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

You mean [0, 1]×{0, 1}?

>>8317143
Nah, nvm, I mixed my question with the proof of this theorem. What I meant was
>square has the same cardinality as the line segment

>>8317148
Right. Check the exponentiation section, https://en.wikipedia.org/wiki/Cardinal_number#Cardinal_arithmetic

>>8317156
I don't see what exponentiation has to do with his/her question.

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>>8317156
What I mean is that this shit is totally counter-intuitive, same as the fact that [0,1] has the same cardinality as [0,2] and I can`t understand how that can be possible. I understand that there is a bijection between them, etc. But I want to know if I can get some sense of intuition on this one.

>>8317165
aleph0^2 dummy

>>8317171
You mean 2^aleph0?

>>8317132
Same reason rationals have the same cardinality as integers. Really the rationals are just Z x Z.

>>8317170
You are thinking of the measure of a set. The cardinality is just the type of infinity it is (in layman's terms).

[0,1] is just as infinite as [0,2]. Just like how infinity*2=infinity.

[0,1] is just as infinite as [0,1]x [0,1]. Just like how infinity*infinity = infinity.

>>8317227
>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.

>>8317314
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.

>>8317314
>>8317327
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).

>>8317314
OP, this guy >>8317327
is talking shit.

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

>>8317254
Okay, this kinda helped a bit. Going to read about layman`s terms too. I hope i can hear more advices.

>>8317347
Yeah I just realized I misinterpreted his question. That's why I added the 2nd response afterwards buddy boi.

>>8317354
Haha "layman's terms" just means "simple english".

I was just giving you a nonrigorous explanation.

>>8317359
Kek, i thought he was some kind of mathematician.

>>8317132
the next cardinality is really fucking big. that's your intuition there.

>>8317327
>>8317341
>>8317347
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...

>>8317132
[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}.

>>8317132
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.

>>8319743
If you adopt the math is a "formal game perspective", you'll never have to fear being wrong again.

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>>8319693
>with a well order for A instead
What if A can't be well-ordered?

>>8319741
>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|.

>>8319970
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.

>>8319987
>R trivially has equal cardinality to 2^N
What's the definition of R, why is this trivial?

>>8320132
>why is this trivial?
>binary expansion of [0,1]