Notes on Chapter 1¶

FA¶

The fundamental axiom of analysis

If $a_{n} \in \mathbb{R}$ for all $n \in \mathbb{N}$, $A \in \mathbb{R}$

$a_{1}\leq a_{2}\leq a_{3} \leq \dots$
$a_n < A$ for all $n \in \mathbb{N}$

Then there exists $a \in \mathbb{R}$ such that $a_n \to a$ as $n \to ∞ $

Theorem 1.2.1 The Axiom of Archimedes¶

The result is trivial in $\mathbb{Q}$. The theorem is proving it in $\mathbb{R}$.

This theorem proves that there is no $\gimel>0$ in $\mathbb{R}$ such that,

$\frac{1}{n} > \gimel$ for all $n$ $\iff$ $\gimel$ is smaller than all strictly positive rationals.

Suppose there exists such a $0<\gimel< \frac{1}{n}$ for all $n$.

Then by preservation of non strict inequality (Lemma 1.6 (vii)) $0<\gimel \leq l = 0$ which contradicts the axiom of Archimedes.

If the theorem was true in $\mathbb{Q}$ but not in $\mathbb{R}$ we would have this exotic $\gimel$.

Exercises¶

In A0 we proved $0_{\mathbb{F}} < 1_{\mathbb{F}}$.

Let $n_{\mathbb{F}} = 1_{\mathbb{F}} + \dots + 1_{\mathbb{F}}$ n times.

In A1 we show by induction that $n_{\mathbb{F}} > 0_{\mathbb{F}}$ and therefore they are not equal.

This means that any ordered field $\mathbb{F}$ has characteristic $∞$.

We will then proceed to show that this characteristic of $∞$ means we contain a copy of the rationals.

Some obvious facts¶

F1¶

$n_{\mathbb{F}} + m_{\mathbb{F}} = (n+m)_{\mathbb{F}}$ for any $n,m \in \mathbb{N}^+$

F2¶

$n_{\mathbb{F}} m_{\mathbb{F}} = (nm)_{\mathbb{F}}$ for any $n,m \in \mathbb{N}^+$ due to distributivity.

F3¶

For any $n \in \mathbb{N}^+$ we have $$ -n_{\mathbb{F}} = -(\overbrace{1_{\mathbb{F}} + \dots + 1_{\mathbb{F}} }^{n}) = \overbrace{-1_{\mathbb{F}} + \dots + -1_{\mathbb{F}}}^{n} $$

Hence, for any $n,m \in \mathbb{N}^+$

\[ \begin{align*} n_{\mathbb{F}} - m_{\mathbb{F}} &= \overbrace{1_{\mathbb{F}} + \dots + 1_{\mathbb{F}} }^{n} + \overbrace{-1_{\mathbb{F}}+ \dots + -1_{\mathbb{F}}}^{m} \\ &= \begin{cases} 0_{\mathbb{F}} &&\text{if } n=m \\ \overbrace{1_{\mathbb{F}} + \dots + 1_{\mathbb{F}}}^{n-m} &= (n-m)_{\mathbb{F}} &\text{ if } n>m \\ \overbrace{-1_{\mathbb{F}} + \dots + -1_{\mathbb{F}}}^{m-n} &= -(m-n)_{\mathbb{F}} &\text{ if } n < m \end{cases} \end{align*} \]

Exercise A3¶

Let $\tau : \mathbb{Z} \to \mathbb{F}$ by,

\[ \tau(n) = \begin{cases} 0_{\mathbb{F}} & \text{if } n = 0 \\ n_{\mathbb{F}} & \text{if } n > 0 \\ -(-n)_{\mathbb{F}} & \text{if } n < 0 \end{cases} \]

i)¶

Show that the map is well defined, injective and

$\tau \left( n+m \right) = \tau(n) + \tau(m)$
$\tau \left( nm \right) = \tau(n) \tau(m)$
$\tau(n) > 0_{\mathbb{F}}, \text{ whenever } n>0.$

for all $n,m \in \mathbb{Z}$.

The case definition is already well defined.

Additive

If one or more $n,m$ equal 0 then additivity is trivial. Therefore assume either both positive OR one positive and one negative OR both negative.

