Hello, this post will be introducing a small (and presumably the most successful) portion of the fascinating subject of combinatorial game value or surreal googology. This is the Veblen function \(\phi:\mathrm{On}\times\mathrm{No}\mapsto\mathrm{No}\), defined in two different ways by me and PlanetN9ne, and later shown to be equivalent (by me!).

Note, there is a veblen function \(\phi:\mathrm{No}\times\mathrm{No}\mapsto\mathrm{No}\) but it’s cursed. Most of this is due to the fact that \(\phi(\frac{1}{2})\) enumerates some fixed points of \(x\mapsto\omega^x\) which are already values of \(\phi(1)\).

## My definition

\(\phi(x,y)=\{0,r\phi(x,y^L),\phi(x^\prime,\phi(x,y)^L)|r\phi(x,y^R),\phi(x^\prime,\phi(x,y)^R)\}\), where \(x^\prime\) ranges over ordinals less than \(x\), \(r\) ranges over all positive real numbers, and \(y^L\) and \(y^R\) range over left and right sets of \(y\).

## Planet’s definition

First, we need to discuss sign expansions. Look it up.

Secondly, we need to discuss a method that interpolates all “hypernormal” ordinal functions. A function f is hypernormal if it is normal and \(\alpha<\beta\) implies \(f(\alpha)+f(\beta)=f(\beta)\).

For any surreal \(x\), let \(s_1,s_2,\ldots\) be the sign expansion of \(x\). Then we can define the interpolated function \(f\) by \(f(x)=+^{f(0)}s_1^{f(1)}s_2^{f(2)}\ldots\). This agrees with \(f\) on all ordinal values. Proof by induction: obviously true for 0, and \(f(\alpha+1)\)’s sign-expansion is just that of \(f(\alpha\)\) with \(f(\alpha+1)\) pluses added. So it is the ordinal sum \(f(\alpha)+f(\alpha+1)\) which is \(f(\alpha+1)\) by hypernormality. The limit case follows from normality.

Planet’s veblen is obtained by interpolating the second argument of veblen in this manner. We use \(\varphi\) to denote it.

## Proof of equivalence

The following properties hold of Planet’s veblen function:

- \(\varphi(\alpha,x)\) is increasing, because \(\varphi(\alpha)\) repeats every sign some number of times and adds some signs to the front, which clearly preserves comparison.
- \(\varphi(\alpha^\prime,\varphi(\alpha,x))=\varphi(\alpha,x)\), given \(\alpha^\prime<\alpha\).. Proof: let the sign expansion of \(x\) be \(stu\ldots\). Then that of \(\varphi(\alpha,x)\) is \(+^{\phi(\alpha,0)} s^{\phi(\alpha,1)}\ldots\), and that of \(\varphi(\alpha^\prime,\varphi(\alpha,x))\) is \(+^{\phi(\alpha^\prime,0)}\)\(+^{\phi(\alpha^\prime,1)+…+\text{up to but not including }\phi(\alpha^\prime,\phi(\alpha,1))} \)\(s^{\phi(\alpha^\prime,\phi(\alpha,1))+…+\text{up to but not including }\phi(\alpha^\prime,\phi(\alpha,2))}\ldots\)\(=+^{\phi(\alpha,0)} s^{\phi(\alpha,1)}\ldots\).
- \(\varphi(\alpha,x)\) is simpler than \(\varphi(\alpha,y)\) if \(x\) is simpler than \(y\), because it applies \(\phi(\alpha)\) to the birthday.
- Every common fixed point of \(\varphi(\alpha^\prime)\) for \(\alpha^\prime<\alpha\) is of the form \(\varphi(\alpha,x)\). Proof: let the fixed point be \(y=+^b -^c +^d -^e\ldots\). (It has to start with a plus because every value is positive.) Then it’s easy to see (similarly to (2)) that \(b,c,d,\ldots\) are all fixed points of \(\phi(\alpha^\prime)\), thus they are values of \(\phi(\alpha)\) making \(y\) a values of \(\varphi(\alpha)\).
- \(\varphi(\alpha,x)\) is the simplest common fixed point of \(\varphi(\alpha^\prime,y)\) between \(\varphi(\alpha,x^L)\) and \(\varphi(\alpha,x^R)\), which trivially follows from the previous.
- \(\varphi(0,x)=\omega^x\). Proof to be given at a later date.

But \(\phi\) is, by construction, the unique function satisfying (5) and (6). Thus, \(\phi=\varphi\).

Mooooseybtw, the definition of the cursed $\textrm{No}\times\textrm{No}\mapsto \textrm{No}$ one was $\phi(0,x)=\{0,r\phi(0,x^{L})|r\phi(0,x^{R})\}$ where $r$ ranges over $\mathbb{R}$ if $x>0$; $\phi(x,y)=\{0,\phi(x,y^{L})+1,\phi(x^{L},y),\phi(x^{L},\phi(x,y)^{L})|\phi(x,y^{R})-1,\phi(x^{R},y)-1,\phi(x^{L},\phi(x,y)^{R})\}$ otherwise

EvinWhere \(r\) is positive.

Mooooseyoops! it’s been pointed out that there’s a mistake here. r must be a *positive* real number. I *think* that domain restriction was somewhere along the way a typo of (r>0) or something, so probably the first line should be something like: $\phi(0,x) = \{0,r\phi(0,x^{L})|r\phi(0,x^{R})\}$ where $r$ ranges over $\mathbb{R}$ and $r>0$. If someone knows better please tell me.