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We can embed $\mathbb{N}$ in a larger number system, such as $\mathbb{Z}$, $\mathbb{Q}$ or $\mathbb{R}$, for convenience. Now since $\mathbb{N}$ is extended by $\mathrm{Ord}$ and $\mathrm{Card}$, the ordinal and cardinal number systems, I was wondering, can we embed $\mathrm{Ord}$ and $\mathrm{Card}$ into larger number systems, also for convenience?

Note that the surreal numbers do not achieve this: if you've heard that $\mathrm{Ord}$ and $\mathrm{Card}$ embed into the surreal numbers, this statement is deceptive, because for example the addition operations do not coincide. (Observe that, since the surreal numbers form a field, thus addition is commutative; whereas addition in $\mathrm{Ord}$ is not. On the other hand, in the surreal numbers we have $x+1 \neq x$, whereas in cardinal arithmetic it holds that $\aleph_0 + 1 = \aleph_0$.)

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    $\begingroup$ What sort of "convenience" are you looking for? $\endgroup$
    – user642796
    May 7, 2013 at 7:31
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    $\begingroup$ Note that the cardinals do not embed into the ordinals as an arithmetic system. $1+\omega=\omega\neq\omega+1$ whereas $1+\aleph_0=\aleph_0=\aleph_0+1$. $\endgroup$
    – Asaf Karagila ♦
    May 7, 2013 at 7:36
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    $\begingroup$ Note that an extension to "integral ordinals" would require either (1) some ordinals do not have unique "additive inverses", or (2) "integral ordinal addition" is non-associative. (Otherwise if $\delta > 0$ is an (additively) indecomposable ordinal, then for any $0 < \alpha < \delta$ we would have $\alpha = \alpha + 0 = \alpha + ( \delta + ( - \delta ) ) = ( \alpha + \delta ) + ( - \delta ) = \delta + ( - \delta ) = 0$, which is absurd!) $\endgroup$
    – user642796
    May 7, 2013 at 8:13
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    $\begingroup$ I would argue that the Surreals are a nice extension of the ordinals. It's true that the ordinal operations don't coincide perfectly, but where they don't coincide, they're not interesting because the ordinal operation reduces to "max" or something. Ordinal operations are only nice on one side, and Surreal arithmetic just gives the "nice" answer no matter the order. $\endgroup$
    – Mark S.
    May 14, 2013 at 20:57
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    $\begingroup$ @MarkS. I agree entirely. The addition and multiplication operations for ordinals are actually defined very poorly, and the surreal number versions are much better. There is no hope of embedding the ordinals in any sort of nice algebraic structure if addition is not cancellative. $\endgroup$
    – Jim Belk
    May 14, 2013 at 22:39

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I would argue that the Surreals are a nice extension of the ordinals. It's true that the ordinal operations don't coincide perfectly, but where they don't coincide, they're not interesting because the ordinal operation reduces to "max". Ordinal operations are only nice on one side, and Surreal arithmetic on ordinals just gives the "nice" answer no matter the order.

Furthermore, as user18921 and Jim Belk pointed out, there are analogies to be made: $\mathbb N$ is extended to $\mathbb N'$, the ordinals. And then $\mathbb Z$ is analogous to $\mathbb Z'$, "the abelian subgroup generated by the ordinals", that is, "the ordinals and their surreal negatives". Then $\mathbb Q$ is analogous to $\mathbb Q'$, "the subfield generated by the ordinals", that is, "values of surreal fractions formed with elements of $\mathbb Z'$".

Finally, there is the question of what corresponds to $\mathbb R$. $\mathbb R$ sits inside the surreals as the set of numbers $r$ bounded (on both sides) by elements of $\mathbb Z$ that are "the simplest* number between $r-q$ and $r+q$ for all nonzero $q\in\mathbb Q$". We can then make an analogous definition: $\mathbb R'$ is the set of all surreals $r$ bounded by elements of $\mathbb Z'$ that are simplest between all shifts by a nonzero element of $\mathbb Q'$.

However, every surreal number is bounded by its birthday ordinal (and the negative of its birthday ordinal), so the first condition is satisfied for all surreals. Also, given any surreal $s$, let $r$ be the simplest number between $s-q$ and $s+q$ for all $q\in\mathbb Q'$. If $r\ne s$ then $r-s\ne0$ and picking $q$ equal to the reciprocal of the birthday of $r-s$ contradicts the definition of $r$. Thus, $\mathbb R'$ really is all of the surreals.

*simplest has a technical meaning in the context of the Surreals.

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  • $\begingroup$ Can you elaborate a bit on how the ordinal operations "reduce to max"? Obviously this is the case for cardinal arithmetic with infinite numbers, e.g. $\aleph_0 + \aleph_1 = \aleph_1$, however its not clear to me in what sense this kind of thing applies to ordinal arithmetic. $\endgroup$
    – goblin GONE
    Dec 30, 2013 at 8:42
  • $\begingroup$ @user18921 For ordinals $\gamma=\omega^\delta$, we have $\alpha<\gamma\Rightarrow\alpha+\gamma=\gamma$. (see, the definition of additively indecomposables ). Because of CNF, this is the main thing at play when adding in the wrong order. For multiplication, the main thing is the addition of the exponents of $\omega$, and so you're in a similar situation. $\endgroup$
    – Mark S.
    Dec 31, 2013 at 2:09

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