Validation of Proof that Nested Interval Property implies Axiom of Completeness I know this is is about the millionth "proof that N.I.P implies A.o.C" questions, but I am wondering where my proof fails (I'm asking where because I'm quite confident it is wrong somewhere).
Proof:
Let there be some arbitrary set $A$. Let $a_n$ symbolize some element from this set $A$. Let $b$ be any upper bound on the set $A$.
Pick some arbitrary element from the set $A$, and label this element $a_1$. Pick an arbitrary upper bound and label this $b_1$. These two create an interval $[a_1, b_1]$. The distance between these points is $|a_1-b_1|$. Now, pick some new $a_2$ and $b_2$ that creates an interval $[a_2, b_2]$ such that the $|a_2-b_2| = \frac{|a_1-b_1|}{2}$ (half the size of the old interval) nested inside the interval $[a_1, b_1]$. Keep on making smaller and smaller nested intervals $[a_n, b_n]$ where $a_n$ and $b_n$ get larger, and the distance between them approaches 0 (since the sequence $\frac{1}{2^n}$ has a limit of 0). Because of nested interval property, there exists some point $x$ such that $a_n \le x \le b_n$ for all n.
The definition of $s = \text{sup} A$ is that $s \le b_n$ for all b, and $s \ge a_n$ for all a. This $x$ clearly satisfies both of these properties, and thus is the least upper bound of A. Since least upper bounds are unique, there is no different least upper bound.
--- end proof
It feels... not rigorous enough, but I can't exactly put my finger on where or why. I think I'm close-ish, though.
Thanks!
 A: The main issue with your proof is that you keep assuming certain things are true, even though you never specified that they be true, or showed that they must hold.
First, you should know the exact statement you are trying to prove. That is,

If $A \subseteq \Bbb R$ is bounded above and non-empty, then $A$ has a least upper bound $s$.

$A$ is not just "some arbitrary set". It has three properties it must satisfy for the Axiom of Completeness (aka, the supremum property) to be applicable:

*

*It must be a subset of $\Bbb R$.

*It must be bounded above.

*It must be non-empty.

Possibly the first could be considered implicit from the context you are working in. But even so, the other two should be explicitly stated. Arbitrary sets of real numbers can be the empty set, or be unbounded. You used these two properties several times, even though you didn't require them of $A$.
The next logical shortcoming is here:

Now, pick some new $a_2$ and $b_2$ that creates an interval $[a_2, b_2]$ such that the $|a_2-b_2| = \frac{|a_1-b_1|}{2}$ (half the size of the old interval) nested inside the interval $[a_1, b_1]$.

First, you need to require that $a_2 \in A$ and $b_2$ is an upper bound of $A$. (Those confused sentences immediately following your introduction of $A$ are not adequate for this - just say something like "pick $a_2 \in A, b_2$ an upper bound of $A$ such that ...".)
But the bigger problem is: how do you know that such $a_2, b_2$ exist? Why does there have to be an element $a_2 \in A \cap [a_1, b_1]$ and upper bound $b_2 \in [a_1, b_1]$ such that $|b_2 - a_2|$ is exactly half of $|b_1 - a_1|$?
(You don't need $|b_n - a_n|$ to decrease by exactly half each time. It is enough that it be bounded above by a sequence that goes to $0$. As a hint on how to do this more easily, consider the midpoint of $[a_1, b_1]$, and the two cases where the midpoint is an upper bound, and where it is not.)
Finally, this is wrong:

The definition of $s = \text{sup} A$ is that $s \le b_n$ for all b, and $s \ge a_n$ for all a.

That is not the definition of the supremum. $a_n, b_n$ are things you defined solely for this proof. The concept of the supremum is not defined in terms of these sequences. You must prove every element $a \in A$ satisfies $a \le s$. And you must prove that every upper bound $b$ of $A$ satisfies $s \le b$. It is not sufficient that it be true just for the ones you picked for your sequences.
