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Munkres Supplementary Exercises to Chapter 1

* : Supplementary Exercises $1.X$

Lemma 2.a
Let $J$ and $E$ be well-ordered sets.

Let $h: J \to E$ be a mapping.

Let $S_\alpha$ denote an initial segment determined by $\alpha$.


 * $(1):\quad$ $h$ is order-preserving and its image is either all of $E$ or an initial segment of $E$


 * $(2):\quad$ $\forall \alpha \in J: h\left({\alpha}\right) = \operatorname{smallest} \left({E\setminus h\left[{S_\alpha}\right]}\right)$

where $h\left[{S_\alpha}\right]$ denotes the image of $S_\alpha$ under $h$.

Hint: show that each of these conditions implies that $h\left[{S_\alpha}\right]$ is an initial segment of $E$; conclude that it must be $S_{h\left({\alpha}\right)}$

Proof
Define the mapping:


 * $\phi: J \to E:$


 * $\phi(\alpha) = \operatorname{smallest}\left({E \setminus h\left[{S_\alpha}\right] }\right)$

where $h$ is the function considered in the exposition of the theorem.

By the Principle of Recursive Definition for Well-Ordered Sets, $\phi$ is thus uniquely determined.

Suppose $h$ satisfies:


 * $(1):\quad$ $h$ is order-preserving and its image is either all of $E$ or an initial segment of $E$

By Strictly Monotone Mapping with Totally Ordered Domain is Injective, $h$ is then injective.

Let $x,y$ be elements of $J$ such that $y \prec x$.

Then $h(x) \prec h(y)$ in $E$.

By the definition of initial segment, $h(x) \in S_{h(y)}$.

Consider $s(x) = \operatorname{smallest}\left({E \setminus h\left[{S_{h(y)}}\right] }\right)$

Because $h$ is injective, $E \setminus h\left[{S_{h(y)}}\right] \ne E$, as $h[J]$ is, by hypothesis, $E$ or an initail segment of $E$.

Suppose $h = \phi$.

Let $x,y$ be elements of $J$ such that $y \prec x$.

Then:

By recursion, this relation holds for all $x,y \in J$.

Thus $h$ is an order-preserving mapping.

Hence each $y \in J$ corresponds to an initial segment $S_{h(y)}$ in $E$.

By Union of Initial Segments is Initial Segment or All of Woset, $f[J]$ is then an initial segment or all of $E$.

Construction of $\Omega$ without using choice
Munkres supplementary exercise 1.8

By the axiom of powers, there exists the power set $\mathcal P \left({\N}\right)$.

By Power Set of Natural Numbers Not Countable, this set is uncountable.

We construct a well-ordering $\left({P \left({\N}\right), \preccurlyeq}\right)$ that has the desired defining properties of $\Omega$.

Proof
Let $<$ denote the usual strict ordering on $\N$.

From the Well-Ordering Principle, $<$ is a well-ordering.

Denote:


 * $\mathcal A = \left\{ { \left({A,\prec}\right) : A \in\mathcal P(\N) }\right\}$

That is, the set of ordered pairs, such that:


 * the first coordinate is a (possibly empty) subset of $\N$


 * the second coordinate is any well-ordering on $A$.

We observe that there is at least one pair of this form for each $A$, taking $\prec \, = \, <$.

Define the relation:


 * $(A, <) \sim (A',<')$




 * $(A, <)$ is order isomorphic to $(A',<')$.

By Order Isomorphism is Equivalence Relation, $\sim$ is an equivalence relation.

Let $E$ be the set of all equivalence classes $\left[\!\left[{\left({A,<}\right)}\right]\!\right]$ defined by $\sim$ imposed on $\mathcal P(\N)$

Define:


 * $\left[\!\left[{\left({A,<_A}\right)}\right]\!\right] \ll \left[\!\left[{\left({B,<_B}\right)}\right]\!\right]$


 * $(A, <_A)$ is order isomorphic to an initial segment of $(B,<_B)$.

We claim that $\left({E,\ll}\right) = \Omega$.

Steps of proof:

$(a.1)$: this is well-defined

$(a.2)$: this is an ordering relation

$(a.2)$ $E$ has a smallest element $\left[\!\left[{\left({\varnothing,\varnothing}\right)}\right]\!\right]$