Knuth arrow notation and transfinite arithmetic

Question

While I was trying to apply the concept of transfinite numbers to deontic logic, a construction showed up of an infinite tower of beth-zero exponents, i.e. $\beth_0^{{\beth_0}^{{\beth_0}^\beth{0...}}}$. I was troubled by having to write it out like that and felt that there should be an easier, known kind of way.

Enter Knuth arrow notation. It occurred to me upon analysis that the tower-expression is equivalent to $\beth_0\uparrow\uparrow\aleph_0$, which is also equal to $\beth_0\uparrow\uparrow\beth_0$. And these equal $\beth_{\beth_0}$. Now it seems to me true that $\beth_0\uparrow\uparrow\aleph_0$ = $\aleph_0\uparrow\uparrow\aleph_0$, and that by implication:

$\beth_{\beth_0}$ = $\beth_{\aleph_0}$ = $\aleph_{\beth_0}$ = $\aleph_{\aleph_0}$

The general formula, here, would appear [with caveats to be discussed below] to be $\beth_0\uparrow\uparrow{n} = \beth_{n}$. Now, this gives us the antecedent in Shelah's inequality $\aleph_0^{\aleph_\omega} < \aleph_{\omega_{4}}$ (the antecedent being, "If two to the power of aleph-n is less than aleph-omega," which is proven in that any finite segment of the tower will be less than the omega-segment). I know it isn't necessary to suppose that the formula that gives the Continuum is a parallel of some formula that would preserve Shelah's inequality; in fact the inequality is trivially true on this model of transfinite arithmetic, as is the traditional application of König's theorem. However, I don't want Shelah's result to go to waste so I'm using it as an indicator of the formula for the Continuum, namely $\aleph_0\uparrow\uparrow2 = \mathfrak{c}$ and $\mathfrak{c} < \aleph_4$.

Again, neither necessary nor, therefore, provable, here. However, I want to show that the three options, here, are each intuitively plausible, not on the basis of what we usually might think, but owing to the explicit use of the Knuth arrow notation. It is my conviction, btw, that one of the main problems with constructing a convincing model of transfinite arithmetic is, indeed, the notation: the preoccupation with the transfinite ordinal subscripts (of the omega-form, usually) makes the computational procedures unwieldy, since the background mismatch of cardinal and ordinal arithmetic affects the clarity of the presentation.

But anyway, the three options are:

$(A)\aleph_0\uparrow\uparrow2 = \aleph_1 :: \aleph_0\uparrow\uparrow n = \aleph_{n-1}$

$(B)\aleph_0\uparrow\uparrow2 = \aleph_2 :: \aleph_0\uparrow\uparrow n = \aleph_{n}$

$(C)\aleph_0\uparrow\uparrow2 = \aleph_3 :: \aleph_0\uparrow\uparrow n = \aleph_{n+1}$

I want to emphasize before going any further that I am claiming that we can directly compute $\aleph_0\uparrow\uparrow\aleph_0$ to the value of $\aleph_\omega$, before/independently on knowing how to directly compute the Continuum. That is, it is not a theorem deduced from a model of infinitary languages (shout out to Woodin: a valiant effort but imagine if we told someone we needed an entire theory of language to prove that 0 + 0 = 0, for example!) or an axiom arbitrarily added as with Cohen pluralism. The spirit of calling the Continuum Hypothesis, by that name, would be violated on the Cohen meta-model, since we would otherwise thence tend to refer to such "axioms" as postulates.

If there is an axiom I am invoking, here, that does not clearly (to my knowledge) show up in ZFC (even as a theorem?), it might be "the axiom of simplicity." Actually, I will specify it as the axiom of coherent simplicity and couple it with the axiom of transcension, "There is some true, non-trivial formula of the general form $\aleph_n^{\aleph_m} = \aleph_{f(n,m)}$," and given the equation and inequality $\aleph_0{^\mathfrak{c}} = \mathfrak{c}^\mathfrak{c}$ & $2^{\mathfrak{c}} > \mathfrak{c}$

... it follows that the function of n and m increases from m if m > n and increases from either n or m if n = m and if n > m, it goes back to the n-case.

Now the notion of coherent simplicity is that the formula "ought" to be "as simple as possible." I would give better accounts of those assertions if I had the time [just want to throw this out there: I've only the time to do this in detail on library computers as I am homeless atm haha] but anyway, to explain what I mean by example, I will give the following argument:

1. Suppose that $\aleph_0^{\aleph_0} = \aleph_{n+2} = \aleph_2$.
   2. The simplest case that gives the same value, in Knuth notation, is for $\aleph_0\uparrow\uparrow2 = \aleph_2 :: \aleph_n\uparrow\uparrow m = \aleph_{n+m}$.
   3. Now, the exponential case is consistent with $\aleph_0{^\mathfrak{c}} = \mathfrak{c}^\mathfrak{c}$. However, it is easy to show that the arrow-case is not:

$\aleph_0\uparrow\uparrow3 = \aleph_3$

$\aleph_2\uparrow\uparrow2 = \aleph_4$

A fortiori, no simple formula higher than an "n + 1" kind of formula, will work. I did find that a formula like this works if we set $\aleph_0\uparrow\uparrow2 = \aleph_{\aleph_1}$ or $\aleph_{\omega+1}$, though, so those (and similar cardinals) have to be ruled out by the argument for "aleph-omega = beth-omega." So, then, the "axiom of coherent simplicity" just says that the simple transcension formula for exponentiation has to fit the one of tetration, and applied, here, this fit is only achieved for the case where the Continuum = $\aleph_1$. QED

So what?!

