0
$\begingroup$

I understand that:

In any metric space X, every convergent sequence $\{p_n\}$ is a Cauchy sequence, which can be shown by noting that, for all $\epsilon >0$, there exists an integer $N$ such that $n,m \geq N$ implies

$d(p_n,p_m) \leq d(p_n,p) + d(p, p_m) < 2\epsilon$.

Since $\epsilon$ was arbitrary, we conclude that $\{p_n\}$ is Cauchy. But also, it is true that in $R^k$, which is also a metric space, every Cauchy sequence converges (which is a little bit harder to prove).

I have a couple questions:

  1. Following from the statements above, can we say WLOG, a sequence $\{p_n\}$ in $R^k$ is convergent if and only if it is a Cauchy sequence?
  2. Also, I don't think I could infer that the statement is true in every metric space. Is there a counterexample to see this?

Thanks!

$\endgroup$
  • 2
    $\begingroup$ 1) Yes. 2) In $\mathbb{Q}$, convergence $\ne$ Cauchy. A simpler example is $(0, 1)$, which is also incomplete. $\endgroup$ – user296602 Apr 10 '18 at 23:13
  • 1
    $\begingroup$ you may want to have a look at complete metric spaces: en.wikipedia.org/wiki/Complete_metric_space $\endgroup$ – fonfonx Apr 10 '18 at 23:14
  • $\begingroup$ Thanks for your comments. So in this sense, would convergence of a sequence always be more encompassing than simply the Cauchy criterion being satisfied? $\endgroup$ – T J. Kim Apr 10 '18 at 23:26
2
$\begingroup$

In $\mathbb{R}^n$, every Cauchy sequence converges. This is a property called completeness; a metric space $X$ is complete if every Cauchy sequence converges. Thus, in a complete metric space, which $\mathbb{R}^n$ is, a sequence is Cauchy if and only if it converges.

For your second question, just take a non-complete metric space, say, $\mathbb{Q} \subset \mathbb{R}$, and consider a sequence of rational numbers that are converging to $\sqrt{2}$ in $\mathbb{R}$. Since $\sqrt{2}$ is not a rational number, this sequence is Cauchy, but it does not converge in $\mathbb{Q}$.

$\endgroup$
  • $\begingroup$ I assume there are easier ways to see this, but since every closed subset of a complete metric space is also complete, does this mean that $\mathbb{Q}$ is not a closed subset of $\mathbb{R}$? $\endgroup$ – T J. Kim Apr 10 '18 at 23:37
  • 1
    $\begingroup$ It does indeed; every real number is a limit point of $\mathbb{Q}$, but clearly not every real number is contained in $\mathbb{Q}$. $\endgroup$ – Chris Apr 10 '18 at 23:45

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.