On the proof of "Show that there is an uncountable number of Cauchy sequences of rational numbers equivalent to any given Cauchy..."

Question

Show that there is an uncountable number of Cauchy sequences of rational numbers equivalent to any given Cauchy sequence of rational numbers.

Proof.

UPDATE: previous choice of $x_k^\prime$ did not work as it is suggested also by the answer below, so I'm trying with a different $x_k^\prime$.

Let $x_1,x_2,\ldots,$ be a given Cauchy sequence of rationals with limit $r$, and consider the set of Cauchy sequences of rational numbers $A = \{(x_1^{(1)},x_2^{(1)},\ldots), (x_{1}^{(2)},x_{2}^{(2)},\ldots),\ldots\}$, which are all equivalent to $x_1,x_2,\ldots$, i.e. for all $n\in\mathbb{N}$ there exists $m_i$ (depending on $x_1^{(i)},x_2^{(i)},x_3^{(i)}$), such that $|x_j^{(i)}-x_j|<1/n$ for all $j>m_i$. We show that the set $A$ is uncountable. Assume for the sake of a contradiction that $A$ is countable. Then, there is a bijection $f:\mathbb{N}\to A$ which maps elements of $\mathbb{N}$ with elements of $A$. That is, we can list the elements of $A$ as follows \begin{eqnarray*} f(1) &= &x_{1}^{(1)},x_{2}^{(1)},x_{3}^{(1)},x_{4}^{(1)},x_{5}^{(1)},\ldots\\ f(2)& = &x_{1}^{(2)},x_{2}^{(2)},x_{3}^{(2)},x_{4}^{(2)},x_{5}^{(2)},\ldots\\ f(3)& = &x_{1}^{(3)},x_{2}^{(3)},x_{3}^{(3)},x_{4}^{(3)},x_{5}^{(3)},\ldots\\ f(4)& = &x_{1}^{(4)},x_{2}^{(4)},x_{3}^{(4)},x_{4}^{(4)},x_{5}^{(4)},\ldots\\ f(5)& = &x_{1}^{(5)},x_{2}^{(5)},x_{3}^{(5)},x_{4}^{(5)},x_{5}^{(5)},\ldots\\ \vdots &=& \vdots\\ \end{eqnarray*} Consider the following sequence of rationals $x_{1}^\prime,x_2^\prime,\ldots$, with $$ x_k^\prime = \begin{cases} x_k^{(k)}/2 & \text{ if } x_k^{(k)} \neq 0\\ 1/k & \text{ if } x_k^{(k)} = 0. \end{cases} $$ We see that $x_k^\prime\not\in A$. To complete the proof we show that $x_k^\prime$ is a Cauchy sequence of rationals and that it is equivalent to $x_k$.

For note that $$ |x_{j}^\prime - x_k^\prime| = \begin{cases} |x_{j}^{(j)}/2-x_k^{(k)}/2| \text{ if } x_j^{(j)}\neq 0,x_k^{(k)}\neq 0\\ |x_{j}^{(j)}/2-1/k| \text{ if } x_j^{(j)}\neq 0,x_k^{(k)}= 0\\ |1/j-x_k^{(k)}/2| \text{ if } x_j^{(j)}= 0,x_k^{(k)}\neq 0\\ |1/j-1/k| \text{ if } x_j^{(j)}= 0,x_k^{(k)}=0. \end{cases} $$

Now, for a given error 1/n, we need to find an $m$, s.t. $|x_{j}^{(j)}-x_k^{(k)}|\leq 1/n$ for all $j,k\geq m$. But I can't figure out how to do it since changing the index implies changing the sequence as well. Any suggestion?

Is it meaningful to proceed this way or would you go by producing a specific set of equivalent Cauchy sequences or rationals?

Answer 1

I believe your proof has bigger problems than the fact that you cannot get rid of $\frac{1}{j}-\frac{1}{k}$. Look at the following Cauchy sequences - all equivalent to $0,0,0,0,0,\ldots$:

$$\begin{matrix}0,&0,&0,&0,&0,&\ldots\\0,&\frac{1}{2},&0,&0,&0,&\ldots\\0,&0,&\frac{2}{3},&0,&0,&\ldots\\0,&0,&0,&\frac{3}{4},&0,&\ldots\\0,&0,&0,&0,&\frac{4}{5},&\ldots\\\vdots&\vdots&\vdots&\vdots&\vdots\end{matrix}$$

Wouldn't your construction end up producing the sequence $1,1,1,1,\ldots$, which is not equivalent to either the starting sequence $0,0,0,0,\ldots$ nor to any of those other sequences?

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To prove the original statement, I would use a different construction. Let $(x_n)_{ n\in\mathbb N}$ be a rational Cauchy sequence. For every subset $X\subset\mathbb N$ we can construct another sequence:

$$y^{(X)}_n=x_n+\frac{1}{n}\mathbf{1}_X(n)=x_n+\begin{cases}\frac{1}{n}&n\in X\\0&n\not\in X\end{cases}$$

where $\mathbf{1}_X$ is the characteristic function of the subset $X$.

It can be proven that:

• Each of those sequences $(y^{(X)}_n)$ is a rational Cauchy sequence,
• Each of those sequences is equivalent to $(x_n)$, and
• All of those sequences are different (for different subsets $X$).

Thus there are at least as many rational Cauchy sequences equivalent to to a given one as there are subsets of $\mathbb N$. With a well-known theorem (which itself uses Cantor's diagonal argument in the proof!) that there are uncountably many subsets of $\mathbb N$, we reach the intended conclusion.

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Having said all this, I wonder if the OP's argument can be fixed - to still produce a new sequence, out of countably many Cauchy sequences equivalent to the original sequence, which will itself be equivalent to the original sequence, using some sort of Cantor's diagonal argument. (Maybe someone would come up with another answer on those lines?)

Answer 2

Two Cauchy sequences $(a_n)_{n\geq0}$ and $(b_n)_{n\geq0}$ of rational numbers are called equivalent if $\lim_{n\to\infty}|a_n-b_n|=0$. In this case the two sequences converge to the same real number $\xi\in{\mathbb R}$.

Assume that $(a_n)_{n\geq0}$ is a Cauchy sequence of rational numbers, and let $n\to\sigma_n\in\{-1,1\}$ $\>(n\geq0)$ be an arbitrary sign sequence. Then $$r_n=a_n+ \sigma_n \>2^{-n}\qquad(n\geq0)$$ defines a sequence $n\mapsto r_n$ $(n\geq0)$ of rational numbers which is equivalent to $(a_n)_{n\geq0}$. Furthermore different sign sequences $(\sigma_n)_{n\geq0}$ lead to different sequences $(r_n)_{n\geq0}$. Since there are uncountably many such sign sequences we obtain uncountably many different rational sequences $(r_n)_{n\geq0}$ equivalent to $(a_n)_{n\geq0}$.