> From: Alberto Monteiro <[EMAIL PROTECTED]>

> The Fool wrote: 
> >  
> > No.  Which exactly my point.  If I can't prove my own 
> > existence I also can't prove god's existence.  
> > Math exists whether god, the universe, 
> > consciousness, I, etc. exist.  Math is the only thing 
> > that is transcendent.  And those math proofs do exist. 
> >  
> > This may all just be some big matrix I am in, but math 
> > works either way. 
> > 
> What about the Axiom of Choice? Does it exist? Or 
> the Continuum Hypothesis? Does it exist? 

AoC: Neither accepting it or NOT accepting it leads to a contradiction. 
Both are valid.

----

Analysis of the Continuum One path to finding out if CH is true is to
look for sets of real numbers that have cardinality greater than aleph0
and less than c. If such sets exist, then CH is false. While looking for
these sets mathematicians have decomposed the reals into different types
of sets and come up with characterizations of the reals, continua, and
continuity. 3.3.1 Decomposing the Reals There are many ways to decompose
the reals. We can split the reals into two sets, such as the rational and
irrational numbers or the algebraic and transcendental numbers, or we can
look at types of subsets of the reals, such as Borel sets and non-Borel
sets. The hope is that these decompositions will help us to characterize
the entire set of reals. For example, the rationals provide a good
approximation to the reals because any real number can be approximated
"as closely as you please" by a rational number. This should serve as a
warning though. Despite the rationals' great ability to represent the
reals they are of a different cardinality than the reals. So, just
because a collection of sets is a good approximation of the reals doesn't
mean that it will give us much useful information about cardinality
questions. Bearing this in mind, let's try to understand CH by looking at
sets that, in some sense, represent the reals. 
CH is true for closed, Borel, and analytic sets, i.e.: 
any infinite closed set either has cardinality aleph0 or c. 
any infinite Borel set either has cardinality aleph0 or c. 
any infinite analytic set either has cardinality aleph0 or c. 
This means that if there is a set that falsifies CH, i.e., a set with
cardinality between aleph0 and c, it will not be one of these types of
sets. Since there are only c of these types of sets and there are
2csubsets of R, there are plenty of sets left which might falsify CH! 

http://www.ii.com/math/ch/
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