# Ordinal subtraction as set difference

Consider the following synthetic definitions for ordinal arithmetic:

1. $\alpha + \beta$ is the order type of $\alpha\times\{0\}\cup\beta\times\{1\}$ ordered by reverse lexicographical ordering.
2. $\alpha \cdot \beta$ is the order type of $\alpha\times\beta$ ordered by reverse lexicographical ordering.
3. $\alpha - \beta$ is the order type of $\alpha\setminus\beta$.

I struggle with the last definition. "Order type of", to my knowledge, means "ordinal isomorphic to". How do these isomorphisms work with set difference? Does this definition satisfy identities like $(\alpha + \beta) - \alpha = \beta$, or particularly $(\alpha + \alpha) - \alpha = \alpha$? Why do we not get $(\alpha + \alpha)\setminus \alpha = \emptyset$ (naive thought: put elements from $\alpha$ behind elements from $\alpha$, then remove all elements that are in $\alpha$)?

Remember that ordinals are sets, linearly (in fact, well) ordered by $\in$. So if I have a subset of an ordinal $A\subseteq\alpha$, then $A$ is also linearly ordered by $\in$, and I can talk about $A$'s order type. In the case of $(\alpha+\alpha)-\alpha$, we get $\alpha$ because $(\alpha+\alpha)\setminus \alpha=\{\alpha+\theta: 0\le\theta<\alpha\},$ which clearly is order-isomorphic to $\alpha$ via the map $\alpha+\theta\mapsto\theta$ (exercise: this map is well-defined).
It might help to do a small concrete example. Take $\alpha=2$. Then (exercise :P) $2+2=4$, which as a set is $4=\{0, 1, 2, 3\}$. Then $4\setminus 2=\{2, 3\}$ (since $2=\{0, 1\}$), which has order type $2$.