# Math Help - sequences and continuity

1. ## sequences and continuity

Here are a few analysis problems I am having trouble with:

1. Let {a_n} be a sequence of real numbers and let r be a real number that satisfies 0 < r < 1. Suppose that | a_n+1 - a_n | <= r*| a_n - a_n-1| for n > 1. Prove that {a_n} is a Cauchy sequence and hence a sequence that converges to a limit.

2. Let {a_n} be a sequence of real numbers. Prove that if {a_n} has two subsequences that converge to different limits, then the sequence {a_n} does not converge.

3. Prove the following theorem: Let {a_n} be a bounded sequence of real numbers and let S be the set of sequential limits of {a_n}. Then the set S contains its greatest lower bound and its least upper bound and
lim inf a_n = inf S & lim sup a_n = sup S.

4. Let S be a bounded nonempty set of real numbers and suppose that supS is not an element of S. Prove that there is a nondecreasing sequence {s_n} of elements of S such that the limit as n approaches infinity of s_n is supS.

Thanks for the help.

2. Originally Posted by eigenvector11
1. Let {a_n} be a sequence of real numbers and let r be a real number that satisfies 0 < r < 1. Suppose that | a_n+1 - a_n | <= r*| a_n - a_n-1| for n > 1. Prove that {a_n} is a Cauchy sequence and hence a sequence that converges to a limit.
Proof

2. Let {a_n} be a sequence of real numbers. Prove that if {a_n} has two subsequences that converge to different limits, then the sequence {a_n} does not converge.
If $a_n$ convergese to $a$ then it means $|a_n-a| < \epsilon$ for $n>N$. If $a_{n_k}$ is a subsequence then $n_k \geq n > N$ thus $|a_{n_k}-a|<\epsilon$. This shows a convergent sequence has all its sequences converge to the same limit.

3. Originally Posted by eigenvector11
3. Prove the following theorem: Let {a_n} be a bounded sequence of real numbers and let S be the set of sequential limits of {a_n}. Then the set S contains its greatest lower bound and its least upper bound and
lim inf a_n = inf S & lim sup a_n = sup S.
Let $a_{n_k}$ be a subsequence having limit $L$. Then $\liminf a_{n_k} = \limsup a_{n_k} = L$ but since $\{ a_{n_k}| k\geq N\}\subseteq \{a_n| n\geq N\}$ it means $\liminf a_n \leq \liminf a_{n_k} = L = \limsup a_{n_k} \leq \limsup a_n$. Thus $\limsup a_n$ is upper bound on $S$ and $\liminf a_n$ is lower bound on $S$ this means $\liminf a_n \leq \inf S \leq \sup S \leq \limsup s_n$. Now there is a theorem that says that for any bounded sequence we can find two subsequences one subsequence convergening to its limsup and the other one converging to its liminf. This means the limsup and liminf belong to the set. Thus $\liminf a_n = \inf S$ and $\limsup a_n = \sup S$.

4. Originally Posted by eigenvector11
4. Let S be a bounded nonempty set of real numbers and suppose that supS is not an element of S. Prove that there is a nondecreasing sequence {s_n} of elements of S such that the limit as n approaches infinity of s_n is supS.
Let $S$ be a bounded non-empty set of real numbers. Then $\sup S = a$ is a real numbers. For $\epsilon = 1$ we have that $a - 1$ is not an upper bound (why is that?) thus there exists $s_1$ so that $s_1 \geq a-1$. Likewise $s_2\geq a - 1/2$, and in general $s_n \geq a - 1/n$. Thus we have $s_n - a \geq - 1/n \implies 0\leq a - s_n\leq 1/n$ and thus $|s_n - a| \leq 1/n$. This means $\lim s_n = a$. Now this constructed sequence is not necessary nondecreasing, try to correct this proof so that it is nondecreasing.

5. Thanks for the help on the first three. I'm not 100% sure how to connect the last proof to a nondecreasing sequence though. I know this means a_n+1 >= a_n, but I don't see how to tie this into the proof.

6. Originally Posted by eigenvector11
Thanks for the help on the first three. I'm not 100% sure how to connect the last proof to a nondecreasing sequence though. I know this means a_n+1 >= a_n, but I don't see how to tie this into the proof.
you said it, make sure that $s_{n + 1} \ge s_n$