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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?

Thanks in advance for your comments

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  • $\begingroup$ By $n\gt1$ I guess you mean $1\lt n\lt\omega$ ? $\endgroup$ – bof Apr 29 '19 at 12:09
  • $\begingroup$ Obviously, yes. I just changes the title of the question. $\endgroup$ – Akerbeltz Apr 29 '19 at 12:16
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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 $n\lt\omega$.

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  • $\begingroup$ That was pretty smart; any comments on the other questions I suggested in my post? $\endgroup$ – Akerbeltz Apr 29 '19 at 12:33
  • $\begingroup$ Too many questions for one post. One post, one question. $\endgroup$ – bof Apr 29 '19 at 12:42

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