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Can somebody give me some intuitive reasoning as to why there should be (and is) more equivalences classes of pairwise Cauchy sequences of rational numbers than there are rational numbers?

Also, from this perspective is there a way to see that the cardinality of this collection of equivalence classes is $2^{\aleph}$? I understand that this is true from then bijecting these equivalence classes with the real numbers, then a bijection with $(0,1)$ and then expressing each decimal in binary, but I was wondering if I could get another perspective.

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For each sequence $(a_n)$ of zeros and ones, define the Cauchy sequence $(b_n)$ by $b_n=\sum_{k=1}^n a_k/3^k$. Then each $(b_n)$ is a Cauchy sequence and different $(a_n)$ yield inequivalent $(b_n)$. There are uncountably many $(a_n)$.

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  • $\begingroup$ Why did you need to have a $3^k$ in the denominator rather than $2^k$? $\endgroup$ – Math is hard Jul 22 '18 at 19:45
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Let $\mathcal Q$ be the set of subsets $A$ of $\mathbb Q$ such that:

  1. $A\neq\emptyset$;
  2. if $a\in A$, then there's a $b\in A$ such that $a<b$;
  3. $A$ has an upper bound.

For each $A\in\mathcal Q$, consider a Cauchy sequence $(q_n)_{n\in\mathbb N}$ of elements of $\mathbb Q$ such that, for each $n\in\mathbb N$,

  1. there is some $a\in A$ such that $a\geqslant q_n$;
  2. $q_n+\frac1n$ is an upper bound of $A$.

Then, if $Q,Q'\in\mathcal Q$ are such that there is a $q\in Q$ which is an upper bound of $Q'$ (or vice-versa), then a Cauchy sequence obtained from $Q$ and a Cauchy sequence obtained from $Q'$ cannot belong to the same equivalence class.

So if, in $\mathcal Q$, you consider the equivalence relation$$Q\sim Q'\iff (\forall q\in Q)(\exists q'\in Q'):q\geqslant q'\text{ and vice-versa,}$$then there are (at least) as many equivalence classes with respect to $\sim$ as there are classes of equivalence of Cauchy sequences. But it's not hard to prove that the number of equivalence classes of $\mathcal Q$ with respect to $\sim$ is, at least, $2^{\aleph_0}$.

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