Bijection between $1+ \alpha$ and $\alpha + 1$

Question

Let $\alpha$ be an ordinal. I want to show that there is a bijection between the ordinals $1 + \alpha$ and $\alpha + 1$.

I tried to proceed by transfinite induction on $\alpha$; in case $\alpha$ is a successor ordinal it is rather easy to show the theorem. However, in case $\alpha$ is a limit ordinal it seems quite hard, as it is not immediately clear how to use the given bijections (because of the induction hypothesis) to construct the "main" bijection. Moreover, even if I would be able to show the result in this manner, the proof seems quite technical and long.

Is there a good (and short) way to show this?

Answer 1

You should be able to explicitly define bijections, and not appeal to an inductive proof. This might rely on certain facts, however:

• If $\alpha < \omega$, then $1 + \alpha = \alpha + 1$ (Prove this by induction).
• If $\omega \leq \alpha$ then $1 + \alpha = \alpha$. (Transfinite induction is probably useful here.)

Perhaps a good starting point is to explicitly define a bijection between $\omega$ and $\omega+1$ and then generalise the idea.

Answer 2

While you can prove this using transfinite induction, you can also note the following thing:

The order type of the lexicographic order on $\{\langle\alpha,0\rangle\}\cup\alpha\times\{1\}$ is $1+\alpha$, and $\alpha+1=\alpha\cup\{\alpha\}$. You should be able to use this to find a fairly explicit bijection now.