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:


 * 1) $$a, b \in S$$, in which case $$a \preceq_1 b$$ or $$b \preceq_1 a$$;
 * 2) $$a, b \notin S$$, in which case $$a \preceq_2 b$$ or $$b \preceq_2 a$$;
 * 3) 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:
 * $$\forall a \in S \cup T: a \in S \or a \in T$$

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.