Understanding the definition of Cauchy sequence

Solution 1:

In the comment you say you know what a converging sequence is. Over the reals a Cauchy sequence is the same thing. So why do we care about them, you might ask. Here is why:

Recall: A sequence $(a_n)$ of real numbers converges to the limit $a\in \mathbb R$ if $\forall \epsilon>0\ \exists N\in\mathbb N:\forall n\geq N\ |a-a_n|<\epsilon$.

You probably have seen examples of converging sequences. $\frac 1n$ is one of them, and the limit is $0$. I assume you have also seen the reason for that (it boils down to the Archimedean axiom). In general the game works as follows: You guess the right limit and then do some algebraic manipulation till you find a suitable $N$ for every possible $\epsilon >0$.

It might bother you that you have to know the limit before you can actually show something. Is there no other way to formulate convergence which doesn't rely on the right guess of the limit? Indeed there is: Cauchy sequences. Note that the limit doesn't show up in the definition and we can start proving things without assuming that some number is the limit. So what makes a limit converging? The difference of subsequent elements of you sequence should definitely be arbitrarily small. So a naive guess for a condition would be $$|a_n-a_{n+1}|<\epsilon$$ for all $n\geq N$. But this is not enough as you can see from the sequence $$s_n=\sum_{i=1}^n\frac 1i.$$ It turns out that the stronger assumption $$|a_n-a_{m}|<\epsilon$$ for all $n,m\geq N$ is enough. So you may not only compare two subsequent elements of your sequence but any two which appear after a certain time.

So we can relatively easy show that every converging sequence is Cauchy. What about the other direction, namely every Cauchy sequence converges? Well, depending on your set up you can define the reals precisely as the completion of $\mathbb Q$. And here lies the true strengh of Cauchy sequences:

Image you don't know yet what the reals are. You just know the rationals $\mathbb Q$ (you define the natural numbers via the Peano axioms and derive first the integers and then the rationals from there). Then you can write down the sequence $$a_1=1,\ a_{n+1}=\frac {a_n}2+\frac1a_n$$ This sequence converges to $\sqrt 2$, but wait: we don't even know what $\sqrt 2$ is. All we know are the rationals. But the seqeunce still converges, or doesn't it? The definition of convergence requires a limit but there is no suitable limit in $\mathbb Q$. But we can take the closest thing to convergence available: Cauchy sequences. We can show that the sequence is Cauchy. We then say in $\mathbb R$ any Cauchy sequence converges and have just defined the reals.

And this idea goes on. You can take any space you like (you probably don't know many examples yet, but you can take the space of all polynomials, or all continuous functions etc.) We can define distances (just as the absolut value on the reals) and can write down sequences. The notion of convergence does not always make sense. The notion of Cauchy sequence does.

Edit: Obviously there are other definitions of the real numbers. If you want try proving the following: Everey bounded sequence of real numbers has a converging subsequence $\Rightarrow$ every Cauchy sequence of real numbers converges.

Solution 2:

This example could clarify things a little bit.

Consider $(x_n)$ where $x_n = \frac{1}{n}$. To prove that this is a Cauchy Sequence, we approach as follows:

Let $\epsilon > 0$ be given. We are interested in showing that there exists an integer $N>0$ such that $m,n <N \Rightarrow |x_m-x_n| < \epsilon$. This would be equivalent to

$$|x_n-x_m| < \epsilon \iff \left| \frac{1}{n} - \frac{1}{m} \right| < \epsilon$$

And now since

$$ \left| \frac{1}{n} - \frac{1}{m} \right| < \frac{1}{n}+\frac{1}{m}$$

If we make $\displaystyle{\frac{1}{n}+\frac{1}{m} < \epsilon}$, the result should follow. This will happen if both $\displaystyle{\frac{1}{n} < \frac{\epsilon}{2}},$ and $\displaystyle{\frac{1}{m} < \frac{\epsilon}{2}}$, i.e. when

$\displaystyle{n > \frac{2}{\epsilon}}$, and $\displaystyle{m > \frac{2}{\epsilon}}$. So, we see that if $N$ is an integer larger than $\displaystyle{\frac{2}{\epsilon}}$ then $\displaystyle{m,n > N \Rightarrow |x_m - x_n| < \epsilon}$

NOTE: Also look at the diagram here to understand how you can determine if a sequence is not Cauchy.

Solution 3:

1) The def. assures you that the inequality will be true for indexes greater than the $\,N\,$ you found, so you're forced to have $\,n,m>N\,$

2) This will depend heavily on the seq. given. It usually depends some algebraic manipulations.

3) Perhaps: draw a straight line representing the real numbers, and an interval $\,I:=(-\epsilon\,,\,\epsilon)\,$ of length $\,epsilon >0\,$ around zero, thus: the existence of that $\,N\,$ and the seq. being Cauchy means that any pair of elements of the seq. with indexes greater than this number have a difference that is a number within $\,I\,$

Solution 4:

Put $$S_n = \{x_k\}_{k\ge n};$$ The sequence $x$ is Cauchy iff the diameter of $S_n$ decreases to 0 as $n\to\infty$. To wit, the "tails" of the sequence shrink to arbitrary smallness.