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I can understand the proof, which I could do myself:

$|s_n - s_m| = |s_n - s + s - s_m|$

$\Rightarrow |s_n - s_m| \leq |s_n - s| + |s_m - s| $

For some $\epsilon > 0, \exists\ \ N(\epsilon) \in \mathbb{N} s.t.$

$ |s_n - s_m| \leq \epsilon + \epsilon \ \ \forall \ \ m,n > N(\epsilon)$

$\therefore |s_n - s_m| \leq 2\epsilon$ where $2\epsilon \in \mathbb{R}$

The fact that the Cauchy sequence gradually closes up, i.e. the elements get closer together in obvious from the last point. $N(\epsilon)$ necessarily is a non-increasing function of $\epsilon$.

But I can't see this physically. I can think of instances in which the sequence may close up, then diverge, then close up again at infinity. What stops the sequence from coming as close as $\epsilon_1$ to the limit, then diverging, then coming back after some elements.

Of course, the Cauchy sequence does not allow this, but I don't see how this instance violates the basic definition of an existence of a limit. The basic definition simply requires that $\forall \epsilon >0\exists N \in \mathbb{N}\ s.t. |s_n -s| < \epsilon$. There is no rule on the $\epsilon$ increasing with $N$.

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Your definition in the last paragraph is wrong. The correct version is $\forall \epsilon>0 \exists N \in \mathbb{N}\forall n\geq N. |s_n-s|<\epsilon$. Notice the difference between the order of $\epsilon$ and $N$: not "there is an $N$ such that for all $\epsilon$" but "for all $\epsilon$ there is an $N$". – JiK Aug 19 '14 at 9:38
@Jik, thank you, I have corrected it. – SPRajagopal Aug 19 '14 at 11:20
up vote 1 down vote accepted

Remember that the definition of a sequence $s_n$ converging to $s$ means for every $\epsilon > 0$ there's some $N(\epsilon)$ so that $$ \lvert s_n - s \rvert < \epsilon $$ for every $n \ge N$.

The important part is that after some fixed point in the sequence all you do is get closer and closer to $s$. Intuitively I hope you can then see since the sequence only gets closer to $s$ all the terms after $s_N$ will be getting closer and closer as well (i.e. it's a cauchy sequence).

To target your specific counter example of a sequence that gets close to $s$ then gets really far from $s$ then converges back to $s$ again note again that the $N$ needs to be so that all terms after $s_N$ need to be getting closer and closer to $s$. So the $N$ that would need to be chosen would need to be at the point in the sequence when it goes back to converge a second time (since otherwise it wouldn't be true that $s_n$ is close to $s$ for every $n \ge N$.

I hope this is clear and helps. Let me know if this is confusing!

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It's not entirely clear. I'm trying to draw a picture so that I can specifically ask you my doubt. Lemme try and photograph one from my notebook. – SPRajagopal Aug 19 '14 at 11:02
Alright, I have a photograph. Here, the y-axis is $s_n$ and the x-axis is the indices. Is this sequence Cauchy? – SPRajagopal Aug 19 '14 at 11:10
@SPRajagopal if it converges to $0$ as its looks it might, then yes. Intuitively you can look at the ending of the sequence, each term is getting closer and closer to each other, which is exactly what we require. – DanZimm Aug 19 '14 at 11:13
But then, that happens only at the ending. The middle part does not have elements that close up. In fact, they diverge. – SPRajagopal Aug 19 '14 at 11:13
@SPRajagopal to diverge means to get unboundedly large or small, so this sequence does not diverge (in fact a sequence can only diverge "at the end" (meaning as it goes off to infinity). Also note that for a sequence to converge or be cauchy we simply care about the end game of the sequence, that is we need to find an $N$ so that after that $N$ everything is close. We dont care about the behavior before the $N$. – DanZimm Aug 19 '14 at 11:35

