Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

My Question: I am not sure about the very last inequality in the proof below; namely, where did we get $\mid a_{n}-a_{N}\mid$ and $\mid a_{N}-b\mid$? I see that $\mid a_{n}-a_{N}\mid<\epsilon/2$ and that $\mid a_{N}-b\mid\leqslant\epsilon/2$, which validates the last inequality, but i just don't see how we chose those two pieces. You can probably skip 2/3 or more of what I have written below to answer my question, I just wrote everything so that it would be clear where I am coming from. Also, I don't see why $\mid a_{n}-b\mid\leqslant\mid a_{n}-a_{N}\mid+\mid a_{N}-b\mid$ is true. Thanks for your time.

Notation let the sequence $a_1, a_2, ..., a_n=(a_n$) for $n\in \mathbb{N}$

Definition: The Cauchy condition states that if a sequence $a_{n}$ is convergent then the following condition must be true as well:

$$ (\forall\epsilon\in\mathbb{R}_{+})(\exists N\in\mathbb{N})(n,m\geqslant N\implies\mid a_{n}-a_{m}\mid<\epsilon)$$

Theorem: $\mathbb{R}$ is complete with respects to the Cauchy sequences in the sense that if $(a_{n})$ is a sequence of real numbers obeying the Cauchy condition then $(a_{n})$ converges to a limit in $\mathbb{R}$.

Proof. Let the set $A$ be the set of all the real numbers comprising the sequence $(a_{n})$, $$ A=\{x\in\mathbb{R}:(\exists n\in\mathbb{N})(a_n=x)\}$$

We first observe that the set $A$ is bounded, by taking $\epsilon=1$ in the Caunchy condition which implies that there must exist a $N_{1}\in\mathbb{N}$ such that for all $n,m\geqslant N_{1}$ , $\mid a_{n}-a_{m}\mid<1$. Now, clearly that the finite set $A^*=\{a_{1},a_{2},...,a_{N_{1}},a_{N_{1}}-1,a_{N_{1}}+1\}$ is bounded, since all finite sets are bounded. Let us say that all the elements of $A^*$ belong to to the interval $[-M,M]$. Since we require that for all $n,m\geqslant N_{1}$, $\mid a_{n}-a_{m}\mid<1$ must be true, it follows that for all $n\geqslant N_{1}$, the inequality $ \mid a_{n}-a_{N_{1}}\mid<1$ is also true. The latter inequality suggest that $A$ is also contained in the interval [-M,M], hence $A$ is bounded.

Next we consider the set, $$S=\{s\in[-M,M]:\exists\mbox{ infinitely many }n\in\mathbb{N},\mbox{ for which }a_{n}\geqslant s\}$$

That is that $a_{n}\geqslant s$ occurs infinitely many times in the sequence $(a_{n})$. Clearly $-M\in S$ and $S$ is bounded above by $M$. According to the least upper bound property of $\mathbb{R}$ we know that $S$ must have a least upper bound, i.e. , $\mbox{l.u.b.}(S)=b$ where $b\in\mathbb{R}$. Thus, we claim that the Caunchy sequence $(a_{n})$ converges to the limit $b$.

Given that for all $\epsilon>0$ we must show that there exists an $N$ such that for all $n\geqslant N$, $\mid a_{n}-b\mid<\epsilon$. The Cauchy condition provides us with a $N_{2}$ such that

$$m,n\geqslant N_{2}\implies\mid a_{n}-a_{m}\mid<\frac{\epsilon}{2}$$

All of the element of $S$ are less than or equal to $b$ so, $b+\epsilon/2\notin S$ . Thus if the element $b+\epsilon/2$ does occur in $(a_{n})$ then it could only occur finitely many times, and thus there must exist a $N_{3}\geqslant N_{2}$ such that $$n\geqslant N_{3}\implies a_{n}\leqslant b+\frac{\epsilon}{2}$$ Since $\mbox{l.u.b.}(S)=b$ , the number $b-\epsilon/2$ cannot be be an upper bound for $S$; thus there exists an element $s\in S$ such that $s>b-\epsilon/2$ which implies that $a_{n}\geqslant s>b-\epsilon/2$ occurs infinitely often. In particular, there exists a $N\geqslant N_{3}$ such that $a_{N}>b-\epsilon/2$ . Since $N\geqslant N_{3}$ , we have $a_{N}\leqslant b+\epsilon/2$ and so we have that $$a_{N}\in(b-\epsilon/2,b+\epsilon/2]$$ Moreover, since $N\geqslant N_{3}\geqslant N_{2}$ $$\mid a_{n}-b\mid\leqslant\mid a_{n}-a_{N}\mid+\mid a_{N}-b\mid<\epsilon$$ which verifies convergence.

