I have seen some examples of metrics on $\mathbb R$ which make it incomplete, like $d(x,y)=|e^{-x}-e^{-y}|$ and $d(x,y)=\arctan x - \arctan y$ where $1,2,3,\ldots$ is a Cauchy sequence which doesn't converge. I am curious about what other examples there are. Is it always infinity which turns out to be 'missing' or are there metrics with Cauchy sequences that do not converge but remain bounded?

  • $\begingroup$ With $\arctan $ you are missing two points, really. $\endgroup$ – Saucy O'Path May 31 '18 at 15:44
  • $\begingroup$ Consider the bijection $f \colon (-\infty, 0] \cup (1, \infty) \to \mathbb{R}$ that simply "puts the pieces next to each other". Then $d(x,y) = |f^{-1}(x) - f^{-1}(y)|$ is a metric that makes $\mathbb{R}$ incomplete because $(1/n)$ is a bounded Cauchy sequence (in both metrics) that does not converge to $0$ in the new metric. $\endgroup$ – Carl Mummert Jun 11 '18 at 17:44

That depends on how much leeway one has in specifying the metric. Consider for example the fact that there exists a bijective map $\phi\colon\mathbb R \to \mathbb R \setminus \mathbb Q$. If you define the metric on $\mathbb R$ by setting $d(x,y)=|\phi(x)-\phi(y)|$, it will be a metric alright, and clearly not complete.


If you restrict yourself to a bounded part of the reals,so some closed interval $[A,B]$ then all equivalent metrics on that part are defined on a compact space, so are always complete.

This is the reason that an equivalent non-complete metric on $\mathbb{R}$ has its non-completeness at infinity, as it were.

  • $\begingroup$ This is true, assuming that the metric has to be equivalent to the original one (not stated in the question). $\endgroup$ – Carl Mummert Jun 11 '18 at 17:45
  • $\begingroup$ @CarlMummert the OP did give examples where this was the case. $\endgroup$ – Henno Brandsma Jun 11 '18 at 17:46

Cauchy sequences always remain bounded:

If $\epsilon = 1$, there is $N$ such that for all $n,m>N$ $$d(x_n,x_m)<1$$

Therefore, $d(x_n,0)\leq d(0,x_{N+1})+d(x_{N+1},x_n)<d(0,x_{N+1})+1$, for $n>N$ and a fixed. This means that the whole sequence is inside the ball with center $0$ and radius $\max(d(0,x_{N+1})+1, d(x_1,0),d(x_1,0),...,d(x_N,0))$.

In your examples, $1,2,3,...$ is bounded.

  • $\begingroup$ The question was vague, but OP probably meant bounded with respect to the standard norm. Cauchy seqs are only bounded with respect to the norm for which it is a Cauchy $\endgroup$ – user160738 May 31 '18 at 15:43
  • $\begingroup$ @user160738 I know what he meant. That is why I proved that what he said is not what he meant. $\endgroup$ – user564402 May 31 '18 at 15:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.