# Please explain the proof of the validity of the Cauchy convergence criterion for real number sequences.

My question pertains to BBFSK, Vol I, Pages 143 and 144.

The following appears in the context of developing the real numbers as limits of sequences of rational numbers.

It is also easy to prove directly that the Cauchy criterion for convergence is valid for a sequence of real real numbers $$\lim a^{\left(n\right)}\left(n=1,2,\dots\right).$$ For let us determine, corresponding to each value of $$n$$, a natural number $$n^{\prime}$$ such that $$\left|a_{n^{\prime}}^{\left(n\right)}-\lim a^{\left(n\right)}\right|<1/n.$$ Then, from the hypothesis of the Cauchy criterion that for $$\epsilon>0$$ there exists a natural number $$n_{0}$$ such that $$\left|\lim a^{\left(n\right)}-\lim a^{\left(m\right)}\right|<\epsilon$$ for $$n,m\ge n_{0},$$ we see that the sequence $$b$$ of $$b_{n}=a_{n^{\prime}}^{\left(n\right)}$$ is fundamental and that $$\lim b$$ is the limit of the sequence of the numbers $$\lim a^{\left(n\right)}.$$

I do not understand the meaning of $$n^{\prime}.$$ Is it a function $$n^{\prime}=\eta\left[n\right]:\mathbb{N}\to\mathbb{N}$$ which changes value in some unspecified way as $$n$$ increases; that is $$b_{n}=a_{\eta\left[n\right]}^{\left(n\right)}?$$ Assuming that to be the case, then are the $$b_{n}=a_{n^{\prime}}^{\left(n\right)}$$ necessarily rational numbers, or necessarily real numbers?

Unfortunately, there's really no other example of the use of this index notation in the book.

The only thing that comes to mind when I read that is: don't try this in Vagus, or you'll get kicked out of the casino.

## 2 Answers

The way I read it, each term in the Cauchy sequence of real numbers is denoted by $$\lim a$$, where $$a$$ is a sequence of rationals; the sequence of reals is indexed by superscripts, and the sequence of rationals that converges to each real is indexed by subscripts. So in what is hopefully slightly clearer notation, you're creating a sequence of reals $$\{(\lim_{m \to \infty}a_m)^{(n)}\}_{n=0}^\infty$$.

So $$n'$$ is an index in the sequence of rationals -- a value of $$m$$. Since $$\{a_m\}_{m=0}^\infty$$ converges to $$\lim a_m$$, you can choose $$n'$$ high enough such that $$a_n'$$ is arbitrarily close to $$\lim a_m$$. If you do that for every term in $$\{(\lim a_m)^{(n)}\}_{n=0}^\infty$$, you get a function that chooses $$n'$$ based on $$n$$, like you said.

And then $$\{b_n\}_{n=0}^\infty=\{a_{n'}^{(n)}\}_{n=0}^\infty$$ is a Cauchy sequence of rationals, so it converges, and it must converge to the same limit as the sequence of reals, meaning that the sequence of reals converges.

• I will have to play with this a bit. Apparently this is a case of nested epsilonics. In particular, my $\eta$ is more properly $n^{\prime}=\eta\left[n,\epsilon\right]:\mathbb{N}\times\mathbb{R}_{+}\to\mathbb{N}$. Also, once the theorem is proven, the rationals $b_n$ could just as well be reals, as could the $a_m$. But it appears the authors intended the subscripted variables to be understood as rationals. – Steven Thomas Hatton May 15 at 2:47
• Would you agree with this? Let $a^{\left(n\right)}={\lim_{m\to\infty}a_{m}^{\left(n\right)}}$ so that the real number $a^{\left(n\right)}$ is the limit of the rational number sequence $\left\{ a_{m}^{\left(n\right)}\right\} _{m=1}^{\infty}.$ In particular, we have $a_{n^{\prime}}^{\left(n\right)}\in\left\{ a_{m}^{\left(n\right)}\right\} _{m=1}^{\infty}.$ – Steven Thomas Hatton May 15 at 15:28
• Well, we know every real can be expressed as the limit of a sequence of rationals, which is how the $a_m$ are defined; that's because the goal is to turn a Cauchy sequence of reals into a Cauchy sequence of rationals, since a priori, we only know Cauchy sequences of rationals converge. But they can be anything after the theorem's proved. Also, $n'$ depends on $1/n$ instead of $\epsilon$ -- so $\eta$ just uses $n$ in two different ways. And yeah, that statement looks right to me! Proving $\{b_n\}_{n=0}^\infty$ is Cauchy is a fun exercise, if you feel like it. – Moriah May 15 at 19:18
• I see. The $n^\prime$ are determined by $\left|a_{n^{\prime}}^{\left(n\right)}-\lim a^{\left(n\right)}\right|<1/n.$ Which is independent of $\epsilon$. I haven't figured out the convergence of $b_n$ yet; but as a first attempt i will consider $\left|\left(a_{n^{\prime}}^{\left(n\right)}-\lim a^{\left(n\right)}\right)-\left(a_{m^{\prime}}^{\left(m\right)}-\lim a^{\left(m\right)}\right)\right|<\epsilon.$ – Steven Thomas Hatton May 16 at 11:02
• My advice would be to start with the end goal -- which distance you want to be less than $\epsilon$ so the sequence is Cauchy -- and then use the triangle inequality to break it down into distances you can make small, and then use that to find a large enough value for $N$ – Moriah May 16 at 17:14

This isn't really an answer. It is an elaboration on Moriah's answer, which I accepted. It's just easier to post in this format than as a comment, while it really doesn't belong in the original post. And posting a new question on the same proof would be redundant.

My proposed starting point for showing that $$\{b_n\}_{n=0}^\infty$$ is a Cauchy sequence may well be virtually synonymous with Moriah's suggestion. I suggested trying:

$$\left|\left(a_{n^{\prime}}^{\left(n\right)}-\lim a^{\left(n\right)}\right)-\left(a_{m^{\prime}}^{\left(m\right)}-\lim a^{\left(m\right)}\right)\right|<\varepsilon.$$

I changed $$\epsilon\to\varepsilon$$ to distinguish it from the following $$\epsilon$$. Write $$\lim a^{\left(n\right)}\equiv\alpha_{n}.$$ Then $$\left|\left(b_{n}-\alpha_{n}\right)-\left(b_{m}-\alpha_{m}\right)\right|<2/n_{\epsilon},$$

where $$n_{\epsilon}$$ is such that $$\forall_{n,m>n_{\epsilon}}\left[\left|\alpha_{m}-\alpha_{n}\right|<\epsilon\right].$$ Rearranging and applying a version of the triangle inequality produces

$$\varepsilon>2/n_{\epsilon}>\left|\left(\alpha_{m}-\alpha_{n}\right)-\left(b_{m}-b_{n}\right)\right|\ge\left|\left|\left(\alpha_{m}-\alpha_{n}\right)\right|-\left|\left(b_{m}-b_{n}\right)\right|\right|.$$

By hypothesis, the first term becomes arbitrarily small, so the second term must do likewise. Thus $$\left\{ b_{n}\right\}$$ is a fundamental sequence.

I believe this is sufficient to satisfy the first part of the original proposition; that is, $$b$$ is a fundamental sequence.

That $$\lim b=\lim\left[\lim a^{\left(n\right)}\right]$$ remains a bit murky, to the extent that I need to review the definition of a real number to convince myself that the limit of a sequence of real numbers satisfies the criteria for being a real number given in BBFSK.