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.
Best Answer
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.