So let me get this straight: There are ACTUALLY people on this

Unless you have a proof to the contrary, attacking people's guesses is a little silly. Nobody claims to *know* it's true.

Yeah man it's easy. Let Z be the integers and R be the reals

S= (Z+R)/2

>>8573077
How can your intuition be so deeply flawed that you honestly think there's not a fucking (uncountable) infinity of cardinalities between any two given cardinalities

This is the computer theoretical equivalent of believing in fucking santa clause

>>8573083
God you're the worst.

>>8573073
go to bed kronecker, it's 10 am and you're drunk

[math]\mathscr{P}( \mathbb{C} )[/math]
Now what, bitch?

>believing in infinities

>>8573073
>>8573127
someone should post wildburger memes

File: 1478002577398.png (664KB, 1421x957px) Image search: [Google]

664KB, 1421x957px

>>8573132

From wiki:

>rational numbers can actually be placed in one-to-one correspondence with the integers, and therefore the set of rational numbers is the same size (cardinality) as the set of integers: they are both countable sets.

Can someone post this one-to-one correspondence from Q to Z ? I'm curious.

>>8573160
https://www.youtube.com/watch?v=DpwUVExX27E&list=PLt5AfwLFPxWKuRpivZd_ivR2EvEzKrDUu&index=9

>>8573175

huh, interesting.

>>8573160
Define a function f from the natural numbers to the integers such that:
f(n)=n/2 if n is even
f(n)=-(n+1)/2 if n is odd
(Note this is easily invertible.)

Now define a function g from natural numbers to rational by writing the natural number in it's unique prime factorization, and applying f to the exponents of each prime factor.

>g(6)=(2^1)(3^1) -> (2^f(1))(3^f(1)) -> (2^-1)(3^-1)=1/6
>g(18)=(2^1)(3^2) -> (2^f(1))(3^f(2)) -> (2^-1)(3^1)=3/2
>g(100)=(2^2)(5^2) -> (2^f(2))(5^f(2)) -> (2^1)(5^1)=10

Extend to all integers by defining g(0)=0 and g(-n)=-g(n)

>There are ACTUALLY people on this board, RIGHT NOW, who ****HONESTLY**** believe the continuum hypothesis?
I'm not sure what you mean, OP.

There are models of ZFC set theory in which the continuum hypothesis is true. There are also models of ZFC set theory in which it is false.

What does it mean to "believe" the continuum hypothesis, in the face of this? What statement is it, exactly, that you expect me to disbelieve?

>>8573079
no way....

>>8573095
>not using a Weierstrass P

>>8573211

That's really nice, anon.

>>8573229
Ive been meaning to make a thread about this.

since we cant prove or disprove the continuum hypothesis, lets say I want to construct some set S which I think may be an example of OPs pic. Does the incompleteness theorem mean that I
>cant construct such a set
>cant prove that this set satisfies N0 < |S| < 2^N0

Can you assume the continuum hypothesis as being true or false and still have a consistent theory to work in?

If I cant prove the above, but I can construct it, will assuming the continuum hypothesis being true then automatically mean this set is larger than 2^N0? what if I assume the hypotheses is false, will I then be able to prove if this set satisfies N0 < |S| < 2^N0?

>>8573226
How?

>>8573218
There is an often implicit assumption that ordinary, day-to-day mathematics should describe something "real", "existing", that there is "one" true model which can then be used, say, in analysis, For example, it is, in some sense, almost absurd talking about infinite-but-Dedekind-finite sets, which is why some people "believe" in Choice Axiom.

And in this pursuit of "true" model we have questions like that. What if, for example, Riemann hypothesis is true with C but false with not C? Which world we actually live in?

>>8573286
>Can you assume the continuum hypothesis as being true or false and still have a consistent theory to work in?
That's the point of something being independent. It can be true, it can be false, you can omit it - consistency doesn't change.