Both positive case is trivial:

\[ \tau(n) + \tau(m) = n_{\mathbb{F}} + m_{\mathbb{F}} \overset{F1}{=} (n+m)_{\mathbb{F}} = \tau(n+m) \]

One positive and one negative case.Assume WLOG $n>0$ and $m<0$.

\[ \tau(n) + \tau(m) \overset{\text{Def}}{=} n_{\mathbb{F}} + -(-m)_{\mathbb{F}} = (*) \]

We can just apply F3 where $-m$ takes the role of $m$.

\[ (*) = \begin{cases} 0_{\mathbb{F}} & & \overset{\text{Def }}{=} \tau(0) &\text{ if } n=-m \\ (n--m)_{\mathbb{F}} &= (n+m)_{\mathbb{F}} & \overset{\text{Def }}{=} \tau(n+m) &\text{ if } n>-m \\ -(-m-n)_{\mathbb{F}} &= -(-(n+m))_{\mathbb{F}} & \overset{\text{Def }}{=} \tau(n+m) &\text{ if } n < -m \end{cases} \]

Multiplicative

If one of $n,m$ is zero then it's easy to show $nm=0_{\mathbb{F}} = \tau(n) \tau(m)$. Assume $n,m \neq 0$, therefore there are cases.

If $n,m>0$.

\[ \begin{align*} \tau(n) \tau(m) \overset{\text{Def}}{=} n_{\mathbb{F}} m_{\mathbb{F}} \overset{F2}{=} (nm)_{\mathbb{F}} \overset{\text{Def}}{=} \tau(nm) \end{align*} \]

If $n>0, m<0$,

\[ \begin{align*} \tau(n) \tau(m) &\overset{\text{Def}}{=} n_{\mathbb{F}} \times -(-m)_{\mathbb{F}}\\ [\text{algebra}] &= - (n_{\mathbb{F}} \times (-m)_{\mathbb{F}}) \\ [\text{F2}] &= -(n \times -m)_{\mathbb{F}} = -(-nm)_{\mathbb{F}} \\ [\text{Def}] &= \tau(nm) \end{align*} \]

If $n,m <0$,

\[ \begin{align*} \tau(n) \tau(m) &\overset{\text{Def}}{=} -(-n)_{\mathbb{F}} \times -(-m)_{\mathbb{F}} = (-n)_{\mathbb{F}} (-m)_{\mathbb{F}} \\ [\text{F2}] &= (-n \times -m)_{\mathbb{F}} = (nm)_{\mathbb{F}} \\ [\text{Def}] &= \tau(nm) \end{align*} \]

Positive Easy since $\tau(n)=(n)_{\mathbb{F}}>0_{\mathbb{F}}$ by A2.

Injective

Prove via contrapositive. Assume $n \neq m$, WLOG let $n \geq m$, therefore $n > m \overset{\text{F7}}{\implies} \tau(n) > \tau(m)$.

ii)¶

Define $\phi: \mathbb{Q} \to \mathbb{F}$ by $\phi(n/m) = \tau(n) \tau(m)^{-1}$

This function is well defined and has properties

$\phi (x+y) = \phi(x) + \phi(y)$
$\phi(xy) = \phi(x) \phi(y)$
$\phi(x) > 0_{\mathbb{F}}$ whenever $x>0$

Proof

Well Defined

Let $x= \frac{n}{m} = \frac{\lambda n}{\lambda m}$ for some $\lambda \in \mathbb{N}$. Then,

\[ \begin{align*} \phi\!\left(\frac{\lambda n}{\lambda m}\right) &= \tau(\lambda n) \tau(\lambda m)^{-1} \\ &= \tau(\lambda) \tau(n) \bigl(\tau(\lambda) \tau(m)\bigr)^{-1} \\ &= \tau(\lambda) \tau(n) \tau(\lambda)^{-1} \tau(m)^{-1} \\ &= \tau(n) \tau(m) \\ &= \phi(x) \end{align*} \]

Additive

Let $x = \frac{n}{m}, y= \frac{p}{q}$ for $n,m,p,q \in \mathbb{Z}$

\[ \begin{align*} \phi(x+y) &= \phi\!\left(\frac{nq + pm}{mq}\right) \\ &= \tau(nq + pm) \tau(mq)^{-1} \\ &= \tau(nq + pm) \tau(m)^{-1} \tau(q)^{-1} \\ &= \bigl( \tau(n) \tau(q) + \tau(p) \tau(m) \bigr) \tau(m)^{-1} \tau(q)^{-1} \\ &= \tau(n) \tau(m)^{-1} + \tau(p) \tau(q)^{-1} \\ &= \phi(x) + \phi(y) \end{align*} \]

