Lecture 5 - Some Consequences of Completeness

Recall from last time that:

\exists R

$\exists$ an ordered field $R$ which contains $Q$ which satisfies the the axiom of completeness.

Here the axiom of completeness says that it satisfies the Least-Upper-Bound Property.

Last time we proved:

Archimedean Property

i. Given any $x \in R$ $\exists n \in N$ where $n > x$ .
ii. Given any real number $y > 0$ , $\exists n \in N$ where $\frac{1}{n} < y$ .

As a corrallary we get:

Corollary

For $x \in R$ , $x = 0$ iff $\forall ε > 0 (| x | < ε)$ .

Proof
Clearly $\to$ so suppose $\forall ε > 0 (| x | < ε)$ . Assume $x \neq 0$ for contradiction:

If $x > 0$ then $| x | = x$ . Then $x > 0$ so by our given then $| x | = x < x$ is false.
If $x < 0$ then $| x | = - x$ . Then $- x > 0$ so by our given then $| x | = - x < - x$ is false.
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Corollary

Let $x \in R$ . Then $\exists n \in Z$ such that $n - 1 \leq x < n$ .

Proof
If $x$ is an integer then we are done. We only have to show $n - 1 < x < n$ . By the archimedian property, then $\exists n$ where $n > x$ . Clearly any $n + 1, n + 2, . . .$ could also work, so say that $n$ is the smallest integer to choose. Thus that requires that $n - 1 < x$ as required.

#todo/school Finish this proof 📅 2024-10-06 ✅ 2024-10-06
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Density of $Q$ in $R$

We say that $Q$ is dense in $R$ . For more info see The Density of Q in R. A review is as follows, but here's the theorem:

Density of

Q

R

$\forall a, b \in R$ where $a < b$ , then $\exists r \in Q$ where $a < r < b$ .

Let's do some scratch work:

We know nothing about $x, y$ , but we do know that since $x < y$ then $0 < y - x$ . Using the archimedean principle, then $\exists n \in N$ where $\frac{1}{n} < y - x$ . At some point we'll hit fractions that hit into the interval:

By the corollary then $\exists a \in Z$ where $a - 1 \leq n x < a$ . As a a result:

$\frac{a}{n} < x$
We want $a < n y$ . Notice that above from $\frac{1}{n} < y - x \Rightarrow 1 < n y - n x \Rightarrow 1 + n x < n y$ . So then:

\begin{array}{r} a \leq 1 + n x < n y \end{array}

So here's the proof:
Proof
Let $x < y$ . By the archimedean property, then $\exists n \in N$ s.t. $\frac{1}{n} < y - x$ . But our corollary says that $\exists a \in Z$ s.t. $a - 1 \leq n x < a$ . Now notice that:

$n x < a \Rightarrow x < \frac{a}{n}$
$a - 1 \leq n x \Rightarrow a \leq 1 + n x < n y$ since $\frac{1}{n} < y - x \Rightarrow 1 + n x < n y$ . Then clearly $a < n y$ so $y > \frac{a}{n}$ .
Therefore:

x < \frac{a}{n} < y

Thus choosing $q = \frac{a}{n}$ finishes the proof.
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We then talked about the Nested Interval Property. It's an alternative to the least upper bound property which was a fundamental axiom for $R$ .

Before we ended, let's look at trying to prove the following

\exists α \in R

s.t.

α^{2} = 2

We can't go through the whole proof here (for time) but here's the idea:

Let $A = {x \in R | x > 0 \land x^{2} < 2}$ . Clearly $A$ isn't empty ( $1 \in A$ ) and bounded ( $2 \notin A$ ). By the LUB property then $α = sup (A) \in R$ .
The goal is to show that $α^{2} = 2$ . The idea is to show $α^{2} ≮ 2$ and $α^{2} ≯ 2$ . If $α^{2} < 2$ then we should show that $α$ is not an UB by finding a small positive $ε > 0$ where $α + ε \in A$ . For this we'd need:

\begin{aligned} α + ε \in A & \Leftrightarrow (α + ε)^{2} < 2 \\ \Leftrightarrow α^{2} + 2 α ε + ε^{2} < 2 \\ \dots \end{aligned}

For the other part showing $α^{2} ≯ 2$ , we want to contradict the least part of being a supremum.