Find Sup and prove it only using the basic definition

[mkln]

Hi everyone. I am trying to solve this problem, but I cannot manage to get a satisfactory solution. It's actually interesting, if only it was for self-knowledge and not for a grade.

Homework Statement

Let A be in R and such that:
$A:=\{a_1, a_2, ...\}$
with
$a_1 = 1;$
$a_{k+1} = 1+ a^{1/2}_k$

find supA. (easy: $supA= \frac{3+5^{1/2}}{2}$)

Homework Equations

prove that what you found is actually the sup of A by using the definition of supremum, that is:
(i) it is an upperbound
(ii) it is smaller than or equal to any other upper bound

or (ii)' any smaller number is not an upper bound

The Attempt at a Solution

I'm stuck! I can show that $\frac{3+5^{1/2}}{2}$ is an upper bound.

Assuming that A has a supremum, and calling it $\gamma$ then I cannot show that $\gamma = \frac{3+5^{1/2}}{2}$

The problem to me is proving that there can be no element of A between $\gamma$ and $\frac{3+5^{1/2}}{2}$

I know there are elements of A just before $\gamma$ and I also know that there are no elements of A after $\frac{3+5^{1/2}}{2}$
I cannot show that if $\gamma < \frac{3+5^{1/2}}{2}$ then there will be a "jump" from one element of A smaller than gamma to an element of A larger than gamma.
I'm left with this "gap".

Constraint: I am supposed to show this using ONLY Rudin, Chapter 1 (so no limits, no convergence, no sequence concepts. Just sets and supremum


Thank you!

M.

 

[HallsofIvy]

If you are not to use limits, how did you find sup(A) so "easily"?

 

[mkln]

Well, how did you determine that the supremum was $(3+ \sqrt{5})/2$? I would bet that you did it by showing that $\{a_n\}$ was an increasing sequence having, say, 3, as upper bound and so converges. Once you know it has an upper bound, it must be true, by taking the limit on both sides of $a_{k+1}= 1+ \sqrt{a_k}$, that the upper bound, $\gamma$ satisfies $a= 1+ \sqrt{a}$. That is the same as $a- 1= \sqrt{a}$ and, squaring both sides, $a^2- 2a+ 1= a$ so $a^2- 3a= -1$. Completing the square, $a^2- 3a+ 9/4= -1+ 9/4= 5/4$. So $a- 3/2= \pm\sqrt{5}$ and $a= (3\pm\sqrt{5})/2$. Since $\{a_n\}$ starts with 1 and increases, it cannot converge to $(3- \sqrt{5})/2< 1$ so it must converge to $(3+ \sqrt{5})/2$.
Is that what you did- or did you use the quadratic formula?

But "converges to" means "for any $\epsilon> 0$ there exist N such that if n> N, $|a_n- (3+\sqrt{5})/2|< \epsilon$. Since $\{a_n\}$ is an increasing sequence, $a_n< (3+ \sqrt{5})/2$ and that reduces to $(3+ \sqrt{5})/2-a_n< \epsilon$.

Okay, if there exist an upper bound, $\gamma< (3+ \sqrt{5})/2$, take $\epsilon= (3+ \sqrt{5})/2- \gamma$ and show that you get a contradiction.

I found the supremum thinking of the expression as being a difference equation, but this does not matter.
The problem with this problem (sic) is that we are supposed to limit ourselves to sets and the definition of supremum.
Therefore we cannot use concepts such as sequences, or limits.
In particular, let me quote:

Note that ... you can use
the monotone convergence theorem ... in order to figure out the answer.

... you need to verify that this number is indeed sup A
using only material from Chapter 1, i.e. you need to verify
the properties (i) and (ii) in Definition 1.8.

In other words, I need to show that that number is an upper bound smaller than or equal to all other upper bounds.

I really wish I could use limits, but the problem explicitly excludes this possibility.

 

[mkln]

If you are not to use limits, how did you find sup(A) so "easily"?

We're not even supposed to show how we find that number. We are only asked to prove it is the supremum.