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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?

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

  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.

    The motivation for this definition is that we want to have "no holes", or "continuity" in the second variable for addition. So at limit steps we take the least value we can, which is the supremum of the previously attained values.

The same thing can be said for multiplication and exponentiation. Although the "direct" definition for exponentiation is a bit awkward in my taste, so I prefer to think about the recursive definition.

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