Is there a contradiction between these two concepts in Cauchy sequence?

Question

I'm studying real analysis from Rudin's Principles of mathematical analysis, and I feel confused about some definition in Cauchy sequences and feel like there is a contradiction in some idea, or maybe there isn't because I'm a beginner..

Here I quote the concerned part from Rudin's book:

If $\{p_{n}\}$ is a sequence in $X$ and if $E_{N}$ consists of the points $p_{N}, p_{N+1}, p_{N+2},...,$ it is clear from the two preceding definitions that $\{p_{n}\}$ is a Cauchy sequence if and only if $\lim_{n\to \infty}\ diam\ E_{N}=0$.

and the mentioned preceding definitions are:

Def. A sequence $\{p_{n}\}$ in a metric space $X$ is said to be a Cauchy sequence if for every $\epsilon>0$ there is an integer $N$ such that $d(p_{n}, p_{m})<\epsilon$ if $n\geq N\ and\ m\geq N.$

Def. Let $E$ be a subset of a metric space $X$, and let $S$ be the set of all real numbers of the form $d(p,q)$, with $p,q\in E$. The sup of $S$ is called the diameter of $E$.

According to the first quoted sentence, it means -as I understand- that the distance between the points in the sequence is approaching and converging to $0$ as you increase the index of the sequence to $\infty$.

But doesn't this idea contradict with an example of a Cauchy sequence that does not converge such as a sequence in the space of all rational numbers (e.g. $1/n$ which will never converge in $\mathbb{Q}$)?

How the distances between its elements will converge to zero while the sequence never converges ?

CORRECTION:

$1/n$ will converge in $\mathbb{Q}$, so the correct example should be $1/n$ in $(0, \infty)$ for example which will not converge.

Answer 1

I don't see where is the problem in your specific example (and in fact, there is none):

For instance, for $x_n=\frac{1}{n},$ we have that $\text{diam}(E_N)=\frac{1}{N}$, which clearly goes to $0$. (and, for instance, $1/n$ converges in $\mathbb{Q}$. Not on $(0,\infty)$ though, but $\text{diam}({E}_N)$ is still $1/N$ and still goes to $0$ as it should. Note that we are seeing this now as a sequence of real numbers. (see below)).

The fact that distances are real numbers and that the diameter is defined in terms of $\sup$ entails nothing about completion(s) of the space(s). You only need to know $\mathbb{R}$, which you must already know to define a metric space. Then the suprema (and infima) are taken when you consider the distances (that is, real numbers), not the elements of the space. This can be a little confusing if you work only on examples of $\mathbb{Q}$. For that, remember the maxim:

We already know the existence and elementary properties of the real numbers when we are talking about metric spaces.

You don't even need to know that $\mathbb{R}$ is the completion of $\mathbb{Q}$.

Bottom line: When we analyze the sequence $a_n:=\text{diam}(E_N)$, we are looking at a sequence in $\mathbb{R}$, no matter what space are the $p_n$ from.

Answer 2

Hint: the distances between points in a Cauchy sequence get arbitrarily small, but, as you noted, the sequence itself need not converge, and the example you gave is an excellent one if you change $\mathbb{Q}$ to $(0,\infty)$ or even $(0,1]$.

A dangerously naive but also very intuitive way of thinking about this is that a Cauchy sequence always does 'converge' in a sense, just not necessarily to something in your set. Again, your example shows this if you change $\mathbb{Q}$ to $(0,\infty)$, simply because the set doesn't include $0$. But $1/n$ converges just fine to $0$ in, for example, $(-1,1)$.