Let $α$ be an ordinal and $A$ be a set of ordinals. Then $\sup\limits_{β∈A} (α+β) = α+\sup\limits_{β∈A}(β)$ My idea is to prove that $\alpha+ \sup\limits_{β∈A}(β)$ is the supremum of $\{α+β\mid β∈A\}$. While I'm able to prove that $\alpha+ \sup\limits_{β∈A}(β)$ is a upper bound of $\{α+β\mid β∈A\}$, I failed to show that $\alpha+ \sup\limits_{β∈A}(β)\le\gamma$ where $\gamma$ is an upper bound of $\{α+β\mid β∈A\}$.
The addition of ordinals is defined as follows $\alpha+0=\alpha$, $\alpha+(\beta+1)=(\alpha+\beta)+1$, and $\alpha+\beta=\sup\limits_{\gamma<\beta}(\alpha+\gamma)$ if $\beta$ is limit.
My attempt:
For all $\beta\in A,\beta\le\sup\limits_{β∈A}(β)$, then $\alpha+\beta\le\alpha+ \sup\limits_{β∈A}(β)$ and thus $\alpha+ \sup\limits_{β∈A}(β)$ is a upper bound of $\{α+β\mid β∈A\}$. Assume that $\gamma$ is an upper bound of $\{α+β\mid β∈A\}$, then $α+β\le\gamma$ for all $β∈A$.
My questions:

*

*Please leave me some hints to complete my proof!


*I meet another proof of this theorem on the Internet:

Could you please elaborate on the sentence Since set inclusion is the ordering, the two are equal. I'm unable to understand what this sentence means.
Thank you so much for your help!


For an ordinal $α$ and a set $A$ of ordinals:

*

*$\sup\limits_{β∈A} (α+β) = α+\sup\limits_{β∈A}(β)$


*$\sup\limits_{β∈A} (α\cdot β)=α\cdot \sup\limits_{β∈A}(β)$

 A: As you say, the $\leq$ inequality follows from the fact that $\forall \beta \in A, \alpha + \beta \leq \alpha + \sigma$, where $\sigma := \sup A$.
For the other inequality, you can argue as follows: if $\sigma \in A$, then the claim is trivial to prove, so assume that $\sigma \not\in A$. Then necessarily $\sigma$ is a limit ordinal, so $\sigma = \underset{\eta < \sigma}{\lim} \eta$ (this doesn't happen if $\sigma$ is a succesor ordinal!).
By definition of ordinal addition, $\alpha + \sigma = \underset{\eta < \sigma}{\lim} \alpha + \eta$. Convince yourself that you can exchange this expression with $\underset{\beta \in A}{\sup} \alpha + \beta$ using the definition of suprenum. So $\alpha + \sigma = \underset{\beta \in A}{\sup} \alpha + \beta$. But that is exactly what you want to prove.
As for the phrase Since set inclusion is the ordering, the two are equal, are you familiar with the basic definitions of ordinals as transitive and well ordered sets? The definition of $\alpha < \beta$ is $\alpha \in \beta$. It follows from the construction of the ordinals that $\alpha \subseteq \beta$ is the same as $\alpha \leq \beta$.
A really nice book explaining the basic stuff is Introduction to Set Theory, by Hrbacek (page 107 in the third edition).
A: On the basis of Guillermo Mosse's answer, I present a proof with detail here.


Lemma: Let $B,C$ be sets, $B⊆C$ such that $∀c∈C,∃b∈B:c≤b$. Then $\sup B=\sup C$.



$$\sup\limits_{β∈A} (α+β)=α+\sup\limits_{β∈A}(β)$$

Let $\sigma=\sup\limits_{β∈A}(β)$. For all $\beta\in A,\beta\le\sigma$, then $\alpha+\beta\le\alpha+ \sigma$ and thus $\alpha+ \sigma$ is a upper bound of $\{α+β\mid β∈A\}$. Assume that $\gamma$ is an upper bound of $\{α+β\mid β∈A\}$, then $α+β\le\gamma$ for all $β∈A$.


*

*If $\sigma\in A$ then $\sigma=\beta$ for some $\beta\in A$ and thus $\alpha+ \sigma=\alpha+ \beta\le\gamma$.

*If $\sigma\notin A$ then clearly $\sigma$ is a limit ordinal. We next prove that $\sup\limits_{\eta<\sigma}(\alpha+\eta)=\sup\limits_{\eta\in A}(\alpha+\eta)$.
Let $C=\{\alpha+\eta\mid \eta<\sigma\}$ and $B=\{\alpha+\eta\mid\eta\in A\}$. It's clear that $B\subseteq C$. For $\alpha+\eta\in C$, then $\eta<\sigma=\sup\limits_{β∈A}(β)$. By property of supremum, there is $\eta'\in A$ such that $\eta<\eta'<\sup\limits_{β∈A}(β)$. Thus $\alpha+\eta<\alpha+\eta'\in B$ for some $\eta'\in A$. Hence the conditions of Lemma are satisfied and thus $\sup B=\sup C$ or equivalently $\sup\limits_{\eta<\sigma}(\alpha+\eta)=\sup\limits_{\eta\in A}(\alpha+\eta)$.
It follows that $\alpha+\sigma=\sup\limits_{\eta<\sigma}(\alpha+\eta)=\sup\limits_{\eta\in A}(\alpha+\eta) \le\sup\limits_{\eta\in A}(\gamma)=\gamma$.


$$\sup\limits_{β∈A} (α\cdot β)=α\cdot \sup\limits_{β∈A}(β)$$

Let $\sigma=\sup\limits_{β∈A}(β)$. We have $\forall\beta\in A:\beta\le\sigma$, then $\forall\beta\in A:\alpha\cdot\beta\le\alpha\cdot\sigma$ and thus $\alpha\cdot\sigma$ is an upper bound of $\{α\cdot β\mid β∈A\}$. Assume that $γ$ is an upper bound of $\{α\cdot β\mid β∈A\}$, then $\forall\beta\in A:α\cdot β\le\gamma$.


*

*If $\sigma\in A$, then $\sigma=\beta$ for some $\beta\in A$ and thus $\alpha\cdot\sigma=\alpha\cdot\beta\le\gamma$.

*If $\sigma\notin A$, then $\sigma$ is clearly a limit ordinal. We next prove that $\sup\limits_{\eta<\sigma}(\alpha\cdot\eta)=\sup\limits_{\eta\in A}(\alpha\cdot\eta)$.
Let $C=\{\alpha\cdot\eta\mid \eta<\sigma\}$ and $B=\{\alpha\cdot\eta\mid\eta\in A\}$. It's clear that $B\subseteq C$. For  $\alpha\cdot\eta\in C$, then $\eta<\sigma=\sup\limits_{β∈A}(β)$. By property of supremum, there is $\eta'\in A$ such that $\eta<\eta'<\sup\limits_{β∈A}(β)$. Thus $\alpha\cdot\eta<\alpha\cdot\eta'\in B$ for some $\eta'\in A$. Hence the conditions of Lemma are satisfied and thus $\sup B=\sup C$ or equivalently $\sup\limits_{\eta<\sigma}(\alpha\cdot\eta)=\sup\limits_{\eta\in A}(\alpha\cdot\eta)$.
It follows that $\alpha\cdot\sigma=\sup\limits_{\eta<\sigma}(\alpha\cdot\eta)=\sup\limits_{\eta\in A}(\alpha\cdot\eta) \le\sup\limits_{\eta\in A}(\gamma)=\gamma$.
