12. More consequences of completeness

Archimedean Property of $\mathbb{N}$

$$\mathbb{N}\text{ is not bounded above}$$

Proof

Suppose for a contradiction, that $\mathbb{N}$ is bounded above
Then $\mathbb{N}$ is non-empty ($1\in \mathbb{N}\subseteq \mathbb{Z}$) and bounded above
So by completeness axiom then $\mathbb{N}$ has a supremum
By the approximation property with $ϵ=\frac{1}{2}$ then

$$\exists n\in N\text{ such that }\sup N-\frac{1}{2}<n\le \sup N$$

As $n+1\in \mathbb{N}$ and $n+1>\sup \mathbb{N}$ and hence a contradiction

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Archimedean Property of the Reals corollary

Let $ϵ>0$ then

$$\exists n\in N\text{such that}0<\frac{1}{n}<ϵ$$

Proof

If not, then $\frac{1}{ϵ}$ would be an upper bound for $\mathbb{N}$
This would contradict with Archimedean Property of ℕ

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Min-Max Property of subsets of $\mathbb{Z}$

Let $S$ be a non-empty subset of $\mathbb{Z}$

1. If $S$ is bounded below, then $S$ has a minimum

2. If $S$ is bounded above, then $S$ has a maximum

Proof - $(1)$

Assume that $S$ is bounded below
By completeness (applied to $\{-s:s\in S\}$), $S$ has an infimum

By approximation property (with $ϵ=1$), there is

$$n\in S\text{ such that }\inf S\le n<\inf S+1$$

If $n=\inf S$ then $\inf S\in S$ so there is a minimum

Otherwise, suppose for a contradiction, that $\inf S<n$
Suppose $n=\inf S+\delta$, where $0<\delta <1$

By the approximation property (with $ϵ=\delta$) there is

$$m\in S\text{ such that }\inf S\le m<\inf S+ϵ=n$$

Now $m<n$ so $n-m>0$ but as $n-m$ is an integer then $n-m\ge 1$
So

$$n\ge m+1\ge \inf S+1$$

which is a contradiction hence

$$n=\inf S\in S\text{ so }\inf S=\min S$$

Proof - $(2)$ - similar to $(1)$

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Properties of $\mathbb{R}$ and $\mathbb{R}\setminus \mathbb{Q}$

Take $a,b\in \mathbb{R}$ with $a<b$ then

1. There is $x\in \mathbb{Q}$ such that $a<x<b$
   (the rationals are dense in the reals)
2. There is $y\in \mathbb{R}\setminus \mathbb{Q}$ such that $a<y<b$
   The irrationals are dense in the reals

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