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Assoli18 [71]
3 years ago
9

Give an example of each of the following, or argue that such a request is impossible.

Mathematics
1 answer:
Masteriza [31]3 years ago
3 0

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.

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We just need to <u>plug in 7 for B! </u>

Let's do it !

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<em>We always start with the parenthesis first ! It's just a rule in math :) </em>

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