If $\mu$ is a limit ordinal, then we define that $$\alpha + \mu = \sup_{\beta < \mu} (\alpha + \beta).$$

Why don't we similarly define that if $\lambda$ is a limit ordinal, then $$\lambda + \beta = \sup_{\alpha < \lambda} (\alpha + \beta)?$$

Or if $\lambda$ and $\mu$ are limit ordinals, then

$$\lambda + \mu = \sup_{\alpha < \lambda, \beta < \mu} (\alpha + \beta)?$$

The same goes for ordinal multiplication and exponentiation - why are they defined that way?


Ordinal addition is really just concatenation of orders, then computing the order type. Because concatenation of well-orders is a well-order, there is a unique ordinal isomorphic to this concatenation. So everything is well-defined.

At limit points we want to have continuity, in the sense that addition and supremums commute, after all if we fix $\alpha$, then concatenating longer and longer ordinals should have the same result (at the limit stage) as concatenating the limit itself.

Remember that we first have the order structure, and we want the arithmetical operations to work with that.

  • $\begingroup$ Can you clarify the relationship between paragraphs 2 and 3? Paragraph 3 seems to be saying, 'ordinal arithmetic is defined that way because it has to be defined that way,' while Paragraph 2 seems to be saying, 'ordinal arithmetic is defined that way because it is convenient for it to be defined that way.' $\endgroup$ Jun 28 '13 at 2:29
  • $\begingroup$ Well, we do it because this way the addition behaves like the order structure is expected to behave. Since we already have an order structure, we want the addition to play nice with it. So by "convenient" we really want something which is compatible with the order structure in the way we feel it should be. $\endgroup$
    – Asaf Karagila
    Jun 28 '13 at 8:02

If you defined

$$\lambda + \beta = \sup\limits_{\alpha < \lambda} (\alpha + \beta)$$

then you would have $\lambda + \beta = \lambda$ for all $\beta \leqslant \lambda$. In particular, no limit ordinal would have a successor.


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