# Deriving least upper bound property from one particular axiom of completeness

Let's have this axiom of completeness for the reals:

Axiom: Let $$(I_n)$$ be a sequence of closed intervals in $$\mathbb{R}$$ such that $$\forall n\in\mathbb{N}:I_{n+1}\subset I_n$$. Also let $$\lim_{n\to\infty} |I_n|=0$$. Then $$\bigcap_{n=1}^\infty I_n\neq \emptyset$$

Now, I wish to prove the least upper bound property from this. That is, each set $$A\subset \mathbb{R}$$ bounded above has a least upper bound (supremum). By definition, $$c$$ is a supremum if $$c$$ is upper bound and for any other upper bound $$c'$$ it holds, that $$c\leq c'$$. We assume that $$(\mathbb{R},+,\cdot,\leq)$$ is given axiomaticaly together with the completeness axiom given. My go:

Let $$A\subset \mathbb{R}$$ be non-empty and bounded above by $$M$$. Choose arbitrary $$a_0\in A$$ and $$b_0:=M$$. Set $$m:=\frac{a_0+b_0}{2}$$. If $$m$$ is an upper bound, then set $$b_1:=m$$ and $$a_1:=a_0$$, else $$b_1:=b_0$$ and then pick $$a_1\in A$$ such that $$a_1>m$$. Now, inductively construct a sequence of self-contained closed intervals $$I_n=\langle a_n,b_n\rangle$$ whose lenghts converge to $$0$$. By Axiom, $$\bigcap I_n$$ is non-empty, so has some element $$c$$. Now, I claim that $$c=\sup{A}$$.

Now, from here I am a bit stuck. I wish to show that this found $$c$$ really is a supremum. By construction, all of $$b_n$$'s are upper bounds for $$A$$ now, how do I proceed? Now, the step "pick '$$a_1$$' is also weird for me, since we do not know anything about the set (it doesnt have to be interval), so I would like a constructive proof too.

• When you say '...pick $a_1 \in A$' it starts sounding a bit spooky (no precise algorithm to sink your teeth into). If you change your question, asking for an alternate construction-proof, I will be able to supply an answer. – CopyPasteIt Oct 16 '18 at 15:23
• I was actually unsure about that step, looked weird for me aswell. I have the supremum part now, but edited the original post. – Michal Dvořák Oct 16 '18 at 15:26
• You don't need the hypotheses $\lim _{n\to \infty} |I_n|=0$ as the conclusion $\bigcap\limits _{n=1}^{\infty} I_n\neq\emptyset$ is true without that assumption. – Paramanand Singh Oct 17 '18 at 0:58

Throughout this post we are analyzing an ordered field $$R$$ satisfying a slightly modified form of the OP's axiom:

$$\text{Axiom (P1) =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=}$$

Let $$(I_n=[a_n,b_n])$$ be any sequence of closed intervals in $$R$$ such that $$\forall n\in\mathbb{N}:I_{n+1}\subset I_n$$. Also assume that for every integer $$n\gt0$$ an interval $$I_k$$ can be found such that

$$\quad b_k−a_k \lt \frac{1}{n}$$

Then the intersection of all the intervals is a singleton.

$$\text{=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=}$$

Proposition 1: The ordered field $$R$$ satisfies the following property:

$$\tag A (\forall a \in R, a \gt 0) \; (\exists n \in \mathbb N, n\gt0) \text{ such that } \frac{1}{n} \lt a$$
Proof
By $$\text{(P1)}$$ the interesection of the intervals $$[-\frac{1}{n},+\frac{1}{n}]$$ can only be equal to $$\{0\}$$.
But this is equivalent to $$\text{(A)}$$. $$\quad \blacksquare$$

So $$R$$ satisfies the axiom of Archimedes and the following is also true.

Proposition 2: Every $$r \in R$$ belongs to exactly one interval of the form $$[n, n+1)$$ with $$n \in \mathbb Z$$.

We are now ready to show that $$R$$ also satisfies the least-upper-bound property.

