Ordered Sum of Tosets is Totally Ordered Set

Theorem
Let $\left({S, \preceq_1}\right)$ and $\left({T, \preceq_2}\right)$ be tosets.

Let $S + T = \left({S \cup T, \preceq}\right)$ be the ordered sum of $S$ and $T$.

Then $\left({S \cup T, \preceq}\right)$ is itself a toset.

General Definition
Let $S_1, S_2, \ldots, S_n$ all be tosets.

Let $T_n$ be the ordered sum of $S_1, S_2, \ldots, S_n$:


 * $\forall n \in \N^*: T_n = \begin{cases}

S_1 & : n = 1 \\ T_{n-1} + S_n & : n > 1 \end{cases}$

Then $T_n$ is a toset.

Proof
By definition of ordered sum, we have that:


 * If $a, b \in S$, then $a \preceq b \iff a \preceq_1 b$.
 * Otherwise, if $a, b \in T$, then $a \preceq b \iff a \preceq_2 b$.
 * If neither of these is the case, then $a \in S, b \in T \iff a \preceq b$.

First we show that $\preceq$ is connected.

We note that as $\left({S, \preceq_1}\right)$ and $\left({T, \preceq_2}\right)$ are both tosets, then $\preceq_1$ and $\preceq_2$ are both connected.

Let $a, b \in S \cup T$.

There are three cases:


 * $a, b \in S$, in which case $a \preceq_1 b$ or $b \preceq_1 a$
 * $a, b \notin S$, in which case $a \preceq_2 b$ or $b \preceq_2 a$
 * 1) Otherwise, in which case either $a \in S, b \notin S$ or $a \notin S, b \in S$. By definition of $\preceq$ it follows that $a \preceq b$ or $b \preceq a$, but clearly not both.

So in all cases $a \preceq b$ or $b \preceq a$, and $\preceq$ is connected.

Now we check in turn each of the criteria for an ordering:

Reflexivity
We have by definition of set union:
 * $\forall a \in S \cup T: a \in S \lor a \in T$

and so either $a \preceq_1 a$ or $a \preceq_2 a$, showing $a \preceq a$.

So $\preceq$ is reflexive.

Transitivity
Suppose $a \preceq b \preceq c$.

We proceed on a case-by-case basis.

There are eight of these, (indirectly) from Cardinality of Power Set.


 * Suppose $a, b, c \in S$. Then $a \preceq c$ by transitivity of $\preceq_1$.


 * Suppose $a, b \in S, c \notin S$. Then $a \preceq c$ by definition of $\preceq$.


 * Suppose $a \in S, b, c \notin S$. Then $a \preceq c$ by definition of $\preceq$.


 * Suppose $a, b, c \notin S$. Then $a \preceq c$ by transitivity of $\preceq_2$.

There are four other cases to consider:
 * $a, c \in S, b \notin S$
 * $b, c \in S, a \notin S$
 * $c \in S, a, b \notin S$
 * $b \in S, a, c \notin S$.

None of these can happen as otherwise one of $a \npreceq b$ or $b \npreceq c$ would be the case.

So we have shown that $\preceq$ is transitive.

Antisymmetry
Suppose $a \preceq b$ and $b \preceq a$.


 * It can not be the case that $a \in S$ and $b \notin S$ because then $b \npreceq a$.
 * It can not be the case that $a \notin S$ and $b \in S$ because then $a \npreceq b$.

So either $a, b \in S$ or $a, b \notin S$.


 * If $a, b \in S$ then $a = b$ by antisymmetry of $\preceq_1$.


 * If $a, b \notin S$ then $a = b$ by antisymmetry of $\preceq_2$.

Thus in all cases it can be seen that $a \preceq b \preceq a \implies a = b$.

So $\preceq$ is antisymmetric.

So we have shown that:
 * $\preceq$ is connected
 * $\preceq$ is reflexive, transitive and antisymmetric.

Thus by definition, $\preceq$ is a total ordering and so $\left({S \cup T, \preceq}\right)$ is a toset.

Proof of General Result
We have that $S_1 + S_2$ is a toset from the main result.

Suppose $T_{n-1}$ is a toset.

Given that $S_n$ is a toset, it follows from the main result that $T_{n-1} + S_n$ is also a toset.

The result follows by induction.