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Munkres Supplementary Exercises to Chapter 1

* : Supplementary Exercises $1.X$

Theorem
Let $E$ be a well-ordered set.

Let $e_0 = \min E$.

Then $S_{e_0} = \varnothing$.

Obviously.

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

Let $k: J \to E$ be an order-preserving mapping.

Then $J$ is order isomorphic to $E$ or an initial segment of $E$.

Proof of Lemma
Let $e_0 = \min E$, the smallest element of $E$.

Define the mapping:


 * $h: J \to E$:


 * $h(\alpha) =

\begin{cases} \min \left({E \setminus h\left[ {S_\alpha} \right]}\right) & \text{ if } h\left[ {S_\alpha} \right] \ne E \\ e_0 & \text{ otherwise } \end{cases}$

By the Principle of Recursive Definition for Well-Ordered Sets, this construction is well-defined and unique.

where $S_\alpha$ is the initial segment determined by $\alpha$ and $h\left[ {S_\alpha} \right]$ is the image of $S_\alpha$ under $h$.

We claim that $h(\alpha) \preceq k(\alpha)$ for all $\alpha \in J$.

there is some $a \in J$ such that $h(a) \not \preceq k(a)$.

Then $k(a) \prec h(a)$ by the trichotomy law.

Because $h(a)$ has an element preceding it, $h(a) \ne e_0$.

Thus $k(a) \prec \min \left({E \setminus h\left[ {S_a} \right]}\right)$.

Then $k(a) \in h\left[ {S_a} \right]$, because it precedes the smallest element that isn't in $h\left[ {S_a} \right]$.

Note that $h\left[ {S_a} \right] \subseteq S_{h(a)}$,

But:

Thus $k(a) \in S_{h(a)}$ as well.

Then $h(a) \prec k(a)$ which is a contradiction.

Theorem
Wosets are Isomorphic to Each Other or Initial Segments

Proof 2
All sets are non=empty yadda yadda

$\left({S, \preceq_S}\right)$ and $\left({T, \preceq_T}\right)$

Let $U = S \cup T$

Define the following relation $\preceq_U$ on $U$:


 * $\forall x, y \in U: x \preceq y$




 * $x, y \in S : x \preceq_S y$

or:


 * $x, y \in T: x \preceq_T y$

or:


 * $x \in S, y \in T$

We claim that this is a well-ordering relation.

First, we show it's an ordering.

Checking in turn each of the criteria for an ordering:

Reflexivity
If $x = y$, they're necessarily both in $S$ or $T$ simultaneously.

Reflexivity then follows from $\preceq_S$ and $x \preceq_T$ being reflexive.

Transitivity
Let $x, y, z \in U$.

If $x, y, z \in S$ or $x, y , z \in T$ simultaneously, then $\preceq$ is transitive by the transitivity of $\preceq_S$ and $\preceq_T$.

Assume $x, y \in S$ and $z \in T$.

Let $x \preceq y$ and $y \preceq z$.

Then $x \preceq z$ because $x \in S$ and $y \in T$.

Assume $x \in S$ and $y, z \in T$.

Then $x \preceq z$ also because $x \in S$ and $y \in T$.

Thus $\preceq$ is transitive.

Antisymmetry
Let $x \preceq y$ and $y \preceq x$.

If $x, y \in S$ then $x = y$ by the antisymmetry of $\preceq_S$.

Likewise if $x, y \in T$.

If $x \in S$ and $y \in T$, then $y \in S$ and $x \in T$ as well.

Thus $x = y$ from the antisymmetry of $\preceq_S$ or $\preceq T$.

To show $\preceq$ is a well-ordering, consider a non-empty $X \subseteq U$.

Then either:


 * $X \cap S = \varnothing$, in which case $U \subseteq T$.

or:


 * $X \cap T = \varnothing$, in which case $U \subseteq S$.

Or:


 * $X \cap T$ is non-empty and $X \cap S$ is non-empty.

In the first case, $X$ has a smallest element defined by $\preceq_T$.

In the second case, $X$ has a smallest element defined by $\preceq_S$.

In the third case, the smallest element of $X \setminus T$ precedes any element of $T$ by the definition of $\preceq$.

The smallest element of $X \setminus T$, which is a subset of $S$, is guaranteed to exist by the well-ordering on $S$.

This smallest element is then also the smallest element of $\left({X \setminus T}\right) \cup T = X$.

Thus $\preceq$ is a well-ordering on $S \cup T$.

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 $\left[\!\left[{\left({\varnothing,\varnothing}\right)}\right]\!\right]$