For any ordinal $\alpha,$ $0+\alpha=\alpha+0=\alpha.$

Question

• For any ordinal $\alpha,$ $0+\alpha=\alpha+0=\alpha.$

By definition we have

$$0+\alpha=(\emptyset\times\left\{0\right\})\cup(\alpha\times\left\{1\right\})=\alpha.$$

So similarly $\alpha+0=\alpha.$

Can you check my proof? Thanks...

Answer 1

In this disjoint sum definition (which works for all order types) this would work as an argument; maybe expand it a bit more by noting that $\emptyset \times \{0\}=\emptyset$ so that $\alpha \times \{1\}$ is just an order-isomorphic copy of $\alpha$ etc. and we identify isomorphic order types.

This statement is often used as part of the recursive definition of addition for ordinals, in which case it needs no proof.