# Tower is Proper Subtower or all of Set

## Lemma

Let $X$ be a non-empty set.

Let $\left({T,\preccurlyeq}\right)$ be a tower in $X$.

Then if $T \ne X$, $\left({T,\preccurlyeq}\right)$ is a proper subtower in $X$.

## Proof

Suppose $T \ne X$.

Then there is an element $x \in X \setminus T$.

The singleton $\{x\}$ can trivially be given a well-ordering, as there are are no elements in $\{x\}$ other than $x$ itself.

Consider the ordered sum on $T \cup \{x\}$ defined by setting $t \prec x$ for all $t \in T$.

By Ordered Sum of Tosets is Totally Ordered Set, this gives a total ordering on $T \cup \{x\}$.

There is a smallest element of any subsetset $T \cup \{x\}$, either the one given by $\preccurlyeq$, or $x$ if the subset considered is the singleton $\{x\}$.

Extend the choice function $c$ defining $\left({T,\preccurlyeq}\right)$ by assigning $c\left({X\setminus S_x}\right) = x$.

Then $T$ is a proper subset of $\left({T \cup \{x\},\text{extension of} \preccurlyeq}\right)$.

So $T$ is a proper subtower in $X$.

$\blacksquare$