1
$\begingroup$

Indeed, this question arises when one wants to prove the well-definedness of an order while constructing $\mathbb{R}$ from equivalence classes of Cauchy sequences in $\mathbb{Q}$.

Definition 1. A Cauchy sequence $a:\mathbb{N}\to\mathbb{Q}$ is positive if and only if it does not converge to $0$ and has a positive tail. More precisly, $\exists \epsilon\in\mathbb{Q}^+,\,\,\forall N\in\mathbb{N},\,\,\exists n\ge N\land |a_n|\ge\epsilon$ and $\exists M\in\mathbb{N},\,\,\forall n\ge M\implies a_n>0.$

One can prove the following, which is also geometrically obvious.

Theorem 1. A Cauchy sequence is positive iff $\exists\omega\in\mathbb{Q}^+,\,\,\exists K\in\mathbb{N},\,\,\forall n\ge K \implies a_n>\omega$.

Definition 2. A sequence $(b_n)$ is said to be co-Cauchy with the Cauchy sequence $(a_n)$ iff they their tails become close together, i.e., $\lim_{n\to\infty}b_n-a_n=0$.

It is also immediate that

Thoerem 2. A sequence which is co-Cauchy with a Cauchy sequence is Cauchy.

My goal is to prove that

Theorem 3. A Sequence which is co-Cauchy with a Positive Cauchy Sequence is Positive.

Proof. Well, let us right our assumptions. Suppose $(a_n)$ is a positive Cauchy sequence and $(b_n)$ is co-Cauchy with it then we have

\begin{align} &\forall \epsilon_1 \in \mathbb{Q}^+, \,\,\exists N_1 \in \mathbb{N}, \,\,\forall n\ge N_1 \implies |a_n-b_n|< \epsilon_1, \tag{1}\\ &\forall \epsilon_2 \in \mathbb{Q}^+, \,\,\exists N_2 \in \mathbb{N}, \,\,\forall m,n\ge N_2 \implies |a_m-a_n|< \epsilon_2, \tag{2}\\ &\exists\epsilon_3\in\mathbb{Q}^+,\,\,\exists N_3\in\mathbb{N},\,\,\forall n\ge N_3 \implies a_n>\epsilon_3, \tag{3} \end{align}

and we are to prove that

$$\exists \epsilon\in\mathbb{Q}^+,\,\,\exists N,\,\,\forall n \ge N \implies b_n> \epsilon.\tag{4}$$

Choose $\epsilon_1:=\frac{\epsilon_3}{2}$ and take $N:=\max\{N_1,N_3\}$. Then for every $n\ge N$ we have that $a_n-\frac{\epsilon_3}{2}<b_n<a_n+\frac{\epsilon_3}{2}$ and $a_n>\epsilon_3$. Since $a_n-\frac{\epsilon_3}{2}>\frac{\epsilon_3}{2}$, this implies that $b_n>\frac{\epsilon_3}{2}$. So we can simply take $\epsilon:=\frac{\epsilon_3}{2}$. This completes the proof.

Is my argument for the last theorem correct? I am a little skeptical as I didn't use the Cauchyness of $(a_n)$ at all. It seems that all these definitions and theorems (except theorem 2) can make sense for arbitrary sequences which are not necessarily Cauchy. Am I right?

$\endgroup$

1 Answer 1

2
$\begingroup$

Your proof is correct. Intuitively, it should make sense--if you have a sequence that stays a bounded distance above $0$, then any other sequence getting close to it will also say above $0$. It doesn't matter whether your first sequence was actually closing in on any fixed positive number, if all you care about is staying above $0$.

Note, though, that the characterization of positiveness which you use from Theorem 1 does use the fact that $(a_n)$ is Cauchy--the original definition of positiveness is weaker for general sequences, and Theorem 3 is not true for general sequences with the original definition of positiviteness. Roughly, you could have a sequence which oscillates between large positive values and positive values approaching $0$, and then a sequence could be co-Cauchy with it while being negative on the part that approaches $0$, so it does not have a positive tail. There are a couple different natural ways you might define a positive general sequence so that Theorem 3 remains true--you could either take the condition of Theorem 1 as the definition, or you could weaken the positive tail condition to say that for any rational $\epsilon>0$, $(a_n+\epsilon)$ has a positive tail.

The real place where the assumption that the sequences are Cauchy will be crucial for the theory of the ordering of real numbers is that this notion of "positivity" gives a total order. In other words, every Cauchy sequence is either positive, converges to $0$, or is negative (meaning if you negate its terms you get a positive sequence). For arbitrary sequences you could have sequences that oscillate between positive and negative values and so are neither positive nor negative. (Relatedly, you also need the Cauchy condition in order to get multiplicative inverses, since you need any sequence which does not converge to $0$ to stay away from $0$ in its tail.)

(Or, in the language of abstract algebra, if you take the set of all sequences modulo the co-Cauchy equivalence relation, you will still get a partially ordered commutative ring. However, you will not get a totally ordered field which is what you want the real numbers to be.)

$\endgroup$
4
  • $\begingroup$ (+1) Thanks for the insight Eric. :) That was really helpful. Well, I was thinking that for Theorem 1 to be true, Cauchyness is also necessary. For example, one can think of a sequence $x_n=\frac{1}{n}$ for odd $n$ and $x_n=1-\frac{1}{n}$ for even $n$. Then the sequence does not converge to zero and also has a positive tail but there is no such $\omega$ that we can get $x_n>\omega$ whenever $n\ge N$. Indeed, I used the result of Theorem 1 in the proof for Theorem 3. $\endgroup$ Mar 15, 2020 at 17:33
  • 2
    $\begingroup$ Oh, that's a good point--Theorem 1 does use the Cauchy assumption. I've added a bit to the answer about that. $\endgroup$ Mar 15, 2020 at 17:36
  • $\begingroup$ By the way, do you know any note or reference which covers these tiny details of constructing Reals from Rationals using Cauchy sequences? $\endgroup$ Mar 15, 2020 at 17:47
  • $\begingroup$ Not off the top of my head, sorry. $\endgroup$ Mar 15, 2020 at 17:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .