Uncountably many equivalent Cauchy sequence?

Question

RTP

There exists uncountably many Cauchy sequence of rationals that are equivalent.

I am trying to solve the above question, and my understanding is that $\Bbb R$ is a set of equivalent classes of Cauchy sequence of rationals. And two sequences are equivalent if they both have the same limit.

So, having prior knowledge (not really knowing the proof) that there are uncountably many real numbers, I want to somehow connect this idea with this problem.

Can someone help me out ?

I am a beginner at analysis, so it would really help if you could dumb it down.

Answer 1

One can show, for example, that there are uncountably many sequences of rationals tending to zero. It is more convenient to show that there are uncountably many sequences of rationals $(r_n)$ tending to infinity (just invert all the terms). Now take the sequence $(x^n)$, where $x\in\mathbb R$ and $x>1$. This is not a sequence of rationals, but it tends to infinity. Now convert it to a sequence of rationals by setting $r_n = \lfloor x^n \rfloor$. The new sequence is a sequence of integers (in particular, rationals). It tends to infinity, and it is easy to show that two different $x$'s will give two different sequences tending to infinity. This yields uncountably many such sequences because there are uncountably many real $x$.

Answer 2

You’re interested in the size of one equivalence class of Cauchy sequences; you don’t need to know how many equivalence classes there are, and in fact knowing that won’t help you. I’ll get you started.

Let $\sigma=\langle x_n:n\in\Bbb N\rangle$ be any Cauchy sequence of rationals; for instance, you could take $x_n=\frac1{2^n}$. Let $A$ be any infinite subset of $\Bbb N$. You can use $A$ to produce a subsequence $\sigma_A$ of $\sigma$ in the following way: let $A=\{a(k):k\in\Bbb N\}$, where $a(0(<a(1)<a(2)<\ldots\;$, and let $\sigma_A=\langle x_{n(a_k)}:k\in\Bbb N\rangle$.

• Verify that $\sigma_A$ and $\sigma$ are equivalent.
• Show that $\{\sigma_A:A\subseteq\Bbb N\text{ and }A\text{ is infinite}\}$ is uncountable. (In fact it has cardinality $2^\omega=\mathfrak{c}$.)