5

I'm trying to follow a proof that if a sequence is convergent, it is necessarily Cauchy. I think I understand the proof, but I want to be sure that I'm wording the logic correctly. For that reason, this proof will be a bit more wordy than what I would usually write.

Theorem: If a sequence, $(s_n)$, is convergent, it is Cauchy.

Proof: Let $(s_n)$ be a convergent sequence, and denote $\lim s_n$ by $s$. Per the definition of convergence, \begin{align*} \forall \epsilon > 0, \exists N, \forall n > N, \left \lvert s_n - s \right \rvert < \epsilon. \end{align*} (Side note: I've seen alternate variants of this, including with sequences of functions, between textbooks and lecture notes, and I'm unsure on whether we require $N$ to be a natural number. It seems to me that this wouldn't be necessary, as we could take it to be a complicated, non-integer expression involving epsilon, and the $n > N$ component of the sequence would simply apply the ceiling function to $N$. I'd appreciate if someone could shed some light on this, as well.)

From here, take $\epsilon > 0$ to be arbitrary. Since $s_n$ converges, we can take $\epsilon = \frac{\epsilon^*}{2}$, and choose $N$ so that $\forall n > N, \left \lvert s_n - s \right \rvert < \frac{\epsilon^*}{2}$. Based on the existence of this $N$, we can choose any arbitrary $m, n > N$ such that $\left \lvert s_m - s \right \rvert < \frac{\epsilon^*}{2}$ and $\left \lvert s_n - s \right \rvert < \frac{\epsilon^*}{2}$, so this holds $\forall m, n > N$. That $\left \lvert s_m - s \right \rvert < \frac{\epsilon^*}{2}$ also implies that $\left \lvert s - s_m \right \rvert < \frac{\epsilon^*}{2}$ by leveraging the properties of absolute value (factoring out a $-1$, using the fact that the absolute value of a product is the product of the absolute values, and then concluding that $\left \lvert -1 \right \rvert$ is equal to the multiplicative identity, $1$.

From here, we get \begin{align*} \left \lvert s_n - s_m \right \rvert & = \left \lvert s_n + \left(s - s\right) - s_m \right \rvert & & \text{Add and subtract $s$} \\ & = \left \lvert \left(s_n - s\right) + \left(s - s_m\right) \right \rvert & & \text{Rearrange} \\ & \leq \left \lvert s_n - s \right \rvert + \left \lvert s - s_m \right \rvert & & \text{Triangle inequality} \\ & < \frac{\epsilon^*}{2} + \frac{\epsilon^*}{2} & & \text{Choice of $N$} \\ & = \epsilon^* \end{align*} Then, since $\epsilon$, $m$, and $n$ were arbitrary, we have \begin{align*} \forall \epsilon > 0, \exists N, \forall m, n > N, \left \lvert s_m - s_n \right \rvert < \epsilon. \end{align*}

So, aside from whether this proof is sound, my questions are:

(1) After choosing an $N$ that guarantees an arbitrarily small absolute difference for $n > N$, am I simply "choosing" $n = m$? How, exactly, do I add a second variable into this? Or does the mere existence of $N$, which guarantees an infinite number of terms of the sequence, guarantee the existence of such an $m$?

(2) Does $N$ need to be a natural number?

(3) Was the logic behind concluding that $\left \lvert s_m - s \right \rvert < \frac{\epsilon^*}{2} \implies \left \lvert s - s_m \right \rvert < \frac{\epsilon^*}{2}$ sound?

Thanks in advance.

Clemens Bartholdy
  • 14,222
  • Your proof is fine. For 1), no, you're not 'choosing' $n=m$; $m$ and $n$ are distinct natural numbers, both however, are greater than $N$ which is a natural number. The thing is, if a sequence converges to a real number, its terms get closer and closer to that number which can be seen very clearly from a diagram (it's visually very clear). 2), yes, I think it should be a natural number because you're NOT looking at the terms of the sequences itself but rather than the term that corresponds to that natural number. 3), yes, the logic behind it was sound. Hope this helps. – Alea Jun 13 '18 at 19:28
  • Your proof is fine. For 1), no, you're not 'choosing' $n=m$; $m$ and $n$ are distinct natural numbers, both however, are greater than $N$ which is a natural number. The thing is, if a sequence converges to a real number, its terms get closer and closer to that number which can be seen very clearly from a diagram (it's visually very clear). 2), yes, I think it should be a natural number because you're NOT looking at the terms of the sequences itself but rather than the term that corresponds to that natural number. 3), yes, the logic behind it was sound. Hope this helps. – Alea Jun 13 '18 at 19:28
  • $N$ does not have to a natural number. Any real number will do. But if we have, for example,$ N=3/2 $ and $\forall n>N;(|s-s_n|<\epsilon /2$ then we also have $\forall n>1;(|s-s_n|<\epsilon /2).$ – DanielWainfleet Jun 14 '18 at 02:01
  • Does this answer your question? Prove that a Cauchy sequence is convergent – Clemens Bartholdy Aug 02 '22 at 10:52

2 Answers2

3

I would condense the proof to this:

If $s_n \to s$ then, for any $c > 0$ there is a $N(c)$ such that $ n > N(c) \implies |s-s_n| < c$.

Therefore, for $n, m > N(c)$ we have $|s_n-s_m| =|(s_n-s)-(s_m-s)| \le |s_n-s|+|s_m-s| \le 2c $.

Now replace $c$ by $\epsilon/2$ and we are done.

Note: I usually use $c$ instead of $\epsilon$ because it is easier to enter and sort of looks like it.

marty cohen
  • 110,450
1

$n$ and $m$ are any two numbers bigger than $N$.

Not sure why you use $\epsilon ^*$ (might as well just use $\epsilon$).

$N$ should be a natural number, because it tells us how far out in the sequence to go, and sequences are usually indexed by natural numbers...

Finally, $\lvert x\rvert =\lvert -x\rvert $, so what you did is fine.

Looks fine.