If $\alpha < \beta $ then $\exists$ a unique $\gamma$ st. $\alpha+\gamma=\beta$, where '$+$' denotes ordinal addition.

If someone would be so kind to check the following attempt, if only for the case where $\gamma$ is a limit ordinal (the part that I am concerned over)


Consider the class $X:=\{\eta:\alpha+\eta\geq \beta\}$

Since $\alpha+\beta\geq \beta\implies \beta\in X$, so $X\neq\emptyset$

$\therefore \emptyset\neq X\subseteq On$ and since $On$ is well ordered by the elementhood relation, and since $X$ is a non-empty subset of $On$: $X$ has a least element. Call this least element $\gamma$.

So $\forall \ \delta \in X$ st. $ \delta \neq \gamma \implies \delta <\gamma$

Claim: Since $\gamma$ is the least element of $X$ $\implies \alpha+\gamma=\beta$

Since $\gamma\in X \implies \alpha+\gamma \geq \beta$. To prove the claim, assume $\alpha+\gamma\neq\beta$. Then we have $\alpha+\gamma > \beta$

We prove the claim by showing the last mentioned inequality does not hold for any ordinal, forcing the desired equality made in the claim.

If $\gamma=0\implies \alpha \geq \beta $, contradicting the initial assumption $\alpha < \beta $

If $\gamma=\epsilon+1$ for some $\epsilon\in On$

So $(\alpha+\epsilon)+1=\alpha+(\epsilon+1)>\beta $, also, $\alpha+\epsilon<\alpha+(\epsilon+1)$ $\require{cancel}$ Hence, $\beta\leq\alpha+\epsilon<(\alpha+\epsilon)+1$ since $\ \cancel{\exists} \kappa$ st. $\alpha+\epsilon<\kappa<(\alpha+\epsilon)+1$. But, since $\beta\leq\alpha+\epsilon\implies\epsilon \in X$, contradicting the minimality of $\gamma=\epsilon+1$ since $\epsilon<\epsilon+1$

Lastly, if $\lim\gamma$ then $\alpha+\gamma=\bigcup_{\delta<\gamma}{\alpha+\delta}$

For any $\delta'<\gamma\implies \alpha+\delta'<\alpha+\gamma$. So along with $\alpha+\gamma > \beta$. This means $\alpha+\delta'<\beta<\alpha+\gamma$. Since if $\beta<\alpha+\delta'<\alpha+\gamma$ this would once again contradict the minimality of $\gamma$ since $\delta'<\gamma$.

But since $\alpha+\delta'<\beta$ where $\delta'<\gamma$ was arbitrary it means that $\forall \ \delta'<\gamma \implies \alpha+\delta'<\beta$ $\ \text{i.e.}\ \beta$ is an upper bound for the set $Y:=\{\alpha+\delta:\delta<\gamma\}$.

Since $Y$ is a set of ordinals $\bigcup Y$ is also an ordinal and is the $\text{l.u.b.}$ for the set $Y.$ $\text{i.e.} \sup Y=\bigcup Y$. But $\bigcup Y=\bigcup_{\delta<\gamma}{\alpha+\delta}=\alpha+\gamma$ by the definition of ordinal addition in the case of limit ordinals.

Also $\sup Y\leq u$ for any upper bound $u$ for the set $Y$. So, $\alpha+\gamma=\sup Y \leq \beta$ since we showed $\beta$ is a upper bound for the set $Y$. So we have $\alpha+\gamma\leq \beta$ and $\alpha+\gamma>\beta$, and both of these inequalities cannot hold simultaneously, contradiction.

This shows the inequality $\alpha+\gamma> \beta$ yields a contradiction for any ordinal $\gamma$. Hence our claim is shown to be true.

Uniqueness follows easily assuming there is some $\zeta$ st. $\alpha+\gamma=\beta $ and $\alpha +\zeta =\beta$

Since then $\alpha+\gamma=\alpha +\zeta$ iff $\gamma=\zeta$, showing uniqueness.

  • 1
    $\begingroup$ Two basic properties of ordinal addition :(1) $a+e+1>b\implies a+e\geq b$....(2). For $a\in On$ and $S\subset On $ we have $\sup\{a+s: s\in S\}=a+\sup S.$.... You can also find $\gamma$ from below: $\gamma=$ $\sup \{\eta \leq \beta :\alpha+\eta \leq \beta\}$. $\endgroup$ – DanielWainfleet Jun 21 '16 at 17:11

There are a few minor errors and one major one in the proof of the Claim. First, $\gamma=0$ implies that $\alpha\ge\beta$, not that $\alpha>\beta$; this is of course still a contradiction. In the successor case you shouldn’t say that there is no $\color{red}\beta$ strictly between $\alpha+\epsilon$ and $\alpha+(\epsilon+1)$: $\beta$ is already a specific ordinal. What you mean is that there is no ordinal $\delta$ such that $\alpha+\epsilon<\delta<\alpha+(\epsilon+1)$, so in particular it’s not the case that $\alpha+\epsilon<\beta<\alpha+(\epsilon+1)$, and therefore $\beta\le\alpha+\epsilon$. (Note that here again you have $<$ where you really need $\le$.) Thus, $\epsilon\in X$ and $\epsilon<\gamma$, which is a contradiction.

Finally, you’re right to worry about the argument in the limit case: $X$ is not a transitive set. Fortunately, a simple argument is available. If $\delta<\gamma$, then $\delta\notin X$, so $\alpha+\delta<\beta$. Thus, $\beta$ is an upper bound for $\{\alpha+\delta:\delta<\gamma\}$, and it follows that


giving us the desired contradiction.


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