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It confused me for a while. Here is the question.

Prove that given any Cauchy sequence of reals, there exists a Cauchy sequence of rationals that converges to the same value.

Sk8er_ZZZ
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2 Answers2

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Your question is not entirely clear. If the sequence of reals is Cauchy, then it converges to some $L$ and clearly you can find a sequence of rationals converging to $L$. I'm guessing you are looking for something more related to a construction of the reals. So, here goes.

Suppose $x_n$ is the given Cauchy sequence of reals. For each $n$ choose an rational $q_n$ such that $d(x_n,q_n)<1/n$ (using the density of the rationals in the reals). Since $d(q_n,q_m)\le d(q_n,x_n)+d(x_n,x_m)+d(x_m,q_m)$ it easily follows that the sequence $q_n$ is Cauchy. This sequences thus represents a real number which is the limit of the original sequence. Is that what you had in mind?

Ittay Weiss
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  • Isn't it a bit much in this context to assume the density of the rationals in the reals? It seems as if that is most of what is to be proved. – Michael Hardy Sep 15 '14 at 03:27
  • @MichaelHardy you may be right. I don't know. It's a bit of guess work what OP had in mind. E.g., which construction of the reals he is working with as it appears he is interested in some detail of a construction. I don't know. In any case, the density of the rationals is typically rather easy to show while metric completeness is harder. Again, who knows... – Ittay Weiss Sep 15 '14 at 03:30
  • His second paragraph certain states a clear question. But the subject line says something different. – Michael Hardy Sep 15 '14 at 03:32
  • exactly. It's a mystery. – Ittay Weiss Sep 15 '14 at 03:35
  • I think if we could prove the question, the subject will satisfy? – Sk8er_ZZZ Sep 15 '14 at 12:15
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It is enough to show that for every real number $x$, there is a sequence of rationals converging to it.

Among the rationals $\ldots,\ -3/n,\ -2/n,\ -1/n,\ 0,\ 1/n,\ 2/n,\ 3/n,\ \ldots$ there is one rational $k/n$, such that $k/n\le x<(k+1)/n$. Let $x_n=\text{that rational number }k/n=\lfloor nx\rfloor/n$.

Then $0\le x-x_n<1/n$. In order to show that $x_n\to x$ as $n\to\infty$, it is then enough to show that $1/n\to0$. To do that, it is enough to show that there is no $\varepsilon$ satisfying $0<\varepsilon<1/n$ for all positive integers $n$. If there were, then $1/\varepsilon$ would be an upper bound of the set of all positive integers, which would therefore have a least upper bound $c$. Then $c-1$ would not be an upper bound, so for some integer $n$ we have $n>c-1$. But the $n+1>c$ and $n+1$ is an integer, so $c$ is not an upper bound, and we have a contradiction.

Michael Hardy
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