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The sequence $\{a_n\}_{n=1}^\infty$ is cauchy if for every $\epsilon>0$, there is a corresponding natural number $N$ such that

$$ m,n\geq N\Rightarrow |a_m-a_n|<\epsilon $$

I am doing a particular problem where the problem talks about cauchy sequences of rational numbers and I am not sure how that is different from a normal cauchy sequence (defined above).

If $\{a_n\}_{n=1}^\infty$ is a cauchy sequence in rational number and if there is a sub-sequence of this sequence, $\{a_{n_j}\}_{j=1}^\infty$ which converges to a rational number $\frac{p}{q}$, then I need to show that the sequence $\{a_n\}_{n=1}^\infty$ converges to the rational number $\frac{p}{q}$.

How this would be different if we did not talk about rational numbers so the problem was the following:

If $\{a_n\}_{n=1}^\infty$ is a cauchy sequence of real numbers and if there is a sub-sequence of this sequence, $\{a_{n_j}\}_{j=1}^\infty$ which converges to a real number $L$, then I need to show that the sequence $\{a_n\}_{n=1}^\infty$ converges to the real number $L$.

Since this question talks about rational numbers and not real numbers, it confuses me.

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  • $\begingroup$ I think the point of the problem is to show that if you had a Cauchy sequence of rational numbers it does not necessarily approach a rational number e.g.$\root \of 2$ so it doesn't converge when defined on the rational numbers $\endgroup$
    – Ziad Fakhoury
    Commented Apr 2, 2017 at 12:02

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The main difference is that $\Bbb R$ is complete, while $\Bbb Q$ is not, which means that every Cauchy sequence of real numbers is convergent, while Cauchy sequence of rationals does not need to converge in $\Bbb Q$.

Thus, when you have your problem stated for real Cauchy sequences, it is almost trivial: sequence $(a_n)$ is Cauchy, therefore convergent, and since its subsequence has limit $L$, $(a_n)$ converges to $L$ as well.

On the other hand, if you switch back to rationals, you do not know that $(a_n)$ is convergent a priori. However, here comes the most important part: $\Bbb R$ is metric space completion of $\Bbb Q$ - meaning that every Cauchy sequence in $\Bbb Q$ will converge to some real number and any real number is a limit of Cauchy sequence in $\Bbb Q$.

Of course, the whole story about $\Bbb Q$ and $\Bbb R$ can be completely bypassed (but you specifically asked for the difference) and one can show that Cauchy sequence is convergent if and only if it has a convergent subsequence.

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  • $\begingroup$ That was exactly the point of my question. I knew how to prove the question for real numbers but did not see how the proof would be different if we were talking about rational numbers. $\endgroup$
    – user377355
    Commented Apr 2, 2017 at 12:45
  • $\begingroup$ Sir, if we consider a convergent sequence of rationals whose limit is not in $\mathbb{Q}$ then is it Cauchy sequence in $\mathbb{Q}$? Please reply. $\endgroup$
    – Akash Patalwanshi
    Commented Aug 11, 2019 at 4:34
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    $\begingroup$ @Akash, $\newcommand{\Q}{\mathbb Q} \newcommand{\R}{\mathbb R}$ yes, if all of its members are in $\Q$, then it is a sequence in $\Q$. Forget about $\R$ for now. "Convergent sequence of rationals whose limit is not in $\Q$" is oxymoron since by definition convergent sequence in $\Q$ must have limit in $\Q$. Now, considering $\R$, every Cauchy sequence in $\Q$ does converge in $\R$, and this is likely what you meant in the sentence I quoted. Type of sequence you describe is indeed: Cauchy sequence in $\Q$, is not convergent in $\Q$, is convergent in $\R$. $\endgroup$
    – Ennar
    Commented Aug 11, 2019 at 7:50
  • $\begingroup$ Thank you so much sir. Got it... $\endgroup$
    – Akash Patalwanshi
    Commented Aug 11, 2019 at 12:49

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