We also see

\[ \begin{align*} \phi(xy) &= \phi\!\left(\frac{np}{mq}\right) \\ &= \tau(n) \tau(p) \tau(m)^{-1} \tau(q)^{-1} \\ &= \tau(n) \tau(m)^{-1} \tau(p) \tau(q)^{-1} \\ &= \phi(x) \phi(y) \end{align*} \]

And we can assume $x>0$ means that $m,n>0$ since we can multiply both by $-1$ if not,

\[ \phi(x) = \tau(n) \tau(m)^{-1} >0_{\mathbb{F}} \]

since A2 gives us $m>0 \implies \tau(m)>0$ which then implies it's inverse is positive (standard result).

More obvious facts¶

F4¶

From definition, $\phi(1) = \tau(1) \tau(1)^{-1} = 1_{\mathbb{F}}$

This can also be proved as a standard algebra result for homomorphisms.

F5¶

$\phi(-x) = -\phi(x)$, literally add them to get $\phi(0) \overset{\text{Def}}{=}0_{\mathbb{F}}$

F6¶

In A2 we proved $0 < x \in \mathbb{Q} \implies \phi(x) > 0$

The converse is straightforward by contrapositive. Assume $x\leq0$.

If $x=0$, then $\phi(0) \overset{\text{Def}}{=} 0$.

If $x<0$ then $\phi(-x) = -\phi(x)>0$ by forward direction that we proved.

Hence $\phi(x) \leq 0$.

Hence $$ \phi(x) > 0_{\mathbb{F}} \iff x > 0 $$

F7 Inequalities are equivalent accross $\phi$¶

\[ \begin{align*} \phi(x) > \phi(y) \iff \phi(x) - \phi(y) > 0_{\mathbb{F}} \iff \phi(x-y) > 0_{\mathbb{F}} \overset{\text{F6}}{\iff} x-y > 0 \end{align*} \]

F8 Axiom of Archimedes in general field¶

Let $a_n = \phi \left( \frac{1}{n} \right)$. This sequence is bounded and decreasing (F7) and therefore converges by FA, call this limit $l$.

Let

\[ \begin{align*} b_n &= \phi \left( \frac{1}{2} \right) \phi \left( \frac{1}{n} \right) = \phi \left( \frac{1}{2} \right) a_n \to \phi \left( \frac{1}{2} \right) l \\ & \text{ by Algebra of Limits, Lemma 1.6 } \\ &= \phi \left( \frac{1}{2n} \right) = a_{2n} \to l \\ & \text{by subsequence.} \end{align*} \]

Hence,

\[ \begin{align*} l = \phi \left( \frac{1}{2} \right) l &\iff \left( 1_{\mathbb{F}} - \phi \left( \frac{1}{2} \right) \right)l = 0 \\ &\iff \phi \left( \frac{1}{2} \right) l = 0 \\ &\iff l=0 \end{align*} \]

Basically the same as before!

F9¶

For any $x \in \mathbb{F}$ there exists $n_{\mathbb{F}} \in \phi(\mathbb{Z}) \colon x< n_{\mathbb{F}}$.

i $x \leq 0_{\mathbb{F}}$ is trivial, $n=1$ does the job.

ii assume $x > 0_{\mathbb{F}}$.

By Archimedes there exists $n \in \mathbb{N}$ such that $$ 0_{\mathbb{F}} < \phi \left( \frac{1}{n} \right) < \frac{1}{x} \iff \phi ( n ) = n_{\mathbb{F}} > x > 0_{\mathbb{F}} $$

F10¶

For any $x \in \mathbb{F}$ there exists $n_{\mathbb{F}}$ such that $n-1 \leq x_{\mathbb{F}} \leq n_{\mathbb{F}}$

Apply F8 to $x$ and $-x$ to get $a_{\mathbb{F}} < x < b_{\mathbb{F}}$ for some $a,b \in \mathbb{Z}$.

Let $n=b$, if $x < (n-1)_{\mathbb{F}}$ let $n \colon = n-1$, repeat...

