/math/ general

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Oooh another math general. Unrelated to OP, what is everyone researching? I was going to go through pic related.

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A function f:R->R (defined everywhere, finite everywhere) that is everywhere locally unbounded (for any open interval (a,b) the function takes arbitrarily large values) ?

(Answer in picture)

>>8682932
Why don't you give us a proof, anon? Sounds like a suspicious claim.

>>8682914
If by injective you mean literally that f maps all of G to some elements of H, and g maps all of H to some elements in G then it follows because:

f maps G to some subset of H, call it H1 such that |G| = |H1|

g maps H to some subset of G, call it G1, such that |H| = |G1|

From this follows that |H| is smaller or equal to |G| and that |G| is smaller or equal to |H|.

Which implies that |G| = |H|, which implies that both these injections are bijections.

Then as f and g are bijective homomorphisms, that means they are isomorphisms.

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>>8682930
http://perso.ens-lyon.fr/ghys/articles/Promenade.pdf
Very interesting read.

>>8682943
That's an ingenious proof for the finite case. Does this work for the infinite case?

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>>8682936
Shit's fucked up.

>>8682914
Seems straightforward for finite groups. I don't trust anything infinite.

>>8682948
Ok, you got me. I don't really understand why my proof doesn't work infinite groups.

Care to explain?

I mean, suppose that the two groups are infinite but not necessarily the same level of infinity.

If there exist injections to and from each other then doesn't that mean that they are the same infinity?

Like, you can have a injection from a countable infinity to an uncountable one, but you can't have an injection from an uncountable set to a countable one.

And because the all-spanning morphisms (f and g) already exist then the existence of the isomorphis is trivial. Right?

>>8682960
You can be injective but not surjective with the same infinite cardinal.

>>8682964
OOOOOOOOOH

You motherfucker.

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

>>8682960
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.

>>8682973
I do want an invite. I have no math friends irl.

[email protected]

>>8682973
Can I get an invite too, man? I hardly have anyone to talk about math related things.
[email protected]

>>8682914
>>8682969
>>8682960
>>8682964
The answer is "no" for infinite groups.

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

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Does every periodic function have an smallest period?

>>8682914
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.

>>8683001
Nope! Easy counter example: a constant function

>>8683001
>>8683016
Let's exclude that then, does every nonconstant periodic function have an smallest period?

>>8683001
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].

>>8683014
>a subgroup of order 2
a normal* subgroup of order 2

>>8683020
Still no, see >>8683021
However if it is continuous and non-constant---yes it does have a minimum period. Exercise for you. ;)

>>8683014
>>8682914
Very interesting that this fails for groups, but it supposedly works for sets

>>8683024
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?

>>8683024
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?

>>8683112
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

>>8683112
depends on how much effort you put in

>>8683079
have u tried thinking about it

>>8683079
Yes.

>>8683117
>>8683118

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.

>>8683130
Proofs come with time. The best way to learn proofs is to read more proofs.

>>8683130
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.

>>8683128
Yeah, I know the solution, I'm posting this to keep the board engaged.

>>8683129
Why are you so sure? The result is true for a compact domain, but is it really true in all of R?

>>8683079
I say no. Consider f(x) = g(x) = x

>>8683130
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.

>>8683148
It's true on any open set.

>>8683143

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".

>>8683155
Oh I see what you mean. Any bounded set.

>>8683130
to elaborate on >>8683143, 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.

>>8683155
>>8683162
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

>>8683023
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.

>>8683220
Fucked up the LaTeX as usual; it should read "minimal period [math] 2 \pi / k[/math]"

>>8683222
>>8683220
Scratch that "proof," there are continuous functions whose Fourier series do not converge pointwise.

>>8683236
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

>>8683236
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

>>8683024
here's an interesting discussion about CSB in other categories:

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

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>>8682940
It should be obvious anon.

>>8682936
>A function f:R->R (defined everywhere, finite everywhere)
there is literally no need for the elaboration in brackets

>>8685492
stop reminding me ;-;

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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.

>>8682940
You asked "does there exist a group isomorphism?"

The answer is yes.