Axiom of completeness: Every set that is bounded above has a least upper bound.

My attempt:

Let $A = \{a_n : n \in \mathbb{N}\}$, satisfying $a_1 < a_2 < a_3 < \cdots a_n$, $b_n$ be a sequence such that $b_1 > b_2 > b_3 > \cdots$ and consider the following closed intervals in $\mathbb{R}$:

$I_1 = [a_1, b_1],I_2 = [a_2, b_2],I_3 = [a_3, b_3],\cdots, I_n = [a_n, b_n]$. By the nested interval property, there is $x = {\bigcap}_{k=1}^\infty I_k$. Further, since the length of $I_n$ tends to $0$, x is unique.

I claim that $x = \sup A$. Proof:

Let $y \neq x$ be an upper bound for $A$. Then $y \geq a_n \forall n\in\mathbb{N}$. Now, since $y \in \mathbb{R}$, then either $y \in I_k$ for some $k \in \mathbb{N}$ (case 1), or $y \notin I_k, \forall k \in \mathbb{N}$ (case 2).

Case 1:

Assume $y \in I_k$ for some $k \in \mathbb{N}$. Then there is no $I_w$, $w > k$, such that $y \notin I_w$, otherwise we must have $a_w > y$, and then $y$ wouldn't be an upper bound. Further, since $I_k \subseteq I_p$ for all $p < k$, we must have $y \in I_n, \forall n \in \mathbb{N}$. Therefore $y = x$.

Case 2:

Assume $y \notin I_k, \forall k \in \mathbb{N}$. Since $y$ is an upper bound, we must have $y > b_1$. But $x = {\bigcap}_{k=1}^\infty I_k$, so $x \in I_2$, therefore $y > x$.

Thus, if $y$ is an upper bound, then $y \geq x$ (I omitted the proof that $x$ is an upper bound because if follows from a short contradiction argument).

Is all of this correct?


Your first step won’t work for sets that aren’t countable. Your second step won’t work for $A=\{1,1/2,1/3,1/4,\dots\}.$

Strictly speaking your $x$ is defined to be a set but you use it as a number. Your proof for case 1 is incorrect. Your proof for case 2 probably needs a bit more care to make it convincing.

This proof is nearly totally wrong in the details that matter and irrecoverably so.

Here is a sketch of how to prove it:

Correct statement of axiom of completeness:

every nonempty set bounded above has a least upper bound.

Proof that NIP $\Rightarrow$ AOC:

Let $A$ be bounded above and nonempty so $x\in A$ and $y$ an upper bound. Then $x\ne y$ and $[x,y]$ meets $A$. Say $I_1=[x,y]$ and inductively construct $I_{n+1}$ from $I_n=[a,c]$ (with $a< c$ as follows: Let $a<b=\frac12(a+c)<c$. If $b$ is an upper bound for $A$ then set set $I_{n+1}=[a,b],$ which meets $A$ and contains an upper bound of $A$. Otherwise there must be some $\alpha\in A$ such that $b<\alpha\le c$ (note we will have $c$ always an upper bound for $A$) and so set $I_{n+1}=[\alpha,c].$

Here is what’s left: observe each $I_n=[a,c]$ has $a\in A$ and $\alpha\le c\forall\alpha\in A$. Make a statement about the sizes of the $I_n$. Apply the nested interval property and extract a number $u$. Prove that $u$ is an upper bound. Prove that $u$ is the least upper bound.

  • $\begingroup$ Thanks. I did manage to work with all that and write a correct proof, so I think the problem's done. $\endgroup$ – Matheus Andrade Dec 27 '17 at 17:27

Well, the proof really was was completely wrong. Following the steps pointed out, I managed to write a correct proof.

" Let $A$ be a set bounded above by $b$ and let $a$ be some number less than some element of $A$. Consider the interval $[a,b]$. If the midpoint $c=(a+b)/2$ is an upper bound for $A$ then choose $a_1=a,b_1=c$ otherwise choose $a_1=c,b_1=b$. Using same procedure get $a_2,b_2$ from $a_1,b_1$ and repeat the process indefinitely to get a sequence of nested intervals. There is a unique c which lies in all these intervals and one can easily show that $c=\sup A$ (how?)"

Since the length of the intervals $\left(\cfrac{b_1 - a_1}{2^{n-1}}\right)$ go to zero, then $\lim b_n - a_n = 0 \Rightarrow \lim a_n = \lim b_n$. I claim $ c = \lim a_n = \lim b_n = \sup A$. Proof:

Let $a \in A$ be an element of A. Since $a \leq b_n \forall n $, then $a \leq c$ as well. If $y$ is an upper bound for $A$, then $y \geq a_n \forall n$, therefore $y \geq c$. So $c = \sup A$.

  • $\begingroup$ Isn't the argument a bit circular since the claim that the length of intervals $(b_1-a_1)/2^{n-1}$ goes to zero seems to depend on the Archimedean property, which is a consequence of axiom of completeness (but not nested intervals property). $\endgroup$ – Poincare-Lelong Jun 18 '18 at 5:11

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