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∈ 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}|=ℵ_{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∈ N U {0}}.

In other words: Counting infinitely fast is equivalent to keep on diagonalizing at a constant rate over any given counting rate. That's *much* faster than simply counting to infinity, which can plainly be done by diagonalizing over S_{B}.

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.