Double induction on ordinals (If $\ \alpha+1=\beta+1$ then $\alpha=\beta$)

Question

Iv'e encountered the following exercise: Show that for any $\alpha,\beta\in\text{Ord}$ if $\alpha+1=\beta+1$ then $\alpha=\beta$

I think the way to prove this is by induction on, say, $\alpha$. Base $\left(\alpha=0\right)$ is easy, but for the successor step I'm stuck. How can I use the information $$\alpha+1=\beta+1\implies\alpha=\beta$$ to prove $$\left(\alpha+1\right)+1=\gamma+1\implies\alpha+1=\gamma$$ Or can I somehow use some sort of “double induction” on both $\alpha$ and $\beta$?

I also have the same problem with the limit step.

Answer 1

I wouldn't use induction, but the raw definitions:

$\alpha+1$ and $\beta+1$ are both totally ordered by $\in$. A total order has at most one maximal element. $\alpha$ is a maximal element of $\alpha+1=\alpha\cup\{\alpha\}$ ...

Answer 2

If you want induction:

Assume it is true for $\alpha: \forall\beta(\alpha+1=\beta+1\implies \alpha=\beta)$

So we have $\alpha+2=\gamma+1$.

Assume $\alpha+1\ne \gamma$, in this case we need to consider 2 cases:

$\gamma\in\alpha+1$, in this case we have $\gamma+1\le \alpha+1$, if those 2 are equal by assumption $\alpha=\gamma$, add one to both sides you get contradiction. If $\gamma+1<\alpha+1$ we get $\alpha+2<\alpha+1$, contradiction again.

The other case is $\alpha+1\in\gamma$ which implies $\alpha+2\in\gamma+1=\alpha+2$, contradiction.

Hence $\alpha+1=\gamma$.

Now I would argue that there is no need for the limit case. The claim is about $\alpha+1$, so it is given that we are working with successor of $\alpha$