A logical gap in the proof of $\sup\limits_{\beta\in A}(\alpha+\beta)=\alpha+\sup\limits_{\beta\in A}(\beta)$

Question

I found this proof on a notes CONSTRUCTING THE INTEGERS: $\Bbb N$, ORDINAL NUMBERS, AND TRANSFINITE ARITHMETIC by Jeremy Booher. You can download freely it here.

At page 6, he presents Lemma 28 and its proof:

Here is my understanding of his proof.

• Let $X=\{\gamma\mid\gamma<\sup\limits_{\beta\in A}(\alpha+\beta)\}$ and $Y=\{\gamma\mid\gamma<\alpha+\sup\limits_{\beta\in A}(\beta)\}$.

• Clearly, $\sup\limits_{\beta\in A}(\alpha+\beta)$ and $\alpha+\sup\limits_{\beta\in A}(\beta)$ are ordinals. To prove that $\sup\limits_{\beta\in A}(\alpha+\beta)=\alpha+\sup\limits_{\beta\in A}(\beta)$, the author intends to show that $X=Y$.

• For $\gamma\in X$, $\gamma<\sup\limits_{\beta\in A}(\alpha+\beta)$ and thus $\gamma<\alpha+\beta$ for some $\beta\in A$. Then $\gamma<\alpha+\sup\limits_{\beta\in A}(\beta)$ and thus $\gamma\in Y$. It follows that $X\subseteq Y$.

Clearly, the author has proved that $X\subseteq Y$, but he has NOT proved $Y\subseteq X$. In other words, he has shown that all ordinals less than $\sup\limits_{β∈A}(α+β)$ is also less than $α+\sup\limits_{β∈A}(β)$, whereas he has done NOTHING in proving the converse direction that all ordinals less than $α+\sup\limits_{β∈A}(β)$ is also less than $\sup\limits_{β∈A}(α+β)$. As a result, he can NOT claim Therefore the ordinals less than $\sup\limits_{β∈A}(α+β)$ and the ordinals less than $α+\sup\limits_{β∈A}(β)$ are the same. Thus he has not completed the proof.

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My questions:

1. Is my understanding of his proof correct?

2. How do I proceed to prove that $Y\subseteq X$?

Answer 1

You're correct in your understanding, but the remaining part of the proof is just the same thing again:

Let $\gamma < \alpha + \sup_{\beta \in A}\beta$. Then $\gamma < \alpha + \beta$ for some $\beta \in A$. So $\gamma < \sup_{\beta \in A}(\alpha + \beta)$.

Note that really the key thing here is that all the implications are bidirectional - so a more compact proof of both directions at once would result by dropping a couple words and using "if and only if":

$\gamma < \alpha + \sup_{\beta \in A}\beta$ if and only if $\gamma < \alpha + \beta$ for some $\beta \in A$, which holds if and only if $\gamma < \sup_{\beta + A}(\alpha + \beta)$.