Cauchy sequences – Serlo

Motivation Bearbeiten

In the last chapter, we learned about convergence of sequences  . They were defined via the Epsilon-criterion, which says that   must hold for all but finitely many  . I.e. it holds for all   with some  . However, in order to apply this criterion, we need a candidate for the limit  . What if we don't have such a candidate? Or more precisely,

How can we show that a sequence converges, if there is no candidate for the limit?

A sequence converges, if the distance of   to   eventually tends to 0. In that case, also the difference between elements   tends to 0. Conversely, if we know that   goes to 0 if both   and   get large, then the amount of points where a possible limit or accumulation point   might be shrinks together to a point. So the sequence should converge, then

A sequence converges, whenever the difference of two elements eventually gets arbitrarily small.

The interesting question is now, what condition we have to put on   and  . If we just consider neighbouring sequence elements via setting  , we might run into trouble: consider the sequence

 

This is a harmonic series. The difference of neighbouring elements is  . But the sequence itself diverges to  . We need a stronger criterion, i.e. we need to consider more pairs   than just neighbours.

Derivation of Cauchy sequences Bearbeiten

We try to get a condition on  , which suffices for showing that a sequence converges. So let's take a sequence   converging to   and play a bit with it: The epsilon-definition of convergence reads

 

We fix  . Then, there is an index   depending on   with   for all   . What can we say about the difference   for  ? There is

 

So the triangle inequality for   implies:

 

Sequence elements coming after   are all separated by less than   . Visually, all sequence elements are all situated inside the interval   which has width  :

 
An Epsilon-neighbourhood of a

The distance between two points in this interval is less than   . So for a convergent sequence   we have:

 

The  , in here is a bit awkward to most mathematicians. They remove it by defining  . The function   is mapping   bijectively onto   . So instead saying "for all  ", we could also use the term "for all  ":

 

Sequences, which fulfil the above property are called Cauchy sequences . This definition does not require a limit   . A sequence which converges, fulfils the above property, so any convergent sequence is a Cauchy sequence. But seeing that any Cauchy sequence converges is not so easy. Generally, this is even wrong: Not every Cauchy sequence converges! However, it is true that every Cauchy sequence in   converges. In the rest this article , we successively construct a proof for this.

Hint

In the following section, we will again use   instead of  .

Definition of a Cauchy sequence Bearbeiten

An introduction to Cauchy sequences (video in German). (YouTube video by the channel Quatematik)

The definition for a Cauchy sequence reads:

Definition (Cauchy sequence)

A sequence   is called Cauchy sequence, if for any   there is a natural number   , such that   for all   .

Intuitively, a sequence is Cauchy, if the difference between any two elements gets arbitrarily small as the element indices go to  . Beware: it is a common mistake to think that only neighbouring elements must get close to each other. However, Cauchy sequences require all differences between elements to go to 0:

Example (A Cauchy sequence)

The sequence   with   converges, so it should be a Cauchy sequence. We may also directly check the definition: for any   , we need to find an   , such that for all   there is

 

We assume   . The case   works analogously by interchanging   and  . Certainly,

 

Now, we take   such that   (the Archimedean axiom allows us to do so). Then,   and hence   for all  . So for all   with   there is:

 

and we get that   is a Cauchy sequence.

Example (Not a Cauchy sequence)

The sequence   with   is not a Cauchy sequence. And it does not converge. We can check the Cauchy sequence definition directly: For any   and   we can choose   far enough away, such that   holds. This is particularly easy for  : any difference between to distinct elements is then  . So for a given  , we just set   and  . Then

 

And   is by definition not a Cauchy sequence.

Every convergent sequence is a Cauchy sequence Bearbeiten

We essentially already proved this within the "derivation of Cauchy sequences". But it's always a good idea, to write down one's findings in a structured way. So let's do this:

Theorem

Every convergent sequence is a Cauchy sequence.

How to get to the proof?

See "derivation of Cauchy sequences". The idea is to start with the  -definition of convergence and to directly prove that the Cauchy condition with   holds true. This will be the case for  .

Proof

Let   be any convergent sequence and   be given. Then, there is an   with

 

for all  . Let now   be arbitrary. Then,

 

Cauchy sequences are bounded Bearbeiten

Convergent series are bounded. And we can prove the same for Cauchy sequences:

Theorem (Cauchy sequences are bounded)

Any Cauchy sequence is bounded.

