Thursday, December 10, 2009

Quickie: Infinite Sets

 This is just a leftover from the primes stuff, but I thought I'd explain it a bit better (in case it might be of interest to others in relation to Cantor):

If we have the set:

S = { { a1, b1, c1, ...}, {a2, b2, c2, ...}, ....}

Is it mappable onto:

{ a, b, c, d, ....}

Hypothesis: Yes

If we define a pass to be taking the a finite number of elements from a finite set, then adding in a new set, we get an algorithm like:

sets = S{N}

    for each set in sets
        get next element in current set
    sets += S{N}

So we traverse the above as follows:

a1,   b1, a2,    c1, b2, a3,   d1, c2, b3, a4,   ...

We can see that eventually we will get through all elements of S.


a = a1, b = b1, c = a2, d = c1, e = b2, f = a3, ....

And thus there is a simple mapping between an infinite set and an infinite set of infinite sets.

This implies that any number of infinities in a set is identical to a single infinity.

Is this a problem?


The number Pi can be seem as a series of "lower" precision approximations that are approaching the actual infinitely precise version of the number. Thus as a set we can see it as:

Pi ~= { 3, 3.1, 3.14, 3.141, 3.1415, ... , Pi }

And that set is equalivalent to:

{ Pi, 3, 3.1, 3.14, 3.141, 3.1415, ... }

 The same is true for all irrationals they can be represented as a set of increasingly precise approximation to the infinite precision Real representation of the number:

{ 1/3, .3, .33, .333, .3333, .... }

And any two different infinite sets can be traverse in one infinite pass.

Also, since our traversal of the sets essentially has an infinite amount of time and space, we can merge any two duplicate-plagued sets together, being sure to only list each element exact once in the new set construction. It may be unnervingly slow, but it is still computable.