of the $\beta_j$
The construction of the family of almost disjoint subsets of $\textbf{N}$ is very explicit.
Can be found in the book by Thomas Jech, Set Theory, Academic Press 1978 
(Lemma 23.9 in page 242 in my edition that is not the last). 
