Mathematics

Ice Boy

4:01 PM

Are you trying to understand it from Euclid's Elements?

Alizter

@BalarkaSen hi

UserX

A year ago, as a plus he wanted to show us that infinities can be of different size, so he told us to compare all the numbers in the line of $y=x$ and all the lines in the $xy$ plane, and we all(him included) reached in an agreement that the plane has more points.

However, now that I learnt some set theory, $|\Bbb R \times \Bbb R|=(2^{\aleph_0})^2= 2^{2\aleph_0}=2^{\aleph_0}$

The Game

@UserX There's a bijection between the line and the plane, though

Indeed

$|\mathbb{R}|\sim|\mathbb{R}^n|$

Mike Miller

@daOnlyBG I liked Stein and Shakarchi's 'Real analysis' when I was learning it.

UserX

And the points on $xy$ plane are an infinite subset of $R$, bigger than $\Bbb N$ though, so their cardinals are equal

The Game

4:09 PM

Anyone for this question ?

1

Q: Is this group finite?

Let $G$ be a sub-group of the invertible real matrices of size $n$, such that $\forall M\in G,M^2=I_n$ Is $G$ finite ?

@UserX No

UserX

So they have exactly the name numbers. Am I wrong somewhere? How can I formulate the set containing the points of $y=x$ in math notation?

The Game

@UserX They're a subset of $\mathbb{R}^2$, not of $\mathbb{R}$

UserX

Typo

Not typo in fact but yea, I know

The Game

@UserX $\{x\mid y=x\}$

That's the way to write the set

UserX

Oh... Thanks

Wait I got confused

Mike Miller

4:13 PM

@TheGame That notation reads like $y$ is fixed; I assume UserX means the diagonal of the plane. So I would write it either as $\{(x,x) \in \Bbb R^2 : x \in \Bbb R\}$ or $\{(x,y) \in \Bbb R^2 : x=y\}$.

probably the latter

UserX

How do I know there is no cardinal between $\aleph_0$ and $2^{\aleph_0}$

Alizter

@UserX It is undecidable.

Daniel Fischer

@UserX Assume the continuum hypothesis.

Mike Miller

You don't. That's known as the 'continuum hypothesis'.

@TheGame There are uncountably many such matrices even for $n=2$, so there are in general as well.

UserX

Hypothesis? So this is unproven but conjectured?

Alizter

4:14 PM

@UserX no. It cannot be proven.

The Game

@MikeMiller $G$ is finite

UserX

Bu we don't know if it's correct either

?

Daniel Fischer

@UserX You can assume that or the opposite of that. Both assumptions are consistent with ZFC.

Alizter

@UserX Exactly so it is undecidable.

Pedro Tamaroff

@DanielFischer Hello.

Alizter

4:15 PM

It has been proven to be unprovable.

Mike Miller

@TheGame I misread it asking about the size of the set of matrices that square to 1; demanding that it be a subgroup is a different situation entirely, you're right. Sorry!

UserX

So is there something wrong with the ZFC system?

The Game

@MikeMiller It is finite, but I do not now how to show it.

Mike Miller

@TheGame Any such group, you mean?

Alizter

@UserX No. Every axiomatic system will have results that cannot be proven

The Game

4:16 PM

@MikeMiller Right now I have two answers saying it is infinite :c which is false

@MikeMiller indeed

UserX

@Alizter What unprovable results does the Peano axiomatic system have?

Mike Miller

@TheGame Jyrki gave a good answer.

Alizter

@UserX Other than set-theory and a few parts the CH does not make any difference to most results.

The Game

@MikeMiller Yay math.stackexchange.com/a/987681/176015 he got it

Alizter

So no need to worry about it

The Game

4:17 PM

@MikeMiller :)

Alizter

However if you really want to it is generally assumed that there is no cardinal there.

Mike Miller

If you don't read carefully (like I didn't) it's easy to read it as asking about the set of such matrices, which I suppose confused the other posters, too

Well, no, one of them just gave a completely useless hint.

Alizter

@UserX Unprovable in peano arithmetic

Mike Miller

@Alizter This is not true. There are theories that are both consistent and complete; these theories just don't satisfies the conditions of Godel's theorem.

Alizter

@MikeMiller You take over I have read very little on this subject.

Mike Miller

4:20 PM

well, just don't ask me to state the conditions of godel's theorems. :P

Chris's sis

@Chris'ssis :P

This is seriously bad

How comes two people are allowed to have the same names ?

Chris's sis

@Chris'ssis lol, yeah, that is really bad (deceiving). :-)

UserX

Also, what's the correct way to formulate that a set $X$ is not a subset of $\Bbb N$ but a subset of $\Bbb R$ thus it has the cardinality of $\Bbb R$ ? How do I prove it?

The Game

@Chris'ssis do @Chris'ssis

And look at the tabs

Double picture :D

Mike Miller

@Huy I liked your old gravatar better.

Random Variable

4:24 PM

@robjohn This answer of mine math.stackexchange.com/questions/780913/…, which I remember you were talking about with someone on chat a while ago and seemingly didn't find any issues with, is bothering me. Does the integral on the first line actually exist as a Cauchy principal value? I'm not familiar with a definition that pertains to non integrable singularities at both endpoints.

UserX

@MikeMiller two requests. One, please answer my question above, two, set theory is interesting, can you suggest a solid book dealing with cardinals, ZFC, and what knowledge do I need to understand that book?

Huy

@MikeMiller: Not my fault.

Mike Miller

@UserX what was your question above? And unfortunately I'm not all that knowledgeable about the subject. I learned what little I know from a book by Enderton, "Elements of Set Theory".

Chris's sis

@Chris'ssis Don't worry I changed my name and pic

UserX

The reference request for set theory is for @robjohn too

Mike Miller

4:30 PM

@Huy it prolly happened when you changed your email.

Huy

@MikeMiller: I never changed it.

Mike Miller

Hmm. The gravatar is generated by stuff in your profile, so you must have changed something.

UserX

@MikeMiller what's the correct way to formulate that a set $X$ is not a subset of $\Bbb N$ but a subset of $\Bbb R$ thus it has the cardinality of $\Bbb R$ ? How do I prove it?

Mike Miller

@UserX You can't, as mentioned above! It's entirely possible from the axioms that there are subjects of $\Bbb R$ that are in bijection with neither of those sets.

And even more practically, you use the Cantor-Schroeder-Bernstein theorem to show there's an injection from each one to the other.

UserX

@MikeMiller Assume I assume the continuum hypothesis

And I didn't mean robjohn

@DanielFischer can you suggest a solid book dealing with cardinals, ZFC, and what knowledge do I need to understand that book?

Mike Miller

4:34 PM

@UserX Suppose you can prove $X$ is not countable, and $X \subset Y$, where $|Y| = \Bbb R$. Then $\aleph_0 < |X| \leq |Y|$, but the continuum hypothesis says $|X|=|Y|$.

Transcript for

$Mathematics$ Mathematics