Give an example of each of the following, or argue that such a request is impossible. a. A sequence that has a subsequence that is bounded but contains no subsequence that converges.
b. A sequence that does not contain 0 or 1 as a term but contains subsequences converging to each of these values.
c. A sequence that contains subsequences converging to every point in the infinite set {1, 1/2, 1/3, 1/4, 1/5,...}.
d. A sequence that contains subsequences converging to every point in the infinite set {1, 1/2, 1/3, 1/4, 1/5,...}, and no subsequences converging to points outside of this set.

Answer and Step-by-step explanation:

Solution:

(a)Let (bn) be a bounded sequence of (an) then, Bolzano- Weierstrass theorem,

(bn) contains a subsequence that is converges.

(b)The sequence :

Xn = 1 + (-1) n / 2 + 1/n

Clearly, (xn) does not contain 1 or 0.

If we take sequence (x2n), then it converges to 1 and when we take sequence (x2n+1),

Then it converges to 0.

(c)The sequence:

1

1,1/2

1, 1/2, 1/3

1, 1/2, 1/3, 1/4

i.e

(xn) = (1,1/2, 1, 1/2, 1/3, …)

Then, the subsequence (xn) that is identically ½ in second column and 1/3 is third column and so on.

It has sub sequence that converges to every point in infinite set {1, 1/2, 1/3, 1/4, 1/5…}.

(d)The sequence

{1, 1/2, 1/3, 1/4, 1/5…}

if we take sequence of given set, then, this set converges to zero. This is not contained in it.

If (xn) converges to every point in set {1, 1/2, 1/3 …} then there exists subsequence which converges to 0. And 0 does not belong to given set.

answer

false

explanation

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