Given a sequence ((xn,yn)) is R2 .prove that the following,
1. ((xn,yn)) is convergent implies ((xn,yn)) is bounded.
2. If ((xn,yn)) is a bounded sequence the ((xn,yn)) has a convergent subsequence
3. ((xn,yn)) is convergent if and only if ((xn,yn)) is bounded and every convergent subsequence of ((xn,yn)) has the same limit.
4. ((xn,yn)) is caught if and only if ((xn,yn)) is convergent
Solution:
1.
Statement: Every convergent sequence is a bounded sequence, that is the set is bounded.
Proof : Suppose a sequence converges to x. Then, for , there exist N such that
This implies for all . If we let
then for all n. Hence is a bounded sequence.
2.
Statement: Every bounded sequence contains a convergent subsequence.
Proof. Let be a bounded sequence. Then there exists some M>0 such that for all . This implies that for all .
- We bisect the interval [-M, M] into two closed intervals of equal length. One of these intervals must contain infinitely many terms of the sequence . Let be that interval, and let be any point in .
- Next we bisect into two closed intervals, and note that one of these intervals must contain infinitely many terms of the sequence . Let be that interval, and choose inside this interval which satisfies
- In general, we bisect into two closed intervals, one which must contain infinitely many terms of . Let be this closed interval, and choose such that . Therefore we have obtained a subsequence of and a sequence of nested intervals
By the Nested Interval Property (i), the intersection is nonempty, and therefore contains some element
We claim that converges to . Let be arbitrary. Then the length of the interval is . By (i), the sequence converges to 0. Therefore, by the Algebraic Limit Theorem, the sequence converges to 0 as well. Hence, there exists some such that if , then But then , hence . Therefore converges to .
3.
Statement: Let be a bounded sequence all of whose convergent proper subsequences converge to . Prove that converges to .
Proof: A bounded sequence of real numbers must have at least one convergent subsequence. All of those converge to , and there is at least one of those; call it . So for every , all except finitely many terms of are in the interval with endpoints . If infinitely many terms of the original sequence lie outside that small interval, then that is a bounded sequence, which therefore has at least one convergent subsequence. That subsequence converges to some point outside that interval small interval. But that contradicts the hypothesis.
4.
Statement: Cauchy Convergence Criterion: A sequence \left(x_{n}\right) is Cauchy if and only if it is convergent.
Proof: Suppose is a convergent sequence, and . Let We can find such that for all . Therefore, by the triangle inequality, for all . So is Cauchy.
Conversely, suppose is Cauchy. Let . By a result proved in class, is bounded. By Bolzano-Weierstrass, it has a convergent subsequence with for some . We can find such that for all . We can also find M such that for all . Let . Then since for all k, if , we have that k, . Therefore, for all by the Triangle inequality.
Therefore, is Cauchy.
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