Proving basic fact about ordinals

Question

In a set of online lecture notes, I saw the following proposition.

Let $C$ be a set of ordinals. Then $\sup \left\{ \alpha +\beta:\beta\in C \right\} =\alpha +\sup C$.

How can I prove this?

Answer 1

For brevity, we'll write $\beta^+ := \beta+1$ for the successor ordinal of $\beta$.

If $\sup(C) = 0$ then it's trivial.

If $\sup(C) = \beta^+$ a successor ordinal, then $\beta^+ \in C$ so we get that the LHS is $\alpha + \beta^+$ immediately.

If $\sup(C) = \lambda$ a nonzero limit, then if $\lambda \in C$, we're done as above, so suppose it's not in $C$. By definition of ordinal addition, $\alpha + \sup(C) = \alpha + \lambda = \sup \{ \alpha + \gamma : \gamma < \lambda \}$. That sup is the same as $\sup \{ \alpha + \beta : \beta \in C \}$: containment one way is because everything in $C$ is $< \lambda$, while containment the other is because $\lambda$ is the sup of $C$ so for any ordinal less than $\lambda$ we can find a larger ordinal which is also less than $\lambda$.