The real field

There exists an ordered field $\mathbb{R}$ which has the LUB property. Also, $\mathbb{R}$ contains $\mathbb{Q}$ as a subfield.

$\mathbb{R}$ having the LUB property is also referred to as the axiom of completeness.

Info

Remember the structural stack on which $\mathbb{R}$ rests:
Field → Ordered Field → Complete ordered field with $\mathbb{Q}$ as subfield

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Consequences of completeness

Nested interval property

Theorem 1(Theorem).

For each $n\in \mathbb{N}$, consider the closed interval, $I_n=[a_n,b_n]=\{x\in \mathbb{R}|a_n\le x\le b_n\}$. Assume that each $I_n$ contains $I_{n+1}$. Then, the resulting nested sequence of closed intervals $I_1\supseteq I_2\supseteq \dots$ has a nonempty intersection.

Proof
Consider the set $A=\{a_n|n\in \mathbb{N}\}$. $A$ is clearly non empty, and bounded above (by every $b_n$). Thus, $x=\sup A$ exists.
Now consider a particular interval $I_n=[a_n,b_n]$. Because $x$ is an upper bound for $A$, be have $a_n\le x$. The fact that each $b_n$ is an upper bound for $A$ and that $x$ is the supremum of $A$ implies that $x\le b_n$.
So, we have $a_n\le x\le b_n$ for every $n$. Hence, $x\in {\bigcup }_{n=1}^{\infty }{I}_n$. ❏

This property fails for open intervals. Consider the sequence of nested open intervals $I_n=\left(0,\frac{1}{n}\right)$ as a counter example.

Theorem 2(Theorem).

In the above theorem, the intersection is a singleton set if and only if ${\lim }_{n\to \infty }(a_n-b_n)=0$.

Refer 2(b) for proof.

Note that the nested interval property is equivalent to the LUB property, and could have been used in place of it to establish the completeness of $\mathbb{R}$, with the LUB property being derived from it.

The Archimedean property

Theorem 3(Theorem).

Given any $x\in \mathbb{R}$, there exists an $n\in \mathbb{N}$ satisfying $n>x$.

Why do we need to prove this?

What we have said so far is that R is an
extension of Q that maintains the algebraic and order properties of the rationals
but also possesses the least upper bound property articulated in the Axiom of
Completeness. In the absence of any other information about R, we have to
consider the possibility that in extending Q we unwittingly acquired some new
numbers that are upper bounds for N. In fact, as disorienting as it may sound,
there are ordered field extensions of Q that include “numbers” bigger than every
natural number. This theorem asserts that the real numbers do not contain
such exotic creatures. The Axiom of Completeness, which we adopted to patch
up the holes in Q, carries with it the implication that N is an unbounded subset
of R.

Proof
FTSOC, assume $\mathbb{N}$ is bounded above. By AoC, $\mathbb{N}$ should have a supremum. Set $\alpha \equiv \sup \mathbb{N}$.
Now, consider $\alpha -1$. This should not be an upper bound of $\mathbb{N}$. Thus, $n>\alpha -1$, $n\in \mathbb{N}$ must exist. But that is the same as saying $\alpha <n+1$. Since $\mathbb{N}$ is closed under addition, $n+1\in \mathbb{N}$. $\Rightarrow \Leftarrow$ ❏

Density of Q in R

Theorem 4(Theorem).

If $a\in \mathbb{R}$, $b\in \mathbb{R}$, and $a<b$, then there exists a $r\in \mathbb{Q}$ such that $a<r<b$.

Proof
We have $b-a>0$. Pick an integer $n$ such that $n(b-a)>1$, existence of which is guaranteed by the Archimedean property. Thus, we have $na+1<nb$. Now, consider $m$ such that

$$m-1\le na<m,$$

the existence of which is also ensured by the Archimedean property. So, we have

$$m\le na+1<nb⟹na<m<nb.$$

Therefore, $a<\frac{m}{n}<b$. ❏

Corollary

Given any two real numbers $a<b$, there exists an irrational number $t$ satisfying $a<t<b$.

Proof
We have $b-a>0$. Pick an integer $n$ such that $n(b-a)>\sqrt{2}$, existence of which is guaranteed by the Archimedean property. Thus, we have $na+\sqrt{2}<nb$. Now, consider $m$ such that

$$\sqrt{2}m-\sqrt{2}\le na<\sqrt{2}m,$$

the existence of which is also ensured by the Archimedean property. So, we have

$$\sqrt{2}m\le na+\sqrt{2}<nb⟹na<\sqrt{2}m<nb.$$

Therefore, $a<\sqrt{2}\frac{m}{n}<b$. ❏

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Cut property

Theorem 5(Theorem).

If $A$ and $B$ are nonempty, disjoint sets with $A\cup B=\mathbb{R}$ and $a<b$ for all $a\in A$ and $b\in B$, then there exists $c\in \mathbb{R}$ such that $x\le c$ whenever $x\in A$ and $x\ge c$ whenever $x\in B$.

Proof
$A$ is nonempty and bounded above (by every $b\in B$). Thus, $\alpha =\sup A$ must exist. Similarly, $\beta =\inf B$ must exist.
Now consider the possibilities:

• If $\alpha <\beta$,
  there must exist $\gamma \in \mathbb{R}$ such that $\alpha <\gamma <\beta$. Clearly, $\gamma \notin A$ and $\gamma \notin B$, i.e, $\gamma \notin \mathbb{R}$. $\Rightarrow \Leftarrow$
• If $\alpha >\beta$,
  there must exist $\gamma \in \mathbb{R}$ such that $\beta <\gamma <\alpha$.
  Since $\alpha$ is the supremum of $A$, there must exist $a\in A$ such that $\gamma <a<\alpha$.
  Similarly, since $\beta$ is the infimum of $A$, there must exist $b\in B$ such that $\beta <b<\gamma$.
  Thus, we have $b<a$. $\Rightarrow \Leftarrow$

Thus, it must be the case that $\alpha =\beta$.
Now, since $\alpha$ is a real number, $\alpha \in A$ or $\alpha \in B$.
If $\alpha \in A$, we have $x\le \alpha$ for all $x\in A$ and $x>\alpha$ for all $x\in B$.
If $\alpha \in B$, we have $x<\alpha$ for all $x\in A$ and $x\ge \alpha$ for all $x\in B$.

Thus, $\alpha$ satisfies the properties for $c$ described in the theorem. ❏

Note that the cut property could be used in place of the Axiom of completeness as the fundamental axiom that distinguishes $\mathbb{R}$ from $\mathbb{Q}$.

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Existence of nth roots of real numbers

Theorem 6(Theorem).

For every real $x>0$, and every integer $n>0$, there is one and only one positive real $y$ such that $y^n=x$.

Refer Rudin, p9 for proof.
Refer Abbot, p23 for proof of existence of square roots, which can be extended to a proof of this theorem.