share|cite|improve this question
It's a common trick: $|a_n-b|=|\color{maroon}{a_n-a_N}+\color{darkgreen}{a_N-b}|$. Note we just wrote $0=a_N-a_N$. Now apply the triangle inequality. – David Mitra Jul 15 '13 at 21:13
@DavidMitra Thanks, it makes sense now. – JimmyJackson Jul 15 '13 at 21:17
@JimmyJackson : normally people just say a metric space is "complete" as opposed to "complete with respect to Cauchy sequences". All that the theorem does is show that if you assume $\mathbb{R}$ satisfies the least upper bound property, then $\mathbb{R}$ is complete. It is also possible to prove the converse. I'm no logician, but I think the "truth" of either property must be taken as an axiom. – Stefan Smith Jul 16 '13 at 1:51
up vote 3 down vote accepted

The fact that


is just the triangle inequality, $|u+v|\le|u|+|v|$, with $u=a_n-a_N$ and $v=a_N-b$.

In the proof we start with an arbitrary $\epsilon>0$, and the goal is to show that there is some $N$ such that $|a_n-b|<\epsilon$ whenever $n\ge N$. We know that we can make $|a_n-a_m|$ as small as we like by taking $m$ and $n$ big enough, because the sequence is a Cauchy sequence. In particular, we can choose $M$ big enough so that $|a_n-a_m|<\frac{\epsilon}2$ whenever $m,n\ge M$. We could have used any positive number in place of $\frac{\epsilon}2$; the specific choice of $\frac{\epsilon}2$ is a matter of convenience, making the conclusion of the proof a little simpler than it might otherwise be. Suppose that instead of $\frac{\epsilon}2$ we had used some number $\delta_1>0$; then at this point we could assert that $|a_n-a_m|<\delta_1$ whenever $m,n\ge M$.

The bulk of the middle part of the proof is an argument showing that there are infinitely many terms of the sequence in the interval


Here again we could have used any positive number in place of $\frac{\epsilon}2$, and the argument would have worked equally well. Suppose that we’d used some $\delta_2>0$ instead; then at this point we’d have shown that there are infinitely many terms of the sequence in the interval $(b-\delta_2,b+\delta_2]$. In particular, there is an $N\ge M$ such that $a_N\in(b-\delta_2,b+\delta_2]$. This means that if $n\ge N$, then certainly $n,N\ge M$, so $|a_n-a_N|<\delta_1$, and moreover $|a_N-b|\le\delta_2$. Thus,


We wanted to make $|a_n-b|<\epsilon$, and $\delta_1$ and $\delta_2$ could be any positive numbers, so we might as well take $\delta_1=\delta_2=\frac{\epsilon}2$. We could have taken $\delta_1$ to be $\frac{\epsilon}3$ and $\delta_2$ to be $\frac{2\epsilon}3$ and arrived at the desired conclusion. For that matter, we could have taken $\delta_1=\delta_2=\frac13$ and arrived at the desired conclusion: if something is less than $\frac{\epsilon}3+\frac{\epsilon}3=\frac{2\epsilon}3$, then it’s certainly less than $\epsilon$.

The basic idea underlying the argument is that one way to show that two things are close to each other is to show that they are both close to some third thing. Here we want to show that $a_n$ and $b$ are close, so we show that there is an $a_N$ that’s close to both of them. Everything else is the detailed work needed to make ‘close enough’ precise and to show that there really is that third object close to both of them.

share|cite|improve this answer
Thanks, this clarified many things including the floating question in my head of why we chose $\epsilon /2$ (which I didn't even bother asking). I appreciate you taking the time to clarify this. – JimmyJackson Jul 15 '13 at 21:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.