I'm trying to find an example of a metric space in which a Cauchy sequence doesn't converge, i.e., an incomplete space. I can explain in words that I can take a sequence of rational approximations to $\pi$ or to $\sqrt{2}$ which don't converge in $\mathbb{Q}$, but I don't know how to prove that explicitly, because I'm essentially using that convergence (in a larger metric space, $\mathbb{R}$, in which $\mathbb{Q}$ emebeds) implies Cauchy, and limits are unique.
The first question I have is: is such an approach sufficiently rigorous? I may need some form of argument about how limits in $\mathbb{Q}$ "carry over" to limits in $\mathbb{R}$.
The other approach I can use is taking a metric space $\mathbb{R}^{>0}$ and the sequence $\left\{\frac {1}{n}\right\}$. Secretely, we know that this sequence converges to $0$, but it definitely doesn't converge in $\mathbb{R}^{>0}$. I managed to prove that it is Cauchy by taking $N = \frac{2}{\epsilon}$ and using the definition. I want to show, however, that it cannot converge to any element in the space. So given a positive number $M$, I want to find an $\epsilon > 0$ so that for any $N$, for specific $m,n > N$, we have $|a_m – a_n| \geq \epsilon$.
I don't have any intuition for how to do this, however. Does anyone have a hint?
Best Answer
Your approach can be made as rigorous as you like, yes: the magic going on under the surface is that the inclusions $\mathbb{Q}\hookrightarrow\mathbb{R}$ and $\mathbb{R}^{>0}\hookrightarrow\mathbb{R}$ (and such incisions in general) are embedding: that is, they preserve the topology, and hence the limits.
Your approach as written can't work: you're trying to show that the sequence isn't Cauchy. Assuming that the $a_n$ is a typo and should be $M$, it's quite doable. For a hint: once the sequence has gotten smaller than something, it will stay below it, so can't converge to anything above it. There's a full solution under the line below, so stop reading at the line if you don't want spoilers.
For an alternative approach, I'll construct an incomplete space (and a Cauchy sequence evidencing that it is incomplete) that is not (naturally) a subset of a "more obvious" complete space.
We'll do this by redefining "distance" on $\mathbb{R}$. We'll define $d(x,y) = |e^{-x} - e^{-y}|$ (using a different name to avoid confusion with $|-|$. You can show that this behaves nicely as a notion of "distance" if you like.
For our sequence to show this, we'll choose the natural numbers $a_n = n$.
This is Cauchy: for any $\varepsilon > 0$, there is some $N$ such that $e^{-N} < \frac{\varepsilon}{2}$. For any $m,n > N$, then, $d(a_n, a_m) = d(n,m) = |e^{-n} - e^{-m}| \leq e^{-n} + e^{-m} < 2e^{-N} < \varepsilon$.
However, if it converged to some $x\in\mathbb{R}$, then for any $\varepsilon > 0$, there would be some $M \in \mathbb{N}$ such that for all $n > M$, we would have $|e^{-n} - e^{-x}| < \frac{\varepsilon}{2}$. But then, for any $n > \max(N,M)$, we would have $e^{-x} < e^{-n} + \frac{\varepsilon}{2} < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$.
But then $e^{-x}$ has to be smaller than any positive $\varepsilon$, so can be at most $0$. There is no such $x$, though, and so this sequence does not converge.
Spoiler line here.
Given $M$, take $\varepsilon = \frac{M}{2}$. Then for any $n > \frac{2}{M}$, we have $a_n < \frac{M}{2}$, so $|a_n - M| = M - a_n > M - \frac{M}{2} = \frac{M}{2} = \varepsilon$.