Suppose $x_n\to x$. Then, fix $\varepsilon>0$. There exists an $N\in\mathbb N$ such that for all $n\in\mathbb N$ $$n\geq N\implies |x_n-x|<\frac{\varepsilon}{2}.$$

Then, if $n,m\geq N$, $$|x_n-x_m|\leq |x_n-x|+|x_m-x|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.$$


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do you think it would be a mistake to take instead $|x_n -x| <\epsilon$ and show $|x_n -x_m|< 2 \epsilon?$ – Alex Aug 19 '14 at 10:27
This is absolutely no problem. Indeed, if $(x_n)$ is a Cauchy sequence, $(y_n)$ define by $y_n=\frac{1}{2}x_n$ is also a Cauchy sequence and reciprocally. So it's equivalent. – idm Aug 19 '14 at 10:30
That's what I thought, thanks – Alex Aug 19 '14 at 10:31
The question is not the proof. I wanted to know where my understanding of it is wrong. I had already supplied the same proof in the question. – SPRajagopal Aug 19 '14 at 10:46
Your proof is almost "correct" but very badly formulated. For example, it's it not "for some $\epsilon>0$, there is a $N\in\mathbb N$" but "for all $\varepsilon>0$ there exists a $N\in\mathbb N$". To prove it, you just have to take a random $\varepsilon>0$ and prove that for this $\varepsilon>0$, there is an $N\in\mathbb N$ s.t. for all $n,m\geq N, |s_n-s_m|<\varepsilon$. I just gave you an example of how to proceed in this type of proof. Moreover, like said JiK, your last sentence is wrong. Do you understand ? – idm Aug 19 '14 at 10:53

You seem to have a mis-conception about the Cauchy condition. Rather than address it I offer you a different way of looking at things which I hope will clarify everything. For a sequence $x_n$ consider the condition: $$\exists L \forall \varepsilon >0 \exists N s.t. |x_n-L|<\varepsilon \forall n>N$$ which is nothing but the familiar notion that $L$ is the limit of the sequence. Now change the quantifiers to get $$\forall \varepsilon >0 \exists L \exists N s.t. |x_n-L|<\varepsilon \forall n>N$$ which expresses something quite different. In the first you first fix the limit, and then you need to produce $N$ for every $\varepsilon $. In the second one you first get the $\varepsilon $, then you may choose $L$ and then produce the $N$. So if the first expression is the definition of limit, then the second condition is the definition of a sort of varying limit. Now, it's an easy exercise that the second condition is equivalent to the condition of the sequence being Cauchy.

Thus the difference between a sequence being Cauchy and converging is in the order of quantification. Notice that it is now a triviality that if a sequence converges then it is Cauchy.

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This is my counterexample that I'm trying to understand. I have a photograph. Here, the y-axis is $s_n$ and the x-axis is the indices. Is this sequence Cauchy? If it is, then the middle part does not conform to the necessity of Cauchy sequences, and yet this sequence does converge in the end. – SPRajagopal Aug 19 '14 at 11:16
@SPRajagopal Your picture does not give a counterexample. In the definition of a Cauchy sequence, you can pick $N(\epsilon)$ beyond $N_3$. – Tunococ Aug 19 '14 at 11:26
not all the elements in the sequences need to get close to anything. Only from some $N$ onwards. In general, convergence and Cauchyness are unaffected by changing, or discarding, finitely many elements in the sequence, no matter how large the number of changes you make. – Ittay Weiss Aug 19 '14 at 11:26
Understood. Thank you. I made the mistake of assuming that the Cauchy requirement was supposed to hold for every $N$, but it is supposed to hold for every $\epsilon>0$. – SPRajagopal Aug 19 '14 at 12:01

Suppose a sequence converges to a limit. This means that for any $\epsilon$, we can find an segment of length $2\epsilon$ in which the sequence will eventually lie (the limit being on the middle of the segment). Once its in there, the elements of the sequence can't be further than $2\epsilon$ from one another (because the furthest any two elements could get would be one on each end of the segment).

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