# Supremum of set

1. Sep 14, 2014

### nuuskur

1. The problem statement, all variables and given/known data
Find sup A if A = {0.2, 0.22, 0.222, 0.2222, ...}
I'll write elements of a set with low case letters and indexes, e.g an

3. The attempt at a solution
Begin by definition of supremum:
$\sup A = a$ if $\forall x \in A, x \leq a$ and $\forall b \in \mathbb R ((\forall x \in A , x \leq b) \rightarrow a \leq b$ essentially, a supremum is the lowest value of upper bounds of a set.
Also if $\exists \max A \rightarrow \sup A = \max A$

So I'll try to find a max A and prove sup A exists:
Let us have $\exists \max A = M$ such that $\forall x \in A, x \leq M$
Suppose that $a_n = M$. $a_n = \sum_{k=0}^n\frac{1}{5 \cdot 10^k}, n \in \mathbb N$. If $a_n \geq a_{n+1}$ then $\max A = \sup A = a_n$
$a_{n+1}= \sum_{k=0}^{n}(\frac{1}{5 \cdot 10^n})+ \frac{1}{5 \cdot 10^{n+1}} \rightarrow a_{n+1} > a_n$. Therefore max A does not exist.

However, I cannot conclusively prove that sup A does not exist at all for this set A. Intuitively I can see that the elements' difference is becoming smaller and smaller and smaller, hence they should eventually be limited to some specific value. Altho it says if max A exists, it's also the sup A, but it does not say if max A doesn't exist, then there is no sup A.

What am I missing?

EDIT:
Quick and dirty - If I assumed sup A = 0.23, for example, would it be sufficient evidence that sup A does not exist if I show there is a value 0.223 which is still greater than all the elements in the set A, but lesser than the supposed 0.23. Then it follows that I can suppose sup A = 0.222223, which still satisfies the upper bound criteria, however is not the least of the upper bounds. How can I show that there is no sup A in this case?

Last edited: Sep 14, 2014
2. Sep 14, 2014

### pasmith

The least upper bound axiom states that if $A \subset \mathbb{R}$ is bounded above then it has a supremum $M$. If it happens that $M \in A$ then $M$ is the maximum of $A$, but it may be that $M \notin A$ in which case $A$ has no maximum.

To your example: try summing the geometric series $$a_N = \sum_{n=1}^N \frac{2}{10^n} = 2 \sum_{n=1}^N \frac{1}{10^n}$$ for fixed $N$. Then consider what happens if you make $N$ arbitrarily large.

3. Sep 14, 2014

### HallsofIvy

Staff Emeritus
Let x= 0.2222....., never ending.

Then 10x= 2.222...., still never ending.

Subtracting, 9x= 2.

4. Sep 14, 2014

### Ray Vickson

Of course there IS a sup in this case; I don't know why you think otherwise. In fact, it is a property of real numbers that any bounded set of real numbers has a supremum. However, maximum and supremum need not be the same thing. In this case the set has no maximum, but that does not matter for the problem at hand.

Others have already showed you how to find the supremum.