Real Analysis : Connectedness and Convergent Sequence

Show that A set E subset or equal to R is connected if and only if, for all nonempty disjoint sets A and B satisfying E=A U B there always exists a convergent sequence (x_n)-->x with (x_n) contained in one of A or B and x an element of the other.

Solution Summary

The relationship between connectedness and a convergent sequence is investigated. The solution is detailed and well presented.

Give an example of each of the following, or argue that such a request is impossible:
1) A sequence that does not contain 0,1 as a term but contains subsequences converging to each of these values.
2) A monotone sequence that diverges but has a convergent subsequence.
3) A sequence that contains subsequences converging to

29.18
Let f be a differentiable on R with a = sup {|f ′(x)|: x in R} < 1.
Select s0 in R and define sn = f (sn-1) for n ≥ 1. Thus s1 = f (s0), s2 = f(s1), etc
Prove that (sn) is a convergence sequence. Hint: To show (sn) is Cauchy, first show that |sn+1 - sn| ≤ aּ|sn - sn-1| for n ≥ 1.

Let {fn} be a sequence of continuous functions, all defined on [a,b]. Suppose {fn(a)} is a diverging sequence of real numbers. Prove that {fn} does not converge uniformly on (a,b]. (Notice a is not included in this interval.)
Assume {fn} converges unifromly and then use the diverging sequence of real numbers as a contradict

See attached file for all symbols.
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? For each of the follwing statements decide if it is true or false. Justify your answer by proving, or finding a couter-example.
1) every bounded sequence of real numbers is convergent.
2) Every convergentsequence is monotone.
3) Every monotone and bounded sequence of real number

Prove that if {f_n:R→R} is a sequence of continuously differentiable functions such that the sequence of derivatives {f_n^':R→R} is uniformly convergentand the sequence {f_n (0)} is also convergent, then {f_n:R→R} is pointwise convergent. Is the assumption that the sequence {f_n (0)} converges necessary?
Pl

Suppose the lim (subscript n --> infinity) sup a_n = theta is finite. Prove there exists a subsequence that converges to theta. Please view the attachment to see the full question, including necessary symbols.

Theorem: Suppose that a sequence S of real numbers has a subsequence that converges to a real number a. Then S converges to a.
I know this is true as an if and only if statement, but I need a counter example to show that just one converging subsequence isn't enough.
Here are two sequences I'm considering: {1,-1,1,-1,1,-1..

Discuss convergence or divergence of the series whose nth term is
〖(-1)〗^n n^n/((〖n+1)〗^(n+1) ) (b) 〖(-1)〗^n 〖(n+1)〗^n/n^n
(c) n^n/((〖n+1)〗^(n+1) ) (d) 〖(n+1)〗^n/n^(n+1)
Given that ∑▒a_n is a convergent series of real numbers, Prove