/math/ general

You can be injective but not surjective with the same infinite cardinal.

OOOOOOOOOH

You motherfucker.

You got me man. You fucking got me. I guess I only have finite IQ.

They are the same infinities, but infinity gives you a lot of room to move around.

So let me show you a counterexample. Let G be the free group with 2 generators. That is, the group generated by the elements {a,b} with no extra relations. Elements of this group are words consisting of a, b and their inverses. It is an infinite group. Similarly let H be the free group with 3 generators. Same thing as before, the words in the alphabet {x,y,z} and their inverses.

There is a rather natural injection from G to H, just send a to x and b to y. What about the other way around?

There is in fact, an injection from H to G. One is given by

g(x) = aa
g(y) = bb
g(z) = ab

(Can you prove this morphism is injective?)

These groups are also not isomorphic, as an isomorphism would have to send two elements of H to a and b and so the proper subgroup of H generated by those two would be all of H, which is impossible.

We have a small chat where I sometimes discuss things like this, and people talk about things they are studying. If you want an invite, you can create a disposable email (you can google that) and I can send you one.

I do want an invite. I have no math friends irl.

[email protected]

Can I get an invite too, man? I hardly have anyone to talk about math related things.
[email protected]

The answer is "no" for infinite groups.

I'm trying to come up with a counter example.

Does every periodic function have an smallest period?

No. Consider the infinite symmetric group [math]\mathfrak S(\mathbb N)[/math].
We have an injection [math]\mathfrak S(\mathbb N) \hookrightarrow \mathbb Z_2 \times \mathfrak S(\mathbb N)[/math] mapping [math]\sigma[/math] to [math](0, \sigma)[/math].
Conversely, we have a morphism [math]\mathbb Z_2 \times \mathfrak S(\mathbb N) \hookrightarrow \mathfrak S(\mathbb N)[/math] sending [math](k, \sigma)[/math] to [math](1,2)^k \tilde\sigma[/math] where [math]\tilde \sigma(n) = \sigma(n+2)[/math].
These groups are not isomorphic because [math]\mathbb Z_2 \times \mathfrak S(\mathbb N)[/math] has a normal subgroup of order 2, whereas [math]\mathfrak S(\mathbb N)[/math] doesn't. Indeed, a subgroup of order 2 of [math] \mathfrak S(\mathbb N)[/math] is necessarily generated by an element of the center (think about it), but [math]\mathfrak S(\mathbb N)[/math] has trivial center.

Nope! Easy counter example: a constant function

Let's exclude that then, does every nonconstant periodic function have an smallest period?

Let [math]f: \mathbb{R} \rightarrow {0,1}[/math] be the indicator function for the dyadic rationals.

For any [math]n[/math], [/math]1/2^n[/math] is a period of [math]f[/math].

>a subgroup of order 2
a normal* subgroup of order 2

Still no, see However if it is continuous and non-constant---yes it does have a minimum period. Exercise for you. ;)

Very interesting that this fails for groups, but it supposedly works for sets

why? We're asking something stronger for groups. Groups are just sets, and if we ignored the group structure Cantor-Schroeder-Bernstein holds. But when you start requiring that the functions involved all must obey the group structure, it doesn't work anymore.

If f, g: R -> R are uniformly continuous functions, then is the product fg uniformly continuous?

the bijection that you construct for sets just demolishes things around and preserves no structure - just size. as soon as you ask for them to also preserve structure you run into issues and the same trick doesn't work.

Tfw brainlet

I'm taking real Analysis and abstract algebra next fall (with some other random upper divs).

How long until I'm on the level of you Math lads?

depends on how hard you work honestly
the things discussed in this thread are interesting but elementary - all you need is some group theory and some analysis to understand it

depends on how much effort you put in

have u tried thinking about it

Yes.

I work hard. I got straight As in all my lower division maths. I like the subject and will be officially a math major next fall when I start upper division coursework, which includes a proofs class over the summer. My linear algebra course had proofs, and I thought they were hard, but I had never done proofs before that. My school is odd, linear algebra has proofs and is a prerequisite for the actual proofs class on campus, the one that actually teaches proof technique.

Proofs come with time. The best way to learn proofs is to read more proofs.

actual "math", not physics or whatever, is all proof stuff. the relationship between straight As in the classes you've taken and being good at actual math is not straightforward. it's probably a good sign, but certainly not a guarantee. conversely, low grades in those classes is probably a bad sign, but definitely not a guarantee of failure.

Yeah, I know the solution, I'm posting this to keep the board engaged.

Why are you so sure? The result is true for a compact domain, but is it really true in all of R?

