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.
By Wosets are Isomorphic to Each Other or Initial Segments, either:
- $(1): \quad$ 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 $\struct {T_2, \preccurlyeq_2}$.
By Order Isomorphism iff Strictly Increasing Surjection, there exists a strictly increasing mapping:
- $i: \struct {T_1, \preccurlyeq_1} \to \struct {T_2, \preccurlyeq_2}$
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 $T_2$.
$\blacksquare$
Sources
- 2000: James R. Munkres: Topology (2nd ed.) $\S 1.11$ Supplementary Exercise $7$