This terminates eventually since $x<a_{\mathbb{F}}$ is not possible. Termination occurs when $x \geq \mathbb{n-1}$

Exercise A4¶

Let $\mathbb{K} = \phi \left( \mathbb{Q} \right) \subseteq \mathbb{F}$.

Prove that if $x \in \mathbb{F}$ then there exists sequence $x_n \in \mathbb{K}$ such that $x_1 \leq x_2 \leq \dots$ such that $x_n \to x$ as $n \to \infty$.

Proof By F9 there exists $a,b \in \phi(\mathbb{Z})$ such that $a \leq x<b$.

Let $a_0, b_0 = a,b$. Then let $c_i = \phi \left( \frac{1}{2} \right) \left( a_{i-1} + b_{i-1} \right)$

\[ \begin{cases} a_i &= c_i & \text{ and } b_i = b_{i-1} & \text{ if } c_i \leq x \\ a_i &= a_{i-1} & \text{ and } b_i = c_i & \text{ if } x_i > x \end{cases} \]

Then $a_n \leq x \leq b_n$ and $a_n \leq a_{n+1}$ for all $n \in \mathbb{N}$. Hence by FA $a_n \to l$ for some $l \in \mathbb{F}$.

Also, $b_i - a_i = \phi\left( \frac{1}{2} \right)^n \left( b_0 - a_0 \right) \to 0_{\mathbb{F}}$.

Then $0 \leq x - a_n \leq b_n - a_n$ implies $0\leq x -l \leq 0$

F11 Convergence in $\mathbb{Q}$ implies convergence in $\phi(\mathbb{Q})$¶

Let $a_n \in \mathbb{Q}$ and $a \in \mathbb{Q}$. If $a_n \to a$, then $\phi(a_n) \to \phi(a)$ in $\mathbb{F}$.

Proof Let $\eta > 0_{\mathbb{F}}$. By F8 Axiom of Archimedes there exists $m \in \mathbb{N}$ such that $\phi \left( \frac{1}{m} \right) < \eta$. It is therefore sufficient to show

\[ \lvert \phi \left( a_n - a \right) \rvert = \lvert \phi(a_n) - \phi(a) \rvert < \phi \left( \frac{1}{m} \right) < \eta \]

By F7 this is equivalent to

\[ \lvert a_n - a \rvert < \frac{1}{m} \]

Which hold for sufficiently large $N$ since $a_n \to a$ in $\mathbb{Q}$.

Hence $\phi(a_n) \to \phi(a)$. $\blacksquare$

F12 Convergence in $\phi(\mathbb{Q})$ implies convergence in $\mathbb{Q}$¶

Let $q_n \in \phi(\mathbb{Q})$ and $q \in \phi(\mathbb{Q})$. If $q_n \to q$ in $\mathbb{F}$, then $\phi^{-1}(q_n) \to \phi^{-1}(q)$ in $\mathbb{Q}$.

Proof Let $\varepsilon > 0$ in $\mathbb{Q}$. Then $\phi(\varepsilon) > 0_{\mathbb{F}}$ by F6. Since $q_n \to q$, there exists $N$ such that for all $n \geq N$,

\[ \left|q_n-q\right| < \phi(\varepsilon). \]

Because

\[ q_n-q = \phi\bigl(\phi^{-1}(q_n)-\phi^{-1}(q)\bigr), \]

F7 gives

\[ \left|\phi^{-1}(q_n)-\phi^{-1}(q)\right| < \varepsilon. \]

Hence $\phi^{-1}(q_n) \to \phi^{-1}(q)$. $\blacksquare$

F13 Different limits in Q implies different limits in $\phi(\mathbb{Q})$¶

Let $x_n,y_n \in \mathbb{Q}$ with $x_n \to x$ and $y_n \to y$ in $\mathbb{R}$. If $x \neq y$, then

\[ \lim_{n \to \infty} \phi(x_n) \neq \lim_{n \to \infty} \phi(y_n). \]

Proof Suppose $\lim_{n \to \infty} \phi(x_n) = \lim_{n \to \infty} \phi(y_n).$

Then $\phi(x_n-y_n) = \phi(x_n)-\phi(y_n) \to 0_{\mathbb{F}}.$

By F12, $x_n-y_n \to 0$ in $\mathbb{Q}$. Therefore,

$x-y = \lim_{n \to \infty} (x_n-y_n) = 0$,

which contradicts $x \neq y$. Hence the two limits are distinct. $\blacksquare$

Exercise A5¶

(i)¶

Let $x \in \mathbb{R}$. By A4 with $\mathbb{F}=\mathbb{R}$, there exists a sequence $x_n \in \mathbb{Q}$ such that

\[ x_1 \leq x_2 \leq \dots \quad\text{and}\quad x_n \to x. \]

Convergence of $x_n$ implies that it is bounded. Therefore by F7 $\phi(x_n)$ is also bounded increasing and therefore converges by FA to some limit $a \in \mathbb{F}$.