Like I said, I was motivated to this line of thought by deontic logic, of all things. Actually, trying to use transfinite arithmetic in ethics is not unprecedented by any means. I wish I could remember the name of the book but I swear I read through a chapter or section in one, in college, that was exactly about trying to use aleph-numbers in ethics.

That being said, I would back up my "axiom of simplicity" on those grounds, then: moral reasoning ought to be as simple/accessible as possible (since it would be unfair for moral arguments to be overly difficult to understand), and if moral reasoning requires basic transfinite arithmetic, then either it reduces to trivial such arithmetic (e.g., assign all agents $\aleph_0$ cardinal value, and then show that sets of agents do not have more cardinal value than any individual agent, say, by showing that $\aleph_0 + \aleph_0 = \aleph_0$) or it is true that the simplest non-trivial cases must therefore be accessible.

Now, after reading up (not enough!) on the notion of cofinality, and hearing about the notion of "final coalgebras" (in a Stanford Encyclopedia of Philosophy article on nonstandard set theory), I actually decided that deontic cardinality "probably" is a strange thing, and I had to work with a concept of "transfinality" to manage what I was saying, then. This amounts to looking at the aleph-series as starting from the absolutely largest infinity, and the powerset operation going down towards aleph-zero, but never reaching it, just as no transcension operation goes from the absolutely smallest infinity, up to the absolute infinite (the universe of sets itself, and the encompassing logos). Now transfinal arithmetic can be defined (or I think this is so) such that maybe trans_finite_ arithmetic need not enter very much into the procedures (at least for deontic purposes) so anyway...

... there's another area of application I could see transfinite arithmetic being useful in, besides maybe as part of an educational "aleph-catalogue" kind of idea (you know what I mean: like in Conway and Guy's The Book of Numbers, or in The Mathematical Traveler (IIRC that was the name?): lists of aleph-numbers, meant as representative samples of the versatility of the notation). And this application is the question of renormalization in quantum gravity theories. The problem with renormalization altogether is, "as they say," that it involves "subtracting infinities." But just as exponentiation is iterated multiplication, taking-the-root is iterated division, which is iterated subtraction. So what if quantum renormalization involves finding infinite roots of aleph-numbers? For example, with all the forces besides gravity, we start from the Continuum and take it to aleph-zero. (I swear to God, please don't kill me, but attempts to make up systems of alephic logarithms and so on, have been ventured, IIRC; IDK much about them but I (really hope I) am not singing in the dark haha.) But maybe gravitons, because their spin is higher than the other forces, are in a $(k > \mathfrak{c})$-dimensional state, or something like that, so...

... which is my last application-question (in three parts): (a) is it possible to coherently/intelligibly describe "aleph-dimensional" structures, e.g. n-dimensional = $\aleph_0$-dimensional, and so on; (b) is there an "ideal graph" of the pure geometrical shuffling among the glyphsets in the aleph-catalogues, that indicates the form of the arithmetic; (c) and if (a) and (b) are true, is there an ideal aleph-dimensional graph of the transit through the aleph-numbers (by means of simple-to-complex transcension)?

Answer 1

Originally I skimmed through most of your question and mainly answered the question in bold. I have since reread the question and decided to give some more comments.

Ordinal vs Cardinal indices

You claim that the ordinal notation in indices of cardinals makes the computation unwieldy, however there is a very distinct reason for using ordinals and not cardinals. That is because ordinals and cardinals are used for a different purpose.

Where cardinals are used to give the cardinality of a set, ordinals are used to give a well-order to things. In particular, any process that is guided by transfinite induction / recursion of length $\alpha$ (for $\alpha$ some ordinal) is basically an iteration where we take a step for each $\beta<\alpha$.

Therefore we use ordinals $\alpha$ as index of aleph numbers $\aleph_\alpha$, since we can describe $\aleph_\alpha$ as the least cardinality (of a well-ordered set) that is strictly larger that $\aleph_\beta$ for all $\beta<\alpha$. This means that $\aleph_\omega$ and $\aleph_{\omega+1}$ are different cardinals, since $\omega<\omega+1$. However, if we would write $\aleph_{\aleph_0}$ and $\aleph_{\aleph_0+1}$ instead, then it becomes very confusing: we interpret $\aleph_0$ as a cardinal, but the $+1$ should be used to denote ordinal addition.