This should not come as a surprise: if the distance   is bounded, then, for a fixed  , there is now way for   to escape to  . Intuitively,   "catches"  . And if a Cauchy sequence converges (we will find out that this is always the case in  ), then it must also be bounded:

Proof (Cauchy sequences are bounded)

Let   be a Cauchy sequence. We know that for any   there is an   with   for all   . Now, we just fix   (actually, any positive real number is OK, here) and get some   with   for all  . Now, we fix   and get

 

for all  . So all   with   are "caught" inside the interval  . That means, after passing index  , all sequence elements are bounded from above by   and from below by  :

 

Only the sequence elements before   remain. But those are finitely many, namely  . A finite set of numbers is always bounded. So these "early sequence elements" are bounded from above by   and from below by   . For the entire sequence, there is

 

So the sequence   is bounded from above by   and from below by   .

Cauchy sequences with convergent subsequences also converge Bearbeiten

We would like to show that in  , a Cauchy sequence converges. This is a somewhat longer task. So we first take a smaller step and prove the following smaller theorem:

Theorem (Cauchy sequences with convergent subsequences also converge)

Any Cauchy sequence  , with a subsequence   converging to   also converges to  .

How to get to the proof? (Cauchy sequences with convergent subsequences also converge)

Let   be a Cauchy sequence and   a subsequence converging to  . For   , the difference   gets very small and for the convergent subsequence, the elements tend to  . The idea is not to fix an element   from the subsequence, which is close to   and catch all   after it by the Cauchy condition.

How close do we need to get? Suppose,   is given. We need to find some   such that   for all   . So the target inequality is

 

If   is an element of the subsequence   , this is not a problem. We could even get   or smaller. And if   is not part of the convergent subsequence? Then, we need to use the Cauchy-property. If we choose   large enough, there will be elements   of the subsequence with  . So we can bound   using the triangle inequality:

 

Both absolutes can be made arbitrarily small. Their sum must be smaller than  . This is fulfilled if we choose both absolutes to be smaller than   . sIn order to get   that small, we choose some   with   for all  . Such an   exists, as   converges to   .

Now the second absolute value: By the Cauchy sequence property, there must be some   with   for all  . In place of   , we choose  . So we need   . We know that  , since   is an ascending number of positive integers. Therefore, we add the requirement   , which implies  . If we now set  . Note: any   will do that job, no matter how big.

So far, the variable   only appeared within   . Here, we required   . So for   , there is only one condition:   . We satisfy it by choosing  . Now, we are ready to write down the proof.

Proof (Cauchy sequences with convergent subsequences also converge)

Let   be a Cauchy sequence and   a subsequence converging to  . Let   be arbitrary. By the Cauchy property, there is an   with   for all   . In addition, there is an   with   for all  . Let   be any natural number with   and   be arbitrary, as well. Then,

 

Every Cauchy sequence converges Bearbeiten

Converges of Cauchy sequences (video by the channel MJ Education, in German).

Now, that we have the smaller theorem above, we can use it to show the final theorem:

Theorem (Every Cauchy sequence converges)

Let   be a real Cauchy sequence. Then,   converges to some  .

This theorem is of particularly high value, since it allows to show convergence of a sequence without having a candidate for a limit.

Before proving it, we will stress out that there are sets where not every Cauchy sequence converges. For instance, we might want to replace the condition   by   (rational numbers). However, there are some possible limits, which are in  , but not in  . For instance, take   , which is the limit of the rational sequence

 

in  , this sequence converges to  . So it is a Cauchy sequence  . Since the Cauchy condition is sustained under the replacement  , this is also a Cauchy sequence in  . However, in  , the limit   is unique, so there cannot be a further limit  . Hence, we have found a Cauchy sequence, which does not converge. We conclude, there are sets like  , where Cauchy sequences may not converge. The crucial point her is that   is complete, while   is not. We have already encountered completeness within the Bolzano-Weierstrass theorem, where it was necessary for the theorem to hold true. Similarly, completeness is also necessary for a Cauchy sequence to converge.

Proof (Every Cauchy sequence converges)

Let   be a Cauchy sequence. Two subsections above, we proved that this implies boundedness. By the Bolzano-Weierstrass theorem, every bounded sequence has a convergent subsequence. So the Cauchy sequence has a convergent subsequence. One subsection above, we proved that any Cauchy sequence with a convergent subsequence also converges to some  . So   must converge to some  , which finishes the proof.