0.7 The Extended Real Number System

We briefly talk about materials in elementary real analysis. We will not prove those properties (see elementary real analysis textbook such as Principles of Mathematical Analysis for reference).

We form the extended real number system by adding two elements $-\infty$ and $\infty$ on $\mathbb{R}$; $$ \bar{\mathbb{R}} = \mathbb{R} \cup \left\{ -\infty, \infty \right\} $$ Also we extend the usual ordering $\leq$ on $\mathbb{R}$ by defining $$ -\infty < x < \infty, (\forall x \in \mathbb{R}) $$

0.6 More about Well Ordered Sets

We discuss more about Well Ordered sets.

If $X$ is well ordered and $A \subset X$, then $\cup_{x \in A} I_{x}$ is either initial segment or $X$ itself.

Let $ J = \cup_{x \in A} I_{x}$. If $J = X$, nothing more to prove. If $J \neq X$, let $b = \inf(X \setminus J)$. If $y < b$ for $y \in X$, by construction, $y \in J$. This means $I_{b} \subset J$. Now pick $y \in J$. Then $y \in I_{x}$ for some $x \in A$. If $y \not\in I_{b}$, then $y \geq b$ gives $b \leq y < x$, in turn $b \in I_{x} \implies b \in J$; contradiction. So $ y \in I_{b}$. Thus, $J = I_{b}$.

This proposition states that any union of initial segments are also an initial segment (including whole set) when the whole set is well ordered.

If $X$ and $Y$ are nonempty and well ordered, then either

1. $X$ is order isomorphic to $Y$
2. $X$ is order isomorphic to an initial segment of $Y$
3. $Y$ is order isomorphic to an initial segment of $X$

Consider the collection $\mathcal{F}$ of order isomorphisms whose domains are initial segments of $X$ or $X$ itself and whose ranges are initial segments of $Y$ or $Y$ itself. $\mathcal{F}$ is nonempty since $f: \left\{ \inf X \right\} \rightarrow \left\{ \inf Y \right\}$ is a memeber of $\mathcal{F}$. Also $\mathcal{F}$ is partially ordered by set inclusion if functions are viewed as subsets of $X \times Y$. By similar reasoning used in Section 0.4, Proposition 1-2, if $S$ is nonempty linearly ordered subset of $\mathcal{F}$, then $$ f^{\ast} = \bigcup_{f \in S} f $$ is well defined function and upper bound of $S$. By Zorn's Lemma, $\mathcal{F}$ has a maximal element; say $g$ with domain $A$ and range $B$. If $A = I_{x}$ and $B = I_{y}$ for some $x \in X$ and $y \in Y$, then we can extend $g$ by $$ \bar{g}: A \cup \left\{ x \right\} \rightarrow Y,\ \bar{g}(a) = \begin{cases} g(a) & (a \in A) \\ y & (a = x) \end{cases} $$ because $I_{x} \cup \left\{ x \right\}$, $ I_{y} \cup \left\{ y \right\}$ are again initial segments (including whole set $X$ or $Y$). This contradicts to maximality of $g$. Hence either $A = X$ or $B = Y$, and the result follows.

Well Ordering principle gives us the way to construct weird uncountable sets.

There is an uncountable well ordered set $\Omega$ such that $I_{x}$ is countable for every $x \in \Omega$. If $\Omega'$ is another set with the same properties, then $\Omega$ and $\Omega'$ are order isomorphic.

First by well ordering principle, uncountable well ordered set $\Omega_{0}$ exists. Then define $$ A = \left\{ t \in \Omega_{0}: I_{t} \text{ is uncountable } \right\} $$ If $\Omega_{0}$ satisfies the desired property, then nothing more to prove. If not, $A$ has minimal element $a$ by well ordering principle. Then the set $\Omega = I_{a}$ would be the desired set, because for any $x \in \Omega$, $x \not\in A$, so that $I_{x}$ is countable. Any other $\Omega'$ satisfying this property is order isomorphic to $\Omega$ itself, because any initial segments of $\Omega$ and $\Omega'$ are countable. (See Proposition 2).

Every nonempty countable subset of $\Omega$ has an upper bound.

Let $A \subset \Omega$ be countable and nonempty. $\cup_{x \in A} I_{x}$ is countable union of countable sets, so it is countable. By Proposition 1, it should be equal to $I_{y}$ for some $y \in \Omega$. Then for any $a \in A$, $$ I_{a} \subset I_{y} \implies a \leq y $$ which is equivalent to saying $y$ is an upper bound of $A$.

The set $\Omega$ is called a set of countable ordinals. We usually define $\Omega^{\ast} = \Omega \cup \left\{ \omega_{1} \right\}$ and extend the well ordering by declaring $$ x < \omega_{1},\ (\forall x \in \Omega) $$ $\omega_{1}$ is called the first uncountable ordinal. We won't get further deep into theory of ordinals.