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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.

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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.

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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).

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