I have a theory that the axiom of choice is equivalent to the statement that sets of distinct cardinals are well ordered by cardinality. I can prove that the axiom of choice implies this. However I am having trouble proving the other way. I want to prove, without choice, that any set is in bijection with a set of distinct cardinals. I don't know if this is true, but if it is could someone provide a proof for me? Thank you.
Subscribe to:
Post Comments (Atom)
real analysis - How to find limhrightarrow0fracsin(ha)h
How to find lim without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...
-
Ok, according to some notes I have, the following is true for a random variable X that can only take on positive values, i.e P(X \int_0^...
-
Self-studying some properties of the exponential-function I came to the question of ways to assign a value to the divergent sum $$s=\sum_{k=...
-
The question said: Use the Euclidean Algorithm to find gcd (1207,569) and write (1207,569) as an integer linear combination of 1207 ...
No comments:
Post a Comment