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)

Lemma
Let $X$ be a non-empty set.

Let $\left({T_1,\preccurlyeq_1}\right)$ and $\left({T_2,\preccurlyeq_2}\right)$ be towers in $X$.

Then either:


 * $\left({T_1,\preccurlyeq_1}\right) = \left({T_2,\preccurlyeq_2}\right)$

or:


 * $\left({T_1,\preccurlyeq_1}\right)$ is an initial segment of $\left({T_2,\preccurlyeq_2}\right)$

or:


 * $\left({T_2,\preccurlyeq_2}\right)$ is an initial segment of $\left({T_1,\preccurlyeq_1}\right)$

Proof
By the definition of a tower in a set, $\left({T_1,\preccurlyeq_1}\right)$ and $\left({T_2,\preccurlyeq_2}\right)$ are well-ordered sets.

By Wosets are Isomorphic to Each Other or Initial Segments:


 * the towers are order isomorphic to each other, or:


 * one is isomorphic to an initial segment in the other.

, in the second case, assume that $\left({T_1,\preccurlyeq_1}\right)$ is order isomorphic to an initial segment in $\left({T_2,\preccurlyeq_2}\right)$.

By Order Isomorphism iff Strictly Increasing Surjection, there exists a strictly increasing mapping:


 * $i: \left({T_1,\preccurlyeq_1}\right) \to \left({T_2,\preccurlyeq_2}\right)$

such that $i[T_1] = T_2$ or an initial segment of $T_2$.

From Characterization of Strictly Increasing Mapping on Woset, we can characterize $i$ as follows:


 * $\forall t \in T_1: i\left({t}\right) = \min \left({T_2\setminus i\left[{S_t}\right]}\right)$

and $i[S_\alpha] = S_{i(\alpha)}$

Define:


 * $Y= \left\{ {y \in T_1: i(y) = y } \right\}$

By the definition of a tower:

By Induction on Well-Ordered Set, conclude that $Y = T_1$.

Because $i(y) = y$, $Y = T_2$ on the image of $i$.

Thus $T_1$ is equal to $T_2$ or an initial segment of $T_2$.