I say no. Consider f(x) = g(x) = x

In principle you can always work harder: you can enlist in more classes, you can read through books yourself, etc etc.

In practice it's good to keep a well-balanced workload and a well-balanced life. But you should be pushing yourself forward as much as you can without being unhealthy.

It's true on any open set.

I've heard that upper division Math is nothing like lower division. A buddy of mine is a first year phd for math, and he told me to focus on doing well in abstract algebra and real analysis. He said "it teaches you to think like a mathematician. So if you do well there, the rest of your upper divs will be easy".

I'm excited honestly. I'm good at computation, I would like to learn more about proofs and "real math".

Oh I see what you mean. Any bounded set.

to elaborate on , it's 100% possible to do well in lower-level "math" classes without having a legitimate conceptual grasp of what's going on, but you absolutely cannot get away with that in real math. if you've been running on the fumes of memorization and algorithms, then your grades from that have no particular bearing on how you'll do at higher levels. on the other hand, if you understand *why* things work the way they work, and you're capable of recognizing inconsistencies within your understanding, then that's reason to expect to do well.

Yeah, a counterexample for the unbounded case is given by the functions

f(x) = x
g(x) = sinx

where both are seen to be uniformly continuous (bounded derivative) but the product is not

Continuous, periodic functions have Fourier series that converge pointwise, right?

In other words (assuming WLOG that [math]f[/math] has period [math]2 \pi [/math], because otherwise we can scale) there exists coefficients [math]a_k[/math] such that [math]f (x) := \sum_{k = -\infty}^{k = \infty} a_k \exp (i k x)[/math]. If there exists a lowest [math]|k| \neq 0[/math] such that [math]a_k \neq 0[/math], then [math]f[/math] then has minimal period [math]2 /pi / k[/math]. But, [math]f[/math] has no minimal period, so we need all [math]a_k = 0, |k| \geq 1[/math]. This means that [math]f[/math] is constant, which contradicts the hypothesis.

Fucked up the LaTeX as usual; it should read "minimal period [math] 2 \pi / k[/math]"

Scratch that "proof," there are continuous functions whose Fourier series do not converge pointwise.

Hint: Consider the set of all periods of this function. What's it like, topologically? What does f not being constant tell you?

saving this from the depths of page 10

Check out Carleson's theorem, it establishes pointwise convergence almost everywhere of the Fourier series of L2 functions. More generally, Fourier series don't necessarily converge pointwise unless they converge at a sufficiently fast rate, as evidenced by the large wealth of proofs concerning the pointwise convergence of a variety of averages of Fourier coefficients.

daily reminder that CSB is not constructive

here's an interesting discussion about CSB in other categories:

mathoverflow.net/questions/1058/when-does-cantor-bernstein-hold

It should be obvious user.

>A function f:R->R (defined everywhere, finite everywhere)
there is literally no need for the elaboration in brackets

stop reminding me ;-;

This is a kind of cool problem I've been working on. If you have a room, where is the best place to look at it from? You have one observer to see the room, and an arbitrary number of obstructions and that there are twice as many walls, since one wall separates two regions.

The answer involves set theory to an extent.

If you consider a single wall separating two regions, then it is pretty damn safe to conclude that now there are two sets (Let's call them A1 and A2) that are in no way equal to each other because they are being separated by the wall (O1).

"If you are on one side of the wall, then you can't see at least SOME part of the other side, and vice versa. This is because there's a wall between the two regions."

This is all fairly basic, but the cool part happens when you think about how multiple walls behave, as well as the fact that the previously stated rule holds up even when the wall is arbitrarily short, and doesn't separate the two regions completely.

This all leads up to proving that it's impossible to see around a corner, all using set theory.

You have two sets A1 and A2 that are separated because of O1. You also have two sets B1 and B2 that are separated because of O2. Pretend for now that O1 and O2 are two walls touching each other and making a corner, but it really doesn't matter because technically the result still holds up even if the walls aren't touching.

Let's say that Region 1 = the intersection of A1 and B1, R2 = A1 (int) B2 , R3 = A2 (int) B1, and R4 = A2 (int) B2.

We'll also say that O1 lies on (is a member of) B1, and that O2 lies on A2.

If P lies in R1, then the regions where P won't be able to see at least some part of them are the intersection of the region that P wouldn't be able to see if it there was only O1 and the region(s) that O2 lies in, as well as the intersection of the region that P wouldn't be able to see if it was only O2, and the region that region that O1 is a member of.

You asked "does there exist a group isomorphism?"

The answer is yes.