After the Short Tutorial On Big Numbers, let ù 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 ù in finite time. So let's "count" as follows:

`1,2,3,...,ù,``2*ù,``3*ù,``...``ù*ù=ù^2=[ù->2->1],``ù^3,``...,``ù^ù=ù^^2=[ù->2->2],``ù^^3,``...,``ù^^ù =ù^^^2=[ù->2->3],``ù^^^3,``...,``ù^^^ù=ù^^^^2=[ù->2->4]``ù^^^^3,``...,``ù^^^^ù=[ù->2->5],``[ù->2->6],``...,``[ù->2->ù],``...,``[ù->3->ù],``...,``[ù->ù->ù]=([3-ù's with arrows])``([4-ù's with arrows])``...,``([ù-ù'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^ù). In the above counting method, when the author reaches step 8: ù^ù he is already counting faster than you.

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

Suppose that your counting rate R is ù-tetrational. Then R=O(e^^ù). In the above counting method, when the author reaches step 11: ù^^ù 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}(ù) found on row n, is such that f_{n}(ù)>R. Further, for *any* steps m≥n, the following holds: f_{m}(ù)≥f_{n}(ù)>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}(ù) on row n there exists m, such that f_{n}(ù)≤2^2^...(m-2's)...^2^ù. Therefore the (counting) operators of the above list and of each step resolve essentially into the following set of operators:

S_{T}={ù,2^ù,2^2^ù,2^2^2^ù,...}.

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}(ù)≤2^2^...(m-2's)...^2^ù.

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})=ù 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.