Cauchy Sequences and Convergence

Recall: For a bounded sequence :

Def: is Cauchy if for all we can find an such that , for all

Thm: A convergent sequence implies that is Cauchy

Proof: s.t.

For any , we have

So we have, for any ,

  • , since we always have

We proved that bounded and implies convergence, and convergence implies Cauchy, so now we will prove Cauchy implies .


Cauchy Implies Infimum = Supremum

Def: A subsequence of a sequence is a sequence where are positive integers, e.g.

In homework, we proved that


Subsequences Approach Sequence Limit

Question: Suppose , is it true that for any subsequence ?

Yes: s.t.

The intuition is that there exists a that is greater than , at which point the bounds will hold true.

And our sequence is , which contains strictly increasing , so such that . Then, for any , we know since .

The converse is not true. Consider . The odd terms converge to , but the even terms to . Note that and . We will prove that for any bounded sequence, there exists a subsequence that converges to and one that converges to .

Another question: If , does this imply that ? It turns out the answer is yes, and the proof is left as an exercise.


Bolzano Weierstrass

Definition: Any bounded sequence has a convergent subsequence

It suffices to prove any bounded sequence has a monotone subsequence. If we can bound this monotone subsequence, it will be monotone and bounded, hence convergent.

Lemma: Any bounded subsequence has a monotone subsequence.

Proof:

Definition: Given a bounded sequence , we say is big if it is bigger than all the terms after it, i.e. .

For any bounded sequence ,

  1. It has infinitely many big terms, or
  2. It has finitely many big terms.

In the first case, the sequence has infinitely many big terms, so we can just take all the big terms: , the sequence of big terms from . This gives a subsequence of , a strictly decreasing subsequence (since each term is strictly greater than the term after).

In the second case, the sequence has finitely many big terms, so we take . Consider only for . These are not big for all of these s.
Now we define . Since , is not big. This implies that there exists such that . Since , is not big. Using the same argument, we can construct , and construct the remaining inductively. This creates a monotonically increasing subsequence.


Bounded to Limsup and Liminf

Theorem: For a bounded sequence, there exists two subsequences that converges to and , respectively.

To prove this theorem, first we prove a lemma:


Lemma: Subsequence iff Infinite Set Within Neighborhood

Lemma: Having a subsequence of a sequence converging to is equivalent to , the set is infinite.

Proof:

(→): converges to :

s.t.

which implies that

(←):
Idea: We’ll find a subsequence () such that the differences are less than 1, , … .

  • We know that is infinite.
  • We know that is infinite, so s.t. .
  • We know that is infinite, so s.t. .
  • etc.

Now back to the original proof:

Proof:

Let
Want to show: , the set is infinite.

Proof by contradiction. Suppose that is finite.

  1. such that . This implies that for all .
  2. Having a finite set means that such that is false .

Now let . For any , we have

  1. is false.

which implies that , which means . But recall that , which is a contradiction.