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.

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.

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

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 $i[T_1]$, 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 $T_2$.

$\blacksquare$

Sources

• 2000: James R. Munkres: Topology (2nd ed.) $\S 1.11$ Supplementary Exercise $7$