09/06/2016, 03:12 PM
To conclude the final thing about this set theory CH thing.
Although i have given the impression that CH is false there is a problem with that.
Just one big problem.
If one considers just the last term of the taylor ( t_w w^w ) - because all w^i with i finite are countable , then neither the fake last term nor another last term give a correct solution.
Keep in Mind that an iteration of x^n / a = x^n^2 / a^(n+1).
Also things like ln(w) and 1,3^ w are not consistent. ( ln is too small , 1,3 is not a base for the reals that has uniqueness )
Also results ( for either f or f(f) ) like w^w / e^w , 4^w/2^w , w^w/w^w , w - 1 , etc are undefined.
Hence there are no Nice solutions.
It follows that CH Cannot be constructively be disproven by additions and products of sets/Cardinals/ordinals ( taylors ).
Since additions , products and powersets are the only potential way to increase ordinals / Cardinals / sets ,
... This implies CH must be true !
Consider this
Let M be a Set with card between countable and uncountable.
Let there be a way to describe card M by a metaset of integers.
So , Lets say that M is a subset of the reals.
How many countable or finite sets do we need to describe M ?
So we take either a finite amount of sets of integers , finite sets or an infinite amount of them. And we use Sums and products.
Notice there is nothing between taking a finite amount and an infinite amount.
After taking ( internal ) bijections we can always reduce to
Card M = 2^a n^b + 2^c n^d + e.
Where a , B, C , d , e are non-uncountable.
Notice a part of a countable or finite set is also finite or countable.
So card M must be countable or uncountable !!
Qed.
Tommy1729
Although i have given the impression that CH is false there is a problem with that.
Just one big problem.
If one considers just the last term of the taylor ( t_w w^w ) - because all w^i with i finite are countable , then neither the fake last term nor another last term give a correct solution.
Keep in Mind that an iteration of x^n / a = x^n^2 / a^(n+1).
Also things like ln(w) and 1,3^ w are not consistent. ( ln is too small , 1,3 is not a base for the reals that has uniqueness )
Also results ( for either f or f(f) ) like w^w / e^w , 4^w/2^w , w^w/w^w , w - 1 , etc are undefined.
Hence there are no Nice solutions.
It follows that CH Cannot be constructively be disproven by additions and products of sets/Cardinals/ordinals ( taylors ).
Since additions , products and powersets are the only potential way to increase ordinals / Cardinals / sets ,
... This implies CH must be true !
Consider this
Let M be a Set with card between countable and uncountable.
Let there be a way to describe card M by a metaset of integers.
So , Lets say that M is a subset of the reals.
How many countable or finite sets do we need to describe M ?
So we take either a finite amount of sets of integers , finite sets or an infinite amount of them. And we use Sums and products.
Notice there is nothing between taking a finite amount and an infinite amount.
After taking ( internal ) bijections we can always reduce to
Card M = 2^a n^b + 2^c n^d + e.
Where a , B, C , d , e are non-uncountable.
Notice a part of a countable or finite set is also finite or countable.
So card M must be countable or uncountable !!
Qed.
Tommy1729
