Wosets are Isomorphic to Each Other or Initial Segments/Proof Without Using Choice

Proof
If the sets $S$ and $T$ considered are empty or singletons, the theorem holds vacuously or trivially.

Thus assume $S$ and $T$ each contain at least two elements.

Let $U = \struct {S, \preceq_S} \cup \struct {T, \preceq_T}$

Define the following relation $\preceq$ on $U$:


 * $\forall x, y \in U: x \preceq y$




 * $x, y \in S: x \preceq_S y$

or:
 * $x, y \in T: x \preceq_T y$

or:
 * $x \in S, y \in T$

We claim that $\preceq$ is a well-ordering.

First, we show it is a total ordering.

Checking in turn each of the criteria for a total ordering:

Reflexivity
If $x = y$, they're necessarily both in $S$ or $T$ simultaneously.

Reflexivity then follows from $\preceq_S$ and $x \preceq_T$ being reflexive, as they are both orderings.

Transitivity
Let $x, y, z \in U$.

If $x, y, z \in S$ or $x, y, z \in T$ simultaneously, then $\preceq$ is transitive by the transitivity of $\preceq_S$ and $\preceq_T$.

Suppose $x, y \in S$ and $z \in T$.

Let $x \preceq y$ and $y \preceq z$.

Then $x \preceq z$ because $x \in S$ and $y \in T$.

Suppose $x \in S$ and $y, z \in T$.

Then $x \preceq z$ also because $x \in S$ and $y \in T$.

Thus $\preceq$ is transitive.

Antisymmetry
Let $x \preceq y$ and $y \preceq x$.

If $x, y \in S$ then $x = y$ by the antisymmetry of $\preceq_S$.

Likewise if $x, y \in T$.

If $x \in S$ and $y \in T$, then $y \in S$ and $x \in T$ as well.

Thus $x = y$ from the antisymmetry of $\preceq_S$ or $\preceq T$.

Conclude that $\preceq$ is a total ordering.

To show $\preceq$ is a well-ordering, consider a non-empty set $X \subseteq U$.

Then either:
 * $X \cap S = \O$

or:
 * $X \cap T = \O$

or:
 * $X \cap S$ is non-empty and $X \cap T$ is non-empty.

In the first case, $X \subseteq T$, by Intersection with Complement is Empty iff Subset.

Then $X$ has a smallest element defined by $\preceq_T$.

In the second case, $X \subseteq S$, also by Intersection with Complement is Empty iff Subset.

Then $X$ has a smallest element defined by $\preceq_S$.

In the third case, the smallest element of $X \setminus T$ is an element of $S$.

Thus it precedes any element of $T$ by the definition of $\preceq$.

The smallest element of $X \setminus T$, which is a subset of $S$, is guaranteed to exist by the well-ordering on $S$.

This smallest element is then also the smallest element of $\left({X \setminus T}\right) \cup T = X$.

Thus $\preceq$ is a well-ordering on $S \cup T$.

Consider the mapping:


 * $k: \struct {T, \preceq_T} \to \struct {U, \preceq}$:


 * $\map k \alpha = \alpha$

Then $k$ is strictly increasing, by the construction of $\preceq$.

Thus there is a strictly increasing mapping from $T$ to $U$.

From Strictly Increasing Mapping Between Wosets Implies Order Isomorphism, $T$ is order isomorphic to $U$ or an initial segment of $U$.

Let $\II_x$ denote the initial segment in $U$ determined by $x$, according to $k$.

Note that $\II_x = \II_{\map k x}$, because $\map k x = x$.

Suppose $x \in S$.

Then $\II_x \subseteq S$ because $\preceq$ is a well-ordering.

Thus there is an order isomorphism from $T$ to $\II_x$ in $S$.

Suppose $x = \min T$, the smallest element of $T$.

Then every element of $S$ strictly precedes $x$, as $x$ is in $T$.

Also, $x$ precedes every element of $T$, so $\II_x \ne T$.

Thus there is an order isomorphism from $T$ to all of $S$.

Suppose $x \in T$ and $x \ne \min T$.

Then $\II_x$ defines an initial segment in $T$.

Also, every element of $S$ strictly precedes $x$, as $x$ is in $T$.

Thus there is an order isomorphism from an initial segment of $T$ to all of $S$.

The cases are distinct by No Isomorphism from Woset to Initial Segment.