# Equality of Towers in Set

## Theorem

Let $X$ be a non-empty set.

Let $\struct {T_1, \preccurlyeq_1}$ and $\struct {T_2, \preccurlyeq_2}$ be towers in $X$.

Then either:

$\struct {T_1, \preccurlyeq_1} = \struct {T_2, \preccurlyeq_2}$

or:

$\struct {T_1, \preccurlyeq_1}$ is an initial segment of $\struct {T_2, \preccurlyeq_2}$

or:

$\struct {T_2, \preccurlyeq_2}$ is an initial segment of $\struct {T_1, \preccurlyeq_1}$

## Proof

By the definition of tower:

$\struct {T_1, \preccurlyeq_1}$ and $\struct {T_2, \preccurlyeq_2}$ are well-ordered sets.
$(1): \quad$ the towers are order isomorphic to each other

or:

$(2): \quad$ one is order isomorphic to an initial segment in the other.

Without loss of generality, in case $(2)$, assume that $\struct {T_1, \preccurlyeq_1}$ is order isomorphic to an initial segment in $\struct {T_2, \preccurlyeq_2}$.

$i: \struct {T_1, \preccurlyeq_1} \to \struct {T_2, \preccurlyeq_2}$

such that $i \sqbrk {T_1}$, the image set of $T_1$ under $i$, 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:

$\forall t \in T_1: \map i t = \map \min {T_2 \setminus i \sqbrk {S_t} }$

and:

$i \sqbrk {S_t} = S_{\map i t}$

Define:

$Y = \set {y \in T_1: \map i y = y}$

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

 $\ds t$ $=$ $\ds \map c {X \setminus S_t}$ Definition of Tower in Set $\ds$ $=$ $\ds \map c {X \setminus i \sqbrk {S_t} }$ by definition of $Y$ $\ds$ $=$ $\ds \map c {X \setminus S_{\map i t} }$ as $i \sqbrk {S_t} = S_{\map i t}$ $\ds$ $=$ $\ds \map i t$ Definition of Tower in Set

So $\map i y = y$ 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 that the image set of $i$ is $T_2$ or an initial segment of $T_2$.

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

$\blacksquare$