the universe is infinitely discrete

and the set of reals is in one to one correspondence with the set of naturals.

discuss.

No.

>>2

Yes.

>>1
Cantor's diagonalization, bijection from the reals to the powerset of the naturals, etc.

i can biject the naturals onto its powerset FYI

>>5
I didn't realize Chuck Norris posted here.

>>6
Pure win.

How the fuck can a subset of a set be one-to-one with the set that it is a subset of

>>8
Let A be a set, then A is a subset of A and A is one-to-one with A.

>>8
OH I NO

S = 0 to infinity
subset of S = 0 + infinity to infinity

The Universe needs to stop going to upper class ballrooms and tea parties.

>>9
I meant for when the subset doesn't contain all of the members of the set, silly.

>>10 no sense

>>12
go read about cardinality

>>13
No you

>>14
the question is whether or not a set can be put into one to one correspondence with its power set, thats the set of all of its subsets, not whether it can be put into one to one correspondence with each of its subsets.  it obviously cant be done for finite sets, so then the question is whether it can be done for infinite sets, and the answer is no, because of similarity to irrationality.
you can pair a natural number with any finite set of natural numbers, and with any patterned set of natural numbers, the problem is subsets that are infinite but follow no discernable pattern, i.e. {4,86,923,924,925,175312,58971768943,6677777777777...}
if you dont have a finite way to represent it, you cant represent it as a finite natural number.  you would need a natural number that approaches infinity as you write it or something.

>>15
This is a very hand-wavey explanation. A much more straight-forward one:

Let X be a set, and P(X) be the set of all subsets of X. Suppose there exists a one-to-one mapping f from X to P(X). Let S = {x in X such that x is not in f(x)}. S is an element of P(X), so there must be an element y in X such that f(y) = S. We know that either y is in S or y is not in S. If y is in S, then y is in f(y), which means y can not be in S. If y is not in S, then y is in f(y), which means y must be in S. This is a contradiction, and since the only assumption we have made is that the one-to-one mapping f exists, we conclude that it does not exist.

yeah lol i had to do that

well obviously you since cant define f where for every x, x is not in f(x)
S is then X, an element of P(X),
any arbitrary x  sent to it will always being a member of f(x) = X


consider the set N+{0} and P(N)
suppose there exists a one to one mapping f from N+{0} to P(N)
where 0 --> N,
and x in N ---> f(x) in P(N) | x not in f(x)

S = {x in X | x not in f(x)}
S = {x in N+{0} | x not in f(x)}
S = N+{0}
S is not an element of P(N)

>>18
This is incomprehensible and irrelevant.

>>19
one of the subsets is the entire set, so you cant have a map f such that f(x) does not contain x for all x.  some x has to point to the subset that is the entire set.  the definition of the set S is inherently contradictory.

S is analogous to russel's paradox.

>>20
First of all, I don't think you even understood the definition of f. In >>18 you say "[you] cant define f where for every x, x is not in f(x)" which is not necessary or relevant. f, in >>16, was an arbitrary one-to-one mapping, and the fact that, by assuming its existance, we could create the set S using ZFC axioms shows that f does not exist. Your explanation in >>18 is a rambling incoherent mess.

Example:
"and x in N ---> f(x) in P(N) | x not in f(x)"
So x, an element of N, goes to f(x) in P(N).. ok. But your "| x not in f(x)" is non-sensical in any notation I've ever seen.

I want to reiterate that I really do not believe you understood the construction of S. S is the set of elements of X whose image in P(X) does not contain themselves. No one said S was necessarily equal to X or any such thing - if we supposed a map from f:N -> P(N) with f(0) = N, the only thing that tells us is that 0 is not an element of S, since 0 is in f(0).

The only correct thing you've said (at least among the rare statements of yours that I can actually interpret) is that S is equivalent to Russell's paradox - It is, and the fact that we can construct that set using just ZFC and the existance of f is proof that f does not exist.

| x not in f(x)
"such that x is not a member of f(x)"
f is supposed to be a bijective mapping.  all im saying, is obviously youre going to have at least one x that is mapped to a subset that contains itself, since one of the subsets is the entire set.
my point was that if somehow you managed to create a map that satisfied that condition, it would imply that S is equal to the entire set.  thats because you wouldnt have mapped something to the whole set.
my argument is not that S is not contradictory, but that its not a valid method because of its similarity to russel's paradox.  i thought the ZFC axioms denied the ability to construct a set like that?

my other bit about 0 is that if you added one more element, but used the powerset of the set without that element, you could point that to the complete set and not bother with S.  youre obviously not mapping a set to its own powerset then, but the cardinality implications dont get affected, since the set of naturals is countably infinite whether you put {0} in or not.

>>22
You do not understand what S is. Nowhere did I say "no element of X is mapped to a set which contains itself." I said let S be the set of elements x in X such that x is not mapped to a set which contains x. Please try to actually read and comprehend this before you reply again.

Further, in regards to: "my argument is not that S is not contradictory, but that its not a valid method because of its similarity to russel's paradox.  i thought the ZFC axioms denied the ability to construct a set like that?"

As I said in >>21, the fact that ZFC axioms in conjunction with the existence of f can create that set is PRECISELY WHY WE CONCLUDE f DOES NOT EXIST. Seriously, do you not understand proof by contradiction? We cannot construct a Russell's Paradox-like set in ZFC on its own because a parameter used to create a subset using the Axiom of Separation cannot reference the set itself. However, if we assume f exists, then "x is not in f(x)" is an acceptable property and thus the Axiom of Separation ensures that {x in X | x is not in f(x)} exists.

Oh, and as an aside - I know what "| x not in f(x)" means, it is the context in which you used it that was wrong (or at least incomprehensible to me).

YOU WIN

but I do understand what S is, the "no element of X is mapped to a set which contains itself." thing was a condition i was placing on f, that would make S = X, its the only case in which thats true.  I don't think I can be more clear about that.

What the fuck is all this jibber jabber, go make me sammich