Weak monotonicity of ordinal addition I'm trying to prove the weak monotonicity of ordinal addition, i.e. if $\alpha \leq \beta$, then $\alpha + \gamma \leq \beta + \gamma$. The proof is not all that difficult, but I want to make sure I got the last step correctly. Basically, I proceeded by transfinite induction on $\gamma$. As I said, I think I pretty much nailed the base case and the successor case, but I'm still not sure about the limit case. Here's my argument:
$\gamma$ is a limit ordinal. Then $\alpha + \gamma = \text{sup}(\{\alpha + \delta \; |\; \delta < \gamma\})$ and $\beta + \gamma = \text{sup}(\{\beta + \delta \; |\; \delta < \gamma\})$. Hence, what we need to show is that, if $\alpha \leq \beta$, then $\text{sup}(\{\alpha + \delta \; |\; \delta < \gamma\}) \leq \text{sup}(\{\beta + \delta \; |\; \delta < \gamma\})$. If $\alpha = \beta$, the claim is trivial, so assume $\alpha < \beta$. By the induction hypothesis, $\alpha + \delta \leq \beta + \delta$ for every $\delta < \gamma$; hence, $\text{sup}(\{\alpha + \delta \; |\; \delta < \gamma\}) \leq \text{sup}(\{\beta + \delta \; |\; \delta < \gamma\})$.
My doubt relates to the last step in the argument. I figured it obvious, but then again, since I don't know if I should trust my intuition in these cases. Any comment wuld be greatly appreciated.
 A: Recall that if $\eta\leq\zeta$ then $\eta\subseteq\zeta$ as sets; and that $\sup A=\bigcup A$ when $A$ is a set of ordinals.
That should be enough to fill in the gap that you (rightfully) feel exists in your proof.

It should also be noted that if you consider the definition of ordinal addition by partitions, that is $\alpha+\beta=\gamma$ if and only if $\gamma$ is order-isomorphic to $\{0\}\times\alpha\cup\{1\}\times\beta$, then the proof of weak monotonicity is even simpler.
If $\alpha\leq\beta$ then we can easily define an order-embedding from $\alpha+\gamma$ into $\beta+\gamma$. Since no ordinal is order isomorphic to its proper initial segment, it follows that $\alpha+\gamma\leq\beta+\gamma$.
A: I'm not sure you can say that $\alpha + \gamma = \text{sup}(\{\alpha + \delta \; |\; \delta < \gamma\})$ so trivially.
What you know is that $\alpha + \gamma = \alpha + \text{sup}(\{\delta \; |\; \delta < \gamma\})$, which is not the same, right? You are sort of using what you are trying to prove when you affirm the first equality.
