# Equality of Towers in Set

## 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 tower, $\left({T_1,\preccurlyeq_1}\right)$ and $\left({T_2,\preccurlyeq_2}\right)$ are well-ordered sets.

the towers are order isomorphic to each other, or:
one is isomorphic to an initial segment in the other.

Without loss of generality, in the second case, assume that $\left({T_1,\preccurlyeq_1}\right)$ is order isomorphic to an initial segment in $\left({T_2,\preccurlyeq_2}\right)$.

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

such that $i[T_1]$, the image set of $i$, is equal to $T_2$ or to 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 $i[S_t] = S_{i(t)}$.

Define:

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

Then for any $t \in T_1$ and $S_t \subseteq Y$:

 $\displaystyle t$ $=$ $\displaystyle c\left({X \setminus S_t}\right)$ Definition of Tower in Set $\displaystyle$ $=$ $\displaystyle c\left({X \setminus i\left[{S_t}\right]}\right)$ by definition of $Y$ $\displaystyle$ $=$ $\displaystyle c\left({X \setminus S_{i(t)} }\right)$ as $i[S_t] = S_{i(t)}$ $\displaystyle$ $=$ $\displaystyle i(t)$ Definition of Tower in Set

So $i(y) = y$ for any initial segment $S_t \subseteq Y$.

By Induction on Well-Ordered Set, $Y = T$.

Thus $i$ is the identity mapping from $T_1$ onto its image.

Recall the image of $i$ is $T_2$ or an initial segment of $T_2$.

So $T_1 = T_2$ or an initial segment of $T_2$.

$\blacksquare$