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.
