# Find Sup and prove it only using the basic definition

1. Sep 27, 2011

### 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.

1. The problem statement, all variables and given/known data
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: $supA= \frac{3+5^{1/2}}{2}$)

2. Relevant 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

3. 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.

2. Sep 27, 2011

### HallsofIvy

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

Last edited: Sep 27, 2011
3. Sep 27, 2011

### mkln

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:

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.

4. Sep 27, 2011

### mkln

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