need number 11 and 12? xn = (-1)n(1-1/n) xn = sin npi xn = sin(npi/2) + cos npi
ID: 2987507 • Letter: N
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need number 11 and 12?
xn = (-1)n(1-1/n) xn = sin npi xn = sin(npi/2) + cos npi xn = n/n2 + 1 xn = n2/n2 + 1 xn = n3/n2 + 1 xn = n2 - n/n3 + 1 Let S be a bounded set of real numbers. Let A be the set of its points of accumulation. That is. A consists of all numbers a epsilon R such that a is a point of accumulation of an infinite subset of S. Then A is bounded. Assume that A is not empty. Let b be its least upper bound. Show that b is a point of accumulation of S. Usually, b is called the limit superior of S, and is denoted by lim sup S. Let c be a real number. Prove that c is the limit superior of S if and only if c satisfies the following property. For every epsilon there exists only a finite number of elements x epsilon S such that x > c + epsilon, and there exist infinitely many elements x of S such that x > c - epsilon. Let {an} be a bounded sequence of real numbers. Let A be the set of its points of accumulation in R. Assume that A is not empty. Let b be its least upper bound. Show that b is a point of accumulation of the sequence. We call b the limit superior of the sequence, denoted by lim sup an. Let c be a real number. Show that c is the lim sup of the sequence {an} if and only if c has the following property. For every epsilon, there exists only a finite number of n such that an > c + epsilon, and there exist infinitely many n such that an > c - epsilon. If {an} and {bn} are two bounded sequences of numbers, show that Define the limit inferior (lim inf). State and prove the properties analogous to those in Excercise 12.Explanation / Answer
11)a) A is bounded, so it contains its own least upper bound. b is the least upper bound of A, so it is an element of A. Every element of A is a point of accumulation of S. b is a point of accumulation of S.
11) b) Suppose that c is the limit superior of S. Then, it is an accumulation point of S, and every open interval containing it contains at least one other point of S. Suppose that for some e there are an infinite number of points of S in (c, c+e). Then, it follows that there exists a point in (c, c+e) that is an accumulation point of S, and c is not the limit superior. Given that there are finitely many points of S in (c, c+e), it follows that there is a least point of (c, c+e) which is in S, so for any e, there must be infinitely many points of S in (c-e, c).
To prove the inverse relationship:
Suppose there are infinitely many points of S in (c-e, c) and finitely many points of S in (c, c+e) for any e. Then c is an accumulation point of S, and there are finitely many points of S in any interval above c. This implies there are no accumulation points of S above c (as any point s above c has only finitely many points of S in (c, 2s - c), so there is at least one point closest to s in that interval, and an interval that doesn't contain the closest point of S contains no points of S.) Thus proving that c is the limit superior of S.
12) a) Essentially identical to 11)a); it's in the set of points of accumulation, so it must be a point of accumulation.
12)b) Essentially identical to 11)b); the limit superior is the maximum point of accumulation, so there are only finitely many points above it in a finite interval above c, and being the maximum point of an infinite sequence implies it is the maximum point of accumulation, and thus that it is the limit superior.
12)c) Suppose that lim sup (an+bn) > lim sup (an) + lim sup (bn). Then, an and bn must togther contain infinitely many points greater than their own lim sups, which is a contradiction by part b.
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