0

I am writing some notes on introductory set theory, starting from the basic axioms all the way to cardinal arithmetic. Right now I am up to ordinal arithmetic. I have defined ordinal addition and ordinal multiplication but not in the most typical way. Usually, most authors define ordinal addition/multiplication by induction. They define ordinal addition as follows:

$$1. \ \alpha+0=\alpha.$$

$$2. \ \alpha+\beta ^{+}=(\alpha+\beta)^{+} $$

$$3.\ \text{If λ is a limit ordinal, then } \alpha+\lambda=\bigcup_{\gamma\in\lambda}\left(\alpha +\gamma\right).$$

Ordinal multiplication is defined analogously. The reason I am not defining ordinal addition and multiplication in this way is because these definitions involve using recursive class functions and I want to avoid classes and class functions. Instead, I define ordinal addition and multiplication synthetically. If $\alpha$ and $\beta$ are ordinals, then I define $\alpha+\beta$ to be the order type of the well-ordered set $\left[\alpha\times\left\{ 0\right\} \right]\cup\left[\beta\times\left\{ 1\right\} \right]$ (with the ordering inherited from $\alpha$ and $\beta$.) I define $\alpha\beta$ to be the order type of the well-ordered set $\alpha\times\beta$ with the reverse lexicographic ordering. I then proved that the synthetic definitions of addition and multiplications agreed with the induction definitions. What I am having trouble with is defining ordinal exponentiation synthetically. There is way to do this, as shown here, but it involves using facts about finite sets. The problem is that I haven't defined finite sets and want to leave it till the next chapter on cardinal numbers. Is there another way to synthetically define ordinal exponentiation without using results about finite sets?

Eigenfield
  • 1,532
  • 2
    You can talk about decreasing functions. In well-orders they must have finite ranges. Of course, you might have to quotient by some equivalence relation, but I think that if you are being careful enough, you can avoid that too. – Asaf Karagila Dec 13 '18 at 10:16
  • 3
    From my notes here: There is also a non-recursive definition using a last-differences ordering on the functions in $A^B$ that have finite support (functions $f:B \rightarrow A$ such that $f(x)$ differs from $\min A$ for finitely many $x$ in $B),$ where $A$ and $B$ are well-ordered sets with order types $a$ and $b.$ Ordinal exponentiation can also be defined directly from transfinite products, where all the factors are the same. Although this assumes that transfinite products have been dealt with, some texts use this approach. – Dave L. Renfro Dec 13 '18 at 12:51
  • Do you know any of the books which define ordinal exponentiation using transfinite products? – Eigenfield Dec 13 '18 at 18:37
  • 1
    One such book is Kamke's Theory of Sets --- Chapter IV, Article 9 (Powers of Ordinal Numbers), p. 104: The power ${\mu}^{\alpha}$ is now once more defined as a product of equal factors. By specializing the general definition of the product, we obtain the following definition of the power: Put $\ldots$" Incidentally, the use of "is now once more" does not mean that ordinal exponentiation is being defined again in this manner, but rather it is a reference to the analogous way that Kamke defined exponentiation of cardinal numbers back on p. 41. – Dave L. Renfro Dec 13 '18 at 21:44
  • @DaveL.Renfro . The problem is that you have to mention "finitely many", which the proposer is asking us to avoid. – DanielWainfleet Dec 17 '18 at 08:20
  • 1
    Before defining exponentiation, you surely defined $\omega$, right? Then you can describe finite ordinals without using cardinality: It's just the ordinals that are less than $\omega$. – celtschk Dec 17 '18 at 08:38
  • Noah Schweber has written a nice related answer: https://mathoverflow.net/a/358458 (Has the exponentiation of ordinals a nice geometric model?) – C7X May 24 '22 at 06:03

0 Answers0