User:GFauxPas/Sandbox

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Munkres Supplementary Exercises to Chapter 1

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
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 $\prec$ to be the usual (strict) ordering of the natural numbers.

The usual ordering is a well-ordering, from the Well-Ordering Principle.

Define the relation:


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




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

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

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

Define the relation:


 * $\left[\!\left[{\left({A,\prec_A}\right)}\right]\!\right] \ll \left[\!\left[{\left({B,\prec_B}\right)}\right]\!\right]$


 * $(A, \prec_A)$ is order isomorphic to an initial segment of $(B,\prec_B)$.

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

From No Isomorphism from Woset to Initial Segment, no equivalence class $\left[\!\left[{\left({A,\prec_A}\right)}\right]\!\right]$ can bear $\ll$ to itself.

From Wosets are Isomorphic to Each Other or Initial Segments, $\ll$ is connected.

From the definition of an order isomorphism, $\ll$ is an ordering.

Thus $\ll$ is a strict total ordering.

The empty set $\varnothing$ contains no elements that could define an initial segment in it.

Consider the empty mapping: $\nu: \varnothing \to \varnothing$.

Such a mapping $\nu$ is bijective, by Empty Mapping to Empty Set is Bijective.

But $\varnothing$ is itself an initial segment of any non-empty set, from Initial Segment Determined by Smallest Element is Empty.

Thus there is an order-isomorphism from $\varnothing$ to an initial segment in any non empty $A$, as $\nu$ is vacuously order-preserving.

Thus $\left[\!\left[{\left({\varnothing,\varnothing}\right)}\right]\!\right]$ is the smallest element of $E$.

Let $\alpha = \left[\!\left[{\left({A,\prec}\right)}\right]\!\right]$ be an element of $E$.

We claim that $\left({A,\prec}\right)$ is order isomorphic to $S_\alpha(E)$, the initial segment of $E$ determined by $\alpha$.

To see this, define the mapping:


 * $f: A \to E$:


 * $f(x) = \left[\!\left[{\left({S_x(A),{\prec \restriction_{S_x(A)}}}\right)}\right]\!\right]$

CAREFUL WITH THE NEXT LINE

Then for all $x \in A$, $f(x) \in S_\alpha$, as $\alpha = \left[\!\left[{\left({A,\prec}\right)}\right]\!\right]$ is an element of $E$.

where ${\restriction}$ denotes restriction.

Suppose $x \prec y$ in $A$.

Then $S_x$ is an initial segment of $S_y$.

From Wosets are Isomorphic to Each Other or Initial Segments, $S_x \ne S_y$.

Thus $f(x) \ll f(y)$.

Conclude that $f$ is strictly increasing.

Recall that the restriction of a strict well-ordering is a strict well-ordering.