User:GFauxPas/Sandbox

Welcome to my sandbox, you are free to play here as long as you don't track sand onto the main wiki. --GFauxPas 09:28, 7 November 2011 (CST)

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:


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

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


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


 * the second coordinate is the ordering $<$ imposes on $A$.

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