# If $\alpha\le\beta$ are ordinal numbers, then the equation $\xi+\alpha=\beta$ cannot have exactly $n$ solutions in $\xi$, for $1<n<\omega$

If $$\alpha\le\beta$$ are ordinal numbers, then the equation $$\xi+\alpha=\beta$$ can have no solutions (in $$\xi$$); for example, when $$\beta$$ is a limit ordinal, and $$\alpha=1$$.

It can also have one single solution, for example when $$\alpha$$ and $$\beta$$ are finite ordinals, because in that case the sum is commutative.

It may also have infinite distinct solutions; for example when $$\alpha$$ and $$\beta$$ are equal to $$\omega$$ (it is well known that for any natural number $$n$$, $$n+\omega=\omega$$)

However, aside from this simple observations, how can we really prove the statement that this type of solutions will never have exactly $$n$$ distinct solutions, for $$n>1$$? I have thought of this: we could distinguish the following cases:

• $$\alpha=\beta$$

1.1 $$\alpha$$ and $$\beta$$ are finite

Then there is a unique solution, since the sum of natural numbers is commutative, and $$0$$ is the only ordinal for which $$\alpha+0=\beta$$

1.2 $$\alpha$$ and $$\beta$$ are infinite

Then there are infinitely many solutions; at least all the ordinals with lower cardinality than that of $$\alpha$$. Which are all the solutions of the equation? I think that all those solutions should be obtained by adding to the set of ordinals with lower cardinality, all the ordinals less than the greatest ordinal contained in $$\alpha$$, with its same cardinal, that is a union of a set of limit ordinals (such as $$\omega^2,\,\omega^{\omega},\,\varepsilon_0,\dots$$). How could we prove this statement, if true?

• $$\alpha<\beta$$

2.1 $$\alpha$$ and $$\beta$$ are finite

Then there is a unique solution, since the sum of natural numbers is commutative, and we know that in this case there exists one unique ordinal $$\xi$$ such that $$\alpha+\xi=\beta$$ (in aprticular, that $$\xi$$ will be greater than one)

2.2 $$\alpha$$ and $$\beta$$ are infinite

• $$\alpha$$ and $$\beta$$ have different cardinality

Then there are can be no solutions, or one solution, but not infinite solutions. How can we prove this? (for example, $$\xi+\omega\not=\omega_1$$ for any $$\xi$$, but $$\xi+\omega=\omega_1+\omega$$ for $$\xi=\omega_1$$)

• $$\alpha$$ and $$\beta$$ have the same cardinality

Then there can be no solutions, or one solution, but not infinite solutions. Again, How can we prove this? (for example $$\xi+\omega_1\not=\omega_1+1$$ but $$\xi+\omega_1=\omega_{1}2$$ for $$\xi=\omega_1$$)

2.3 $$\alpha$$ is finite but $$\beta$$ is infinite

Every infinite ordinal can be written in the following form: $$\beta=\gamma+n$$ where $$\gamma$$ is a limit ordinal and $$n$$ is a natural number. In this circumstance, the equation can be reduced to $$\xi+m=\gamma+n$$. So, there are two cases:

• If $$m\le n$$, then there exists a unique natural number $$n'$$ such that $$m+n'=n$$, so setting $$\xi=\gamma+n'$$ we find a (unique, for $$n'$$ is unique) solution to the equation.

• If $$m>n$$, then there are no solutions to the equation. How can we prove this?

I think this chaotic configuration of cases is not the optimal way to demonstrate the statement is true. Can there be a shorter, simpler way to understand this result? Any comments on the hightlighted points?

• By $n\gt1$ I guess you mean $1\lt n\lt\omega$ ?
– bof
Commented Apr 29, 2019 at 12:09
• Obviously, yes. I just changes the title of the question. Commented Apr 29, 2019 at 12:16

Suppose $$\xi+\alpha=\beta$$ and $$\eta+\alpha=\beta$$ where $$\xi\ne\eta$$. We may assume that $$\xi\lt\eta$$, and so $$\xi+\alpha\le\xi+1+\alpha\le\eta+\alpha=\xi+\alpha,$$ so $$\xi+\alpha=\xi+1+\alpha$$, so $$\alpha=1+\alpha$$. It follows that $$n+\alpha=\alpha$$ for all $$n\lt\omega$$, so $$\xi+n+\alpha=\beta$$ for all $$n\lt\omega$$.