Ripheus23

This is how my argument proceeds so far:

Suppose that 2 to the power of aleph-aleph-zero, equals [aleph-zero]-aleph-zero. Then evaluate 2 to the power of [aleph-zero]-aleph-zero. Now, if 2 to the power of aleph-zero equals aleph-aleph-zero, then from R to the power of aleph-n we can construct all the alephs after aleph-zero and through aleph-aleph-aleph-aleph-aleph-aleph-...-aleph-n. But this exhausts all the points in the infinite-dimensional array of symbols for these numbers, at this stage. [One of the deeper ideas is that we can symbolize all the aleph numbers for {R to the power of aleph-n} using a single object, an infinite-dimensional super-square, if you will [infinite stacks of squares, in an ascending order per stack, at an infinite number of angles perpendicular to the origin].] Accordingly, 2 to the power of [aleph-zero]-aleph-zero does not map back to any point reached from the first series of aleph-n.

However, therefore 2^[aleph-zero]-aleph-zero, would equal something else if 2^aleph-aleph-zero did not equal [aleph-zero]-aleph-zero. And 2^aleph-aleph-zero would not equal that if 2^aleph-zero, did not equal aleph-aleph-zero. Ergo, 2^aleph-zero does not equal aleph-1 but aleph-aleph-zero, therefore the Continuum Hypothesis is false.

Now, my objection to my own reasoning so far is that it seems to "beg the question," if you will. Or, to put it more kindly, the picture in the previous post is the Supercontinuum Hypothesis, which itself must be solved independently to prove my point. Anyway, if the SH is true, though, we have the following interesting series:

• 2^2^2^2^2... = aleph-zero
• 2^aleph-zero = aleph-aleph-zero
• 2^aleph-aleph-zero = [aleph-zero]-aleph-zero
• 2^[aleph-zero]-aleph-zero = ???