how to prove the addition of transfinite cardinal numbers?

Question

How do you prove the following transfinite cardinal addition?:

$ \alpha + \beta = \max(\alpha,\beta)$?

And as the consequence, $\alpha + \alpha = \alpha$ where $\alpha$ and $\beta$ are transfinite cardinal numbers?

Answer 1

To show that $\alpha + \beta = \beta$ whenever $\alpha \leq \beta$ are cardinals and $\beta$ is infinite, one often goes through the following sequence of deductions:

1. First demonstrate that $\beta \cdot \beta = \beta$ for infinite cardinals $\beta$.
2. As a corollary to (1) demonstrate that $\beta + \beta = \beta$.
3. As a corollary to (2) demonstrate the general case.

The "trick" for the first part is to construct a well-ordering on $\beta \times \beta$ (the cartesian product) which has order-type $\beta$.

For example (and this is how it is done in Jech's text), we could define $\langle \xi , \zeta \rangle \leq \langle \eta , \nu \rangle$ iff one of the following holds:

1. $\max \{ \xi , \zeta \} < \max \{ \eta , \nu \}$; or
2. $\max \{ \xi , \zeta \} = \{ \eta , \nu \}$ and $\xi < \eta$; or
3. $\max \{ \xi , \zeta \} = \{ \eta , \nu \}$ and $\xi = \eta$ and $\zeta \leq \nu$.

By transfinite induction we can then show that for each infinite cardinal $\beta$ the associated well-ordering has order-type $\beta$. (Transfinite induction is possible for the ordering defined here because if $\alpha < \beta$ are infinite cardinals, then the ordering on $\alpha \times \alpha$ is an initial segment of the ordering on $\beta \times \beta$.)

Since there is a well-ordering on $\beta \times \beta$ of order-type $\beta$, it follows that the sets $\beta \times \beta$ and $\beta$ have the same cardinality. As we define the cardinal product $\beta \cdot \beta$ to be $| \beta \times \beta |$, we are done.

Answer 2

How would you prove $\aleph_0 + \aleph_0 = \aleph_0$? If you do that in the obvious way, can you think about how to generalize it to arbitrary cardinals $\alpha$? (It might help to think about Cantor normal form for the latter, but not necessary). Then if you can do that, can you see how you'd get $\alpha + \beta = \max(\alpha, \beta)$?