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