26 Jul '06 09:04>
Originally posted by SPMarswell its not clear to me.
then clearly | R |
why can't it be 2^aleph?
Originally posted by SPMarsok, thats a good proof. so |X| = |R|
R is the set of real numbers.
N is the set of natural numbers.
X is the set of countable subsets of R (including the finite ones).
If C and D are sets let C^D denote the set of functions from D to C.
If C and D are sets then C < D means "there is an injection from C to D" and C=D means "there is a bijection from C to D".
We have
R < X < R^N = (2 ...[text shortened]...
Since we have R < X and X < R it follows from the Schroeder-Bernstein Theorem that X = R.
Originally posted by aginisThe Axiom of Choice is equivalent to the statement that, for any two sets A and B we have A < B or B < A.
ok, thats a good proof. so |X| = |R|
therefore |P(R)| = |T(R)| + |R| but it still doesn't follow that |T(R)| = 2^aleph
Originally posted by aginisI thought that the cardinality of A + B was the max of the cardinalities of A and B? This is assuming the Axiom of Choice which tells us that A < B or B < A. Then the 'max' of A and B is well-defined.
|T(R)| + |R| = T(R) ????
this is only proven for known cardinalties however there could exist a cardinality x such that x < 2^aleph but x + aleph = 2^aleph
Originally posted by aginisYep that's a nice proof!
the proof i used is as follows.
let A = {x < 0 }
B = [0,1]
C = { x > 1}
then |A|=|B|=|C|=|R|
T(R) is a subset of P(R) therefore T(R) < 2^aleph (1)
however if we define S_k = A u K where K is an element of P(B) then
|R|=|A| |R\S_k| = |R| = |S_k|
S = {S_k} is a subset of T(R) therefore
2^aleph = |P(B)| = |S| < |T(R)| (2)
from (1) and (2) it follows that |T(R)| = 2^aleph
Originally posted by SPMarsok i also did a little checking and i see where the communication problem is comming up. I am assumming that the (generalized)continuum hypothesis is false (or at the least i won't use it in a proof).
Yep that's a nice proof!
I still think mine is fine though, too.
Originally posted by aginisHi aginis!
ok i also did a little checking and i see where the communication problem is comming up. I am assumming that the (generalized)continuum hypothesis is false (or at the least i won't use it in a proof).
whereas you are using it to say that
if A,B > aleph_0 then A+B = max {A,B}
that is to say that cardinalities are "known" if
1. it is finite OR
2. it is undecidable anything you can prove with it you should be able to prove without it (i think).
Originally posted by SPMarsi think you are getting axiom of choice mixed with something else.
Hi aginis!
I don't think the continuum hypothesis (generalised or not) has anything to do with it.
What I'm saying assumes the Axiom Of Choice, and that's all. If you believe that (as is often done) then the following theorem is true:
"Let A and B be cardinal numbers, the larger of which is infinite and the smaller of which is non-zero. Then
A + ...[text shortened]... is from the book "Elements of Set Theory" by H. B. Enderton, published by Academic Press.
Originally posted by aginisNo, I'm not, honestly. The Axiom Of Choice says that an arbitrary (no matter how large) product of non-empty sets is non-empty. Although this result seems highly plausible, it cannot be proven from the ZF axioms. So "Choice" is added to form the ZFC axioms. The Axiom of Choice is equivalent to Zorn's Lemma, the form in which it's most used.
i think you are getting axiom of choice mixed with something else.
AoC says that if you have a non-empty set you can choose one of its elements at random.
can you cite a proof of A+B = max {A,B} ?
Originally posted by SPMarsthanks i was looking for that but couldn't find it.
No, I'm not, honestly. The Axiom Of Choice says that an arbitrary (no matter how large) product of non-empty sets is non-empty. Although this result seems highly plausible, it cannot be proven from the ZF axioms. So "Choice" is added to form the ZFC axioms. The Axiom of Choice is equivalent to Zorn's Lemma, the form in which it's most used.
You can loo ...[text shortened]... oned in the section on cardinal addition in the Wikipedia entry on "Cardinal number",