the large cardinals and large ordinals are very axiomatic in nature.
so without proofs of bijections or the lack of bijections it is pretty hard to talk about that.
( although i do like the comments here )
in my not so humble opinion its also a matter of taste because of the above and because of the possible use of ZF©. ( which has not been proven consistant ! )
i already commented my personal large cardinal axioms ( kinda ) , but i feel it is more intresting to consider small cardinalities.
to be specific : what is the cardinality of f(n) where n lies between n and 2^n ?
since cardinalities are not influenced by powers
card ( Q ) = card ( Q ^ finite )
we can write our question as
for n <<< f(n) <<< 2^n
card(f(n)) = ?
the reason i dont want to get close to n or 2^n is the question :
is there a cardinality between n and 2^n ?
in other words : the continuum hypothesis.
in stardard math and standard combinatorics , we usually do not work with functions f : n <<< f(n) <<< 2^n.
but on the tetration forum they occur very often.
card(floor(sexp(slog(n)+1/(24+ln(ln(n)))))) = ?
card(floor(n + n^4/4! + n^9/9! + n^16/16! + ...)) = ?
regards
tommy1729
so without proofs of bijections or the lack of bijections it is pretty hard to talk about that.
( although i do like the comments here )
in my not so humble opinion its also a matter of taste because of the above and because of the possible use of ZF©. ( which has not been proven consistant ! )
i already commented my personal large cardinal axioms ( kinda ) , but i feel it is more intresting to consider small cardinalities.
to be specific : what is the cardinality of f(n) where n lies between n and 2^n ?
since cardinalities are not influenced by powers
card ( Q ) = card ( Q ^ finite )
we can write our question as
for n <<< f(n) <<< 2^n
card(f(n)) = ?
the reason i dont want to get close to n or 2^n is the question :
is there a cardinality between n and 2^n ?
in other words : the continuum hypothesis.
in stardard math and standard combinatorics , we usually do not work with functions f : n <<< f(n) <<< 2^n.
but on the tetration forum they occur very often.
card(floor(sexp(slog(n)+1/(24+ln(ln(n)))))) = ?
card(floor(n + n^4/4! + n^9/9! + n^16/16! + ...)) = ?
regards
tommy1729