Proposition 3: Let a non-empty subset $$A$$ of $$R$$ be bounded above by $$M \in R$$. Then there exists a least upper bound for $$A$$.
Proof
The set $$C = \{n \in \mathbb Z \, | \, n \text{ is an upper bound for the set } A \}$$ must have a least element $$c_0$$. Set $$b_0 = c_0 -1$$, so that $$A$$ has points in the closed interval $$[b_0,c_0]$$.

We continue using recursion/induction.

Let the interval $$[b_k,c_k]$$ satisfy the following:

$$\tag 1 c_k \text{ is an upper bound for the set } A$$
$$\tag 2 \text{The set } A \text{ has points in } [b_k,c_k]$$
$$\tag 3 c_k - b_k = 2^{-k}$$

Let $$\mu = \frac{b_k+c_k}{2}$$. If $$A$$ has points in $$[\mu,c_k]$$, define $$b_{k+1} = \mu$$ and $$c_{k+1} = c_k$$; else, define $$b_{k+1} = b_k$$ and $$c_{k+1} = \mu$$.

So we have a constructed a $$\text{decreasing}_{↓0}$$ chain $$[b_n, c_n]$$ of closed intervals satisfying (1), (2) and (3) with the intersection being a singleton set $$\{\gamma\}$$.

It is immediate that if $$\lambda$$ and $$\kappa$$ are any two distinct numbers, they can't both belong to all the intervals $$[b_n, c_n]$$. So if $$a \in A$$, it must be less than or equal to $$\gamma$$. Otherwise, $$b_n \le \gamma \lt a \le c_n$$ is true for all $$k$$, but then $$a = \gamma$$, a contradiction.

So $$\gamma$$ is an upper bound for the set $$A$$.

Suppose $$\rho$$ is another upper bound for $$A$$ that is stricly less than $$\gamma$$. Then for any $$n$$, $$b_n \le \rho \lt \gamma \le c_n$$. Again, this implies that $$\rho = \gamma$$, a contradiction.$$\quad \blacksquare$$

• This looks good, just to clarify. We assume this $c_0$ to be the integer given by the archimedean property, right? – Michal Dvořák Oct 16 '18 at 15:36
• @MichalDvořák Using the fact that $\mathbb N$ is well-ordered. – CopyPasteIt Oct 16 '18 at 16:15
• @MichalDvořák and axiom (P1). – CopyPasteIt Oct 17 '18 at 0:38
• You can essentially say, that for any $\epsilon>0$ there is $n\in\mathbb{N}$ such that $|I_n|<\epsilon$. There is no need for that $1/n$. – Michal Dvořák Oct 17 '18 at 21:49
• Yes. But I want to get rid of the limit stuff and anyway, looks like we do need singleton. See math.stackexchange.com/questions/1855286/… and other links – CopyPasteIt Oct 17 '18 at 22:16

You forgot to say that you pick $$a_1\in A$$.

In order to prove that $$c=\sup A$$ you have to rule out (i) the existence of an $$a\in A$$ with $$a>c$$ and (ii) the existence of a $$b which is an upper bound of $$A$$. Note that both $$a$$ and $$b$$ here would have a positive distance from $$c$$, whereas $$|I_n|\to0$$. This will allow you to produce a contradiction in both cases (i) and (ii).

• It is clear that $(a_n)$ is non-decreasing and $(b_n)$ is non-increasing. Now, by construction, $b_n$ is bounded below by all of $a_n$'s and $a_n$'s are all bounded above by all of $b_n$, (by monotone convergence theorem, which I am able to prove from the given Axiom) so this implies $\lim a_n=\lim b_n$. So, if we denote this limit $l$, then is bounded by $l$ from above and $b_n$ is bounded by $l$ from below. But all of $a_n$'s are, by construction, in $A$ so are not suprema for $A$... Well, I think I am running in circles :/ – Michal Dvořák Oct 16 '18 at 13:06
• It's much simpler: You get the contradictions by looking at an $[a_n,b_n]$ with sufficiently large $n$. – Christian Blatter Oct 16 '18 at 14:59