After the Short Tutorial On Big Numbers, let ω be the first uncountable ordinal and [a->b->c] be Conway's Chained Arrow Notation. Suppose that through some ingenious counting trick, the author can "count" to ω in finite time. So let's "count" as follows:

`1,2,3,...,ω,``2*ω,``3*ω,``...``ω*ω=ω^2=[ω->2->1],``ω^3,``...,``ω^ω=ω^^2=[ω->2->2],``ω^^3,``...,``ω^^ω =ω^^^2=[ω->2->3],``ω^^^3,``...,``ω^^^ω=ω^^^^2=[ω->2->4]``ω^^^^3,``...,``ω^^^^ω=[ω->2->5],``[ω->2->6],``...,``[ω->2->ω],``...,``[ω->3->ω],``...,``[ω->ω->ω]=([3-ω's with arrows])``([4-ω's with arrows])``...,``([ω-ω's with arrows])``...,``etc.`

Can you estimate the author's *counting rate*?

For *any* counting rate R that you give the author, as long as the given rate is representable by some consistent ordinal notation, the author can show that the above counting method counts *faster* than your rate.

For example: Suppose that your counting rate R is exponential. Then R=O(e^ω). In the above counting method, when the author reaches step 8: ω^ω he is already counting faster than you.

Suppose that your counting rate R is super-exponential. Then R=O(e^e^ω). In the above counting method, when the author reaches step 9: ω^^3; he is already counting faster than you.

Suppose that your counting rate R is ω-tetrational. Then R=O(e^^ω). In the above counting method, when the author reaches step 11: ω^^ω he is already counting faster than you.

Now generalize: Suppose that your counting rate is R. Then, there exists a step n, such that the ordinal f_{n}(ω) found on row n, is such that f_{n}(ω)>R. Further, for *any* steps m≥n, the following holds: f_{m}(ω)≥f_{n}(ω)>R. Hence the method above always counts eventually faster than *any* counting rate R.

This it what it means for someone to be able to count *infinitely fast*.

On the other hand, for each ordinal f_{n}(ω) on row n there exists m, such that f_{n}(ω)≤2^2^...(m-2's)...^2^ω. Therefore the (counting) operators of the above list and of each step resolve essentially into the following set of operators:

S_{T}={ω,2^ω,2^2^ω,2^2^2^ω,...}.

We define the *power* P(T) of the operator T to be the number of 2's n the corresponding upper bounding tower of 2's in T. In other words,

P(T\in S_{T})=m, with f_{n}(ω)≤2^2^...(m-2's)...^2^ω.

We then note that if m<n then P(T_{m} o T_{n})=m+n, and the set S_{T} becomes order isomorphic with the set of Natural numbers. Hence:

Ord(S_{T})=ω and

|S_{T}|=N_{0}.

There is a natural correspondence between the operator T with P(T)=m and the corresponding binary operator B_{m}=1/2^m. Therefore the following set is order-isomorphic to the set S_{T}:

S_{B}={1/2^{k}: k\in N U {0}}.

In other words: Counting infinitely fast is equivalent to being able to understand perfectly well binary counting.

The (binary) operators T_{8} and T_{9} give the resolution on the 397 pixels-wide figure above, which is the resolution limit of the author's computer screen (1px) using a binary decomposition, since 397/2^8-397/2^7~1.55 and 397/2^9-397/2^8~0.7754.