Constructing an interval by uniting smaller intervals

[Whistlekins]

Homework Statement

We have $C_n = [1-\frac{1}{n},2-\frac{1}{2n}]$ and $C = C_1 \cup C_2 \cup C_3 \cup ...$ and are asked to describe the interval C and then prove that it is actually what we say it is.

Homework Equations

The Attempt at a Solution

I am guessing that $C = [0,2)$ and to prove this, I need to show that the infimum and minimum is 0, supremum is 2, maximum does not exist.

Am I right in describing this interval? And what other features of this interval would I need to prove in order to prove my description is valid?

 

[Bacle2]

Why do you think 2 is not included in the interval? What is the limit of 2-(1/2n)? If 2 is not included,

in C, can 2- 1/2n converge to 2?

 

[Whistlekins]

Why do you think 2 is not included in the interval? What is the limit of 2-(1/2n)? If 2 is not included,

in C, can 2- 1/2n converge to 2?

I don't really understand what you mean. 2 cannot be in C because 2-1/2n -> 2 as n -> ∞, so C_n = (1,2) as n -> ∞, wouldn't it?

 

[SammyS]

I don't really understand what you mean. 2 cannot be in C because 2-1/2n -> 2 as n -> ∞, so C_n = (1,2) as n -> ∞, wouldn't it?

I doubt that Bacle2 is saying that 2 is in interval C. He's just asking how to prove that you can have both

2 is not in C.

and

$\displaystyle \lim_{n\to\infty}(2-1/(2n))=2$ ?​

 

[Whistlekins]

I doubt that Bacle2 is saying that 2 is in interval C. He's just asking how to prove that you can have both

2 is not in C.

and

$\displaystyle \lim_{n\to\infty}(2-1/(2n))=2$ ?​

Well, now I have no idea.

How else could I show that 2 is not in C? Unless 2 IS in C...

 

[Whistlekins]

If I make $A$ the set of all maxima of all $C_n = \{ 2-\frac{1}{2n}\}$ for all natural numbers n,

Then the maximum of this set $A$ does not exist since it is infinite. However the least upper bound would be the limit = 2. Am I correct in thinking this?

Then sup(C) = sup(A) = 2. Is this reasoning correct?

 

[Bacle2]

Sorry if my statement was not clear. What I meant was that if 2 were included in the union, then it would be in some of the intervals, by def. of union. Then, re this last, the issue of convergence ,and (2-1/2n) being strictly increasing, come into play.