I have just discovered that square [0,1] X [1, 0] has the same cardinality as the [0,1]...

>the rationals have the same cardinality as integers

Hold up. What? I understand why they are both countably infinite because you can find the bijection with natural numbers, but how do you prove they have the same cardinality?

Pls no bully I'm a cheme with limited math background.

Look up Cantor's diagonalization argument.

Additionally, there's some theorem that says that the cartesian product of two countable sets is countable. And a definition of countable is having the same cardinality as the natural numbers.

So yeah, I guess I should just say that the definition of countable infinite is having the same cardinality as the naturals (which has the same cardinality as the integers).

OP, this guy is talking shit.

If two sets have a bijection, they are the same cardinality BY DEFINITION. You don't need diagonalization for this

Okay, this kinda helped a bit. Going to read about layman`s terms too. I hope i can hear more advices.

Yeah I just realized I misinterpreted his question. That's why I added the 2nd response afterwards buddy boi.

Haha "layman's terms" just means "simple english".

I was just giving you a nonrigorous explanation.

Kek, i thought he was some kind of mathematician.

the next cardinality is really fucking big. that's your intuition there.

Thank you.

>Look up Cantor's diagonalization argument.
I've read it before, but if I recall correctly I don't think I fully understood, but basically real numbers have "higher" carnality than natural numbers because it has more elements than a one-to-one correspondence with natty numbers.

Gonna go read up on it again...

[math] A\times A[/math] and A have the same cardinality for any infinite set A. The proof is the same as for the integers but you arrange your square with a well order for A instead.

Cardinality is one of many notions to capture size.

The predicate "same cardinality" is literally directly "there is a bijection".
There is not more to it.

In set theory alone, there is no notion of dimension - it's mostly of relevance in algebra.
Now if you use set theory to model products and lists, then you can again consider cardinalities of those, namely the lengths, and here you have a naive notion of dimensions.
You may want to read RxR as "R squared", and the size should then be infinitiy squared. It happens to be infinity again, for cardinal arithmetic.

Something we know is bigger than R is the
R->{0,1}
the set of functions from R to an two element set. You may write it as "2^R".

In fact, for any set X, the set X->{0,1} is bigger than X.
For example, N->{0,1}, the functions from the natural numbers to the two element set, is bigger than N.

Interestingly, even if you adopt a lot of strong (set theory) axioms (e.g. Zermelo Frankel set theory), the question whether R is bigger than N->{0,1} isn't solvable from those theories. This is the continuum hypothesis.

The question whether there is a set Y that's bigger and than N and smaller than R is essentially up to the user, it's almost like choosing between different programming languages. (Except not really, because any Turing complete one may simulate another ones, the computable reign isn't as open as the axiomatic set theory one one)

If you adopt the generalized continuum hypothesis, then (up until the very made up large cardinals), you demand that the chain of sizes of the sets
X,
X->{0, 1}
(X->{0, 1}) -> {0,1}
((X->{0, 1}) -> {0,1}) -> {0,1}
...
is all there is to set theory, there is no inbetween mystery.

Assuming the continuum hypothesis (whcih you are free to adopt if you want or not, without fucking up anything about set theory), then if RxR were bigger than R, then it would have to be at least as big as R->{0,1}.

There is no intuition in this. This is a purely theoretic reasoning.
The trick behind every wacky stuff like this is the notion of infinity - Do you believe that right now there is an infinite amount of numbers? Or do you believe that right now there is a finite amount of them but you can expand that amount infinitely?

If you believe the first one, then welcome to the modern mathematics.

If you reject the first one and accept the second then welcome to patrician group of Finitists who are hated by everyone else.

If you adopt the math is a "formal game perspective", you'll never have to fear being wrong again.

>with a well order for A instead
What if A can't be well-ordered?

>the question whether R is bigger than N->{0,1} isn't solvable from those theories. This is the continuum hypothesis.
Holy shit at least learn set theory before you try to explain it to someone else. R trivially has equal cardinality to 2^N. The continuum hypothesis is that |R| = |2^N| is the smallest cardinal larger than |N|.

Take your pedophile cartoons back to you degenerate.

Always can. Pick the first element, then the next, then the next, etc. After the natural numbers pick another one to come after all of them, then the next, etc. Continue until you exhaust all the elements in A. Make this completely rigorous with ordinals and transfinite induction, and don't fall for the rejection of the axiom of choice meme.

>R trivially has equal cardinality to 2^N
What's the definition of R, why is this trivial?

>why is this trivial?
>binary expansion of [0,1]