# Justification for ordinal arithmetic at limit ordinals

Wikipedia tells me that:

$$\alpha + \lambda := \bigcup_{\beta < \lambda} \left ( \alpha + \beta \right )$$ for a limit ordinal $\lambda$. Multiplication and exponentiation, as Wikipedia says, are defined similarly.

My question is: what justifies this definition?

• It's the smallest ordinal bigger than $\alpha + \beta$ for all $\beta < \lambda$. – ogogmad May 19 '15 at 17:19
• @user3491648 doesn't help me one bit, considering I'm a complete newbie. – user132181 May 19 '15 at 17:20
• if $\alpha$ and $\beta$ are ordinals then $\alpha \cup \beta$ is the maximum of $\alpha$ and $\beta$. – ogogmad May 19 '15 at 17:22
• @user3491648 that I understand :) – user132181 May 19 '15 at 17:23
• try to read my first comment again and see if you understand it better now. – ogogmad May 19 '15 at 17:25

1. $$\alpha+\beta=\gamma$$ if and only if $$\gamma$$ is the unique ordinal that can be written as a disjoint union of an initial segment of order type $$\alpha$$, and a tail-segment of order type $$\beta$$.
You can check from this definition that addition indeed behaves this way for limit ordinals (since the tail-segment will be the union of the $$\gamma$$'s).
2. You define ordinal addition by induction: $$\alpha+0=\alpha$$; $$\alpha+s(\beta)=s(\alpha+\beta)$$, where $$s$$ is the successor function; and $$\alpha+\beta=\sup\{\alpha+\gamma\mid\gamma<\beta\}$$ for $$\beta$$ limit.