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?


  • 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

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 $\mathbb{Q}$.

  • $\begingroup$ 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
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    $\begingroup$ 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

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