(ii)¶

Let $x_n' \in \mathbb{Q}$ with $x_n' \to x$. We claim that $\phi(x_n') \to a$ as well.

Since $x_n-x_n' \to 0$ in $\mathbb{Q}$, F11 gives $\phi(x_n -x_n') \to 0_{\mathbb{F}}$

So, $\phi(x_n) \to a$ and $\phi(x_n) - \phi(x_n') \to 0_{\mathbb{F}}$ implies (by AOL) that $\phi(x_n') \to a$.

(iii)¶

Define $\Theta : \mathbb{R} \to \mathbb{F}$ by

\[ \Theta(x) = \lim_{n \to \infty} \phi(q_n), \]

where $(q_n)$ is any rational sequence converging to $x$.

Part A4 gives us such a $q_n$ exists. A5 (i) gives us convergnce. A5 (ii) gives us that this limit is the same regardless of rational sequence chose.

If $x \in \mathbb{Q}$, taking the constant sequence $q_n=x$ shows that $\Theta(x)=\phi(x)$, so $\Theta$ extends $\phi$.

(iv) Injectivity¶

If $x \neq y$, choose rational sequences $x_n \to x$ and $y_n \to y$. Then F13 gives

\[ \Theta(x) = \lim_{n \to \infty} \phi(x_n) \neq \lim_{n \to \infty} \phi(y_n) = \Theta(y), \]

so $\Theta$ is injective.

(iv) Surjectivity¶

Let $q \in \mathbb{F}$. By A4 there exists an increasing sequence $q_n \in \phi(\mathbb{Q})$ such that $q_n \to q$.

Convergence of $q_n$ implies it's bounded. F7 gives us that $\phi(q_n)^{-1}$ is therefore bounded and increasing in $\mathbb{Q}$.

So by FA in $\mathbb{R}$ there exists $\ell \in \mathbb{R}$ such that $\phi^{-1}(q_n) \to \ell.$

Now $\Theta(\ell)$ is the $\phi$-limit of any rational sequence converging to $\ell$, so using the particular sequence $\phi^{-1}(q_n)$,

\[ \Theta(\ell) = \lim_{n \to \infty} \phi\bigl(\phi^{-1}(q_n)\bigr) = \lim_{n \to \infty} q_n = q. \]

Hence $\Theta$ is surjective.

(v)¶

Let $x,y \in \mathbb{R}$, and choose rational sequences $x_n \to x$ and $y_n \to y$.

Then $x_n+y_n \to x+y$, so by definition of $\Theta$,

\[ \Theta(x+y) = \lim_{n \to \infty} \phi(x_n+y_n) = \lim_{n \to \infty} \bigl(\phi(x_n)+\phi(y_n)\bigr) = \Theta(x)+\Theta(y). \]

Also $x_n y_n \to xy$, so similarly

\[ \Theta(xy) = \lim_{n \to \infty} \phi(x_n y_n) = \lim_{n \to \infty} \bigl(\phi(x_n)\phi(y_n)\bigr) = \Theta(x)\Theta(y). \]

Finally, if $x>0$, choose $m \in \mathbb{N}$ such that $\frac{1}{m} < x$. Since $x_n \to x$, we have $x_n > \frac{1}{m}$ for all sufficiently large $n$. Hence by F7,

\[ \phi(x_n) > \phi\left(\frac{1}{m}\right) > 0_{\mathbb{F}} \]

eventually, and therefore

\[ \Theta(x) = \lim_{n \to \infty} \phi(x_n) \geq \phi\left(\frac{1}{m}\right) > 0_{\mathbb{F}}. \]

So $\Theta$ preserves positivity, and therefore order as well. Hence $\Theta : \mathbb{R} \to \mathbb{F}$ is an isomorphism of ordered fields extending $\phi$. $\blacksquare$