Supremum

Let $A$ be an ordered set and $B\subset A,B\ne \varnothing$, and bounded above. If there exists $a\in A$ such that

1. $a$ is an upper bound for $B$
2. $\forall c\in A,c<a,\exists b\in B,c<b$. Then we say $B$ has the least upper bound property, and $a$ is called the least upper bound of $B$. Or,

$$\sup B:=a$$

Here, we force the minimality of $a$ by supposing that any $c$ that exists would have some $b$ be greater than it. We are capturing the “best” upper bound.

Motivation

Can we find the best upper bound for the set

$$A=\{x\in \mathbb{Q}\mid x^2<2\}$$

• Note that $A\ne \varnothing$ because $0,1\in A$.
• $A$ is bounded above. $\forall x\in A,x^2<2⟹x<\sqrt{2}$ by definition.
• Can we find the best upper bound for $A$ in $\mathbb{Q}$. $\exists y\in \mathbb{Q}$ such that:
  1. $\forall x\in A,x\le y$
  2. $y$ is the smallest rational number that is an upper bound for $A$
• Answer: No.

Proof:

We show this by contradiction. That is, suppose $\exists y\in \mathbb{Q}$ that satisfies $(1)$ and $(2)$. By the lemma, $y^2<2$ or $y^2>2$ by trichotomy. WLOG, assume, $1\le y\le 2$.

If $y<1$, then since $1\in A$, $y$ cannot satisfy $(1)$. If $y>2$, then since $2\ge \sqrt{2}$ is an upper bound, then $y$ is not the smallest upper bound satisfying $2$.

Case 1: $y^2<2$.

<-----|-----|-----|-->
     y^2   z^2    2

We have some distance $a=2-y^2>0$. If we can find some $z\in \mathbb{Q}$ where $z^2<2⟹z\in A$ then we have some contradiction. Let $z=y+r$ where $r$ is some “wiggle room”, i.e. $r\in (0,1]$ and $z^2<2$.

If this $z$ existed, then

$$z^2=y^2+2yr+r^2<2⟺2yr+r^2<2-y^2=a$$

From the fact that $1\le y\le 2$, we see that $2yr\le 4r$, and given that $0<r\le 1$, then $r^2<r$. Given these inequalities, we can show:

$$2yr+r^2<4r+r=5r<a$$

So, by finding some $r$ that satisfies this, we are done. Choose $r=a/5$ and to choose it to be smaller, than we can simply multiply by $1/2$ and get $r=a/10$. As we need $r\in (0,1]$,

$$\begin{array}{rl}0<r\le 1& ⟺0<\frac{a}{10}\le 1\\ & ⟺0<2-y^2\le 10\\ & ⟺-2<-y^2\le 8\\ & ⟺-8\le y^2<2\end{array}$$

So, $r$ exists only if $-8\le y^2<2$. So, $1\le y⟹1\le y^2$ and proves the LHS of the inequality, and the assumption of the case is th RHS. As $r$ exists, $z$ exists, $z^2$ exists and there is no upper bound.

Case 2: $y^2>2$.

      2    z^2   y^2
<-----|-----|-----|-->

Again, let $a=y^2-2>0$. The goal is to find some $z^2\in \mathbb{Q}$ where the above inequality holds. So

• $z=y-r$.
• $r\in (0,1]$
• $z^2>2$ So,

$$z^2=y^2-2yr+r^2>2⟺r^2-2yr>2-y^2=a$$

for some $z$ to exist, this inequality must hold. As before,

• $1\le y\le 2⟹2r\le 2yr\le 4r$
• $r\in (0,1]⟹0<r^2<r$ Such that,

$$r^2-2yr>0+-4r>2-y^2=a$$

Then, let $r=-a/8$. So,

$$\begin{array}{rl}0<r\le 1& ⟺0<\frac{-a}{8}\le 1\\ & ⟺-8\le a<0\\ & ⟺-8\le 2-y^2<0\\ & ⟺2<y^2\le 10\end{array}$$

Where the LHS is true by the assumption and the RHS is true by the fact that $y\le 2⟹y^2\le 4$. Therefore it is a contradiction.

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Given a contradiction in Case 1 and Case 2, does that mean no $y\in \mathbb{Q}$ for property $(1)$ and $(2)$…