0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $c = \sup A$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.