Knuth Arrow

There seems to be a difficulty with just copying the Knuth Arrow: its recursive definition does not go through unless we take some care at the limit step. We cannot use the standard recursive definition, as it uses subtraction:

\begin{align} A\uparrow^n 0 &= 1\\ A\uparrow^1 \beta &= A^\beta\\ A\uparrow^n \beta &= A\uparrow^{n-1}(A\uparrow^n (\beta-1)) \end{align}

Note that $\beta$ cannot be limit ordinals in this definition, therefore $\aleph_0\uparrow\uparrow \omega$ does not immediately make sense. We can remedy this by definining an extra clause that $A\uparrow^n \beta$ is the limit $\bigcup_{\alpha<\beta} A\uparrow^n\alpha$ when $\beta$ is a limit ordinal.

Now, again, it is important to use ordinals after the arrows, and not cardinals. For instance, if we assume the Generalised Continuum Hypothesis we have that $\aleph_0\uparrow\uparrow \omega=\aleph_\omega$, while $\aleph_0\uparrow\uparrow\omega+1=\aleph_0\uparrow(\aleph_0\uparrow\uparrow\omega)=\aleph_0^{\aleph_\omega}=\aleph_{\omega+1}$. Hence it is the order-type of the thing after the arrows that is important, not the cardinality.

Fixed Points

You call $\aleph_\omega$ a fixed point of the $\aleph$ function. This is not true: A fixed point would be a cardinal $\aleph_\alpha$ such that $|\alpha|=\aleph_\alpha$. Note that $|\omega|\neq\aleph_\omega$, thus this is not a fixed point.

We can find a fixed point, by taking the limit of the sequence $\aleph_\omega, \aleph_{\omega_\omega}, \aleph_{\omega_{\omega_\omega}},\dots$. The existence of fixed points follows from the aleph function being a normal function.

You state that $\aleph_\omega^{\aleph_0}=\aleph_\omega$. This does not have to be the case, since it is consistent that $2^{\aleph_0}>\aleph_\omega$. Clearly then also $\aleph_\omega^{\aleph_0}>\aleph_\omega$.

The claim that $\aleph_0^{\aleph_\omega}=\aleph_\omega$ is even worse, since it is provably false: this is a direct contradiction to Cantor's theorem, since $2^{\aleph_\omega}= \aleph_0^{\aleph_\omega}=\aleph_\omega<2^{\aleph_\omega}$.

Finally you state that $\aleph_{\omega}^{\aleph_\omega}=\aleph_{\omega_\omega}$, but once again this is provably false: this is a direct contradiciton to König's Theorem. One way to state König's Theorem is that $\kappa<\mathrm{cf}(2^\kappa)$ for all $\kappa$. However, your equation states that $2^{\aleph_\omega}=\aleph_{\omega}^{\aleph_\omega}=\aleph_{\omega_\omega}$, but $\mathrm{cf}(\aleph_{\omega_\omega})=\aleph_0$ and $\aleph_0<\aleph_\omega$, contradiction.

Consistency of the Three Cases

All of the three make sense as a result of Easton's theorem:

For regular cardinals $\kappa$, the values of $2^\kappa$ can be anything as long as they do not contradict monotonicity ($\lambda<\kappa\to2^\lambda\leq 2^\kappa$) and König's theorem ($\kappa<\mathrm{cf}(2^{\kappa})$).

In particular, if we have any (class) function $F$ with only regular cardinals in its domain and cardinals in its range such that $\kappa<\mathrm{cf}(F(\kappa))$ and such that $F$ is increasing (not necessarily strictly), then it is consistent with $\mathsf{ZFC}$ that $2^\kappa=F(\kappa)$ for all regular cardinals $\kappa\in\mathrm{dom}(F)$.

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In case (A) we have the function $F:\aleph_n\mapsto \aleph_{n+1}$. These certainly are increasing and satisfy König's theorem, thus these values are consistent with $\mathsf{ZFC}$. Actually, this is the GCH (at least up to $\aleph_\omega$).

In case (B) we have

• $\aleph_0^{\aleph_0}=\aleph_2$
• $\aleph_0^{\aleph_0^{\aleph_0}}=\aleph_0^{\aleph_2}=\aleph_3$
• $\aleph_0^{\aleph_0^{\aleph_0^{\aleph_0}}}=\aleph_0^{\aleph_3}=\aleph_4$
• etc

So $F:\aleph_n\mapsto\aleph_{n+1}$ for $n\geq 2$ and $F:\aleph_0\mapsto \aleph_2$. This is again consistent. Since $\aleph_1$ is not in the domain of $F$, we have a choice of letting $2^{\aleph_1}$ be either $\aleph_2$ or $\aleph_3$.

Similarly, in case (C) we get $F:\aleph_n\mapsto\aleph_{n+1}$ for $n\geq 3$ and $F:\aleph_0\mapsto \aleph_3$, once again consistent.

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A model for (A) is easily retrieved by taking Gödel's model for the constructible universe $\mathsf{ZFC}+V=L$.

A model for (B) is the result of adding $\aleph_2$ Cohen reals to the model mentioned above. This raises the cardinality of $2^{\aleph_0}$ to become $\aleph_2$ without changing the other cardinals.

A model for (C) is the result of adding $\aleph_3$ Cohen reals to the first model. In this model $2^{\aleph_0}$ and $2^{\aleph_1}$ become $\aleph_3$, while the rest stays untouched.