Mathematics - 2017-11-24

user193319

12:12 AM

Determine the complete ring of quotients of the ring $\Bbb{Z}_n$ for each $n \ge 2$. I'm a little confused...What exactly is this asking? From my understanding, there isn't just one complete ring of quotients. A ring of quotients is $S^{-1} \Bbb{Z}_n$ where $S$ is a multiplicative subset (i.e., closed under multiplication) of $\Bbb{Z}_n$, so, presumably, there could be many different $S$.

MatheinBoulomenos

Complete ring of quotients means that we take $S$ to be all the non-zerodivisors

user193319

So, in other words, take the largest subset $S$?

MatheinBoulomenos

Not the largest subset $S$, but the largest subset such that the map $R \to S^{-1}R$ is injective

You can localize at zero-divisors as well

user193319

Yes; you're right. I should have qualified what I meant by 'largest'. Thanks for the help.

Leaky Nun

12:48 AM

@zxmkn p^n - 1 = (p^d - 1) (1 + p^d + p^2d + p^3d + ... + p^(n-d))

justification by factorization

user193319

Is there another name for ring of quotients? I type into google "ring of quotients" and nothing comes up, except results about the quotient ring which is something different.

anon

1:09 AM

fraction field?

(or field of fractions)

@zxmkn study the finite geometric series formula

if n=dk then (x^n-1)/(x^d-1) = (u^k-1)/(u-1) = u^(k-1)+...+u^2+u+1 where u=x^d

Leaky Nun

@anon sniped :P

Leaky Nun

1:31 AM

if $\pi_1(X) = 0$ then what is the condition for $\pi_1(X/G) = G$?

@anon @BalarkaSen

Mockingbird360

@LeakyNun Should I take the square root of both sides now and integrate or should I double integrate?

Leaky Nun

I can barely see what you wrote

Mockingbird360

@le

@LeakyNun Is it clear now?

Leaky Nun

it's the same paper

studrayght5

1:48 AM

Bought one of my wish list books from amazon after seeing how cheap they made it. As soon as I checked out, the price shooted up nearly 300%. Strange.

Keep_On_Cruising

2:15 AM

Hi

Can someone help me with this question math.stackexchange.com/questions/2534625/…?

have to submit soon

It's about an exact DV

nitsua60

in RPG General Chat, 25 secs ago, by nitsua60

@Keep_On_Cruising It looks like you're saying sin(y)-sin(x)=N; why isn't it cos(y)[sin(y)_sin(x)]=N, rather?

Keep_On_Cruising

He

It is cos(y)[sin(y)_sin(x)]

nitsua60

Where's the 2 in the numerator come from?

Keep_On_Cruising

Well

My = cosycosx right?

Faust

Morning sheeep and @LeakyNun

Keep_On_Cruising

2:22 AM

and Nx = -cosycosx

nitsua60

Why isn't M_y = 0? M is a function solely of x, as it's written.

Keep_On_Cruising

So My -Nx =2cosycoxx

sorry

you're right that's a typo

I'll edit

M_x = sinysinx

nitsua60

Ah, gotcha. (Still doesn't help the sin(y) in your denominator that's bothering you.)

Keep_On_Cruising

M_y I mean :P

idd

You have anny ideas?

nitsua60

brb

Keep_On_Cruising

2:41 AM

This problem

...

Antonios-Alexandros Robotis

Hello!

Keep_On_Cruising

Hi

Could you take a look at this problem math.stackexchange.com/questions/2534625/… ?

would be really appreciated :D

Antonios-Alexandros Robotis

I'll take a look in a moment.

Keep_On_Cruising

Alright :)

Antonios-Alexandros Robotis

3:07 AM

sorry @Keep_On_Cruising It's not obvious how to fix the problem :/

at least not to me.

Keep_On_Cruising

oke

orbit-stabilizer

4:06 AM

hello

Secret

6:50 AM

[Random]

Structure of $\omega$ (cont.)

{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15...}

(Processing next step. Please wait...)

Let L be a partition into two sets such that each set has only one connected component under the order topology of the ordinals

It is easy to see that any such L on $\omega$ consists of a finite set and a set of order type $\omega$

However. $\omega$ is not amorphous for L is only one of the many possible classes of partition on $\omega$. For example, the partition A that split $\omega$ into the even and odd naturals is an example of two infinite subsets

More generally, since $\omega$ is well ordered, it is possible to select some countable subset algorithmically. The complement then forms the other infinite subset

$\omega$ also does not contain any limit ordinals

Hmm... actually what is the order type of the even numbers?

it seemed to be also $\omega$ since we can do the order preserving bijection $(0,1,2,3,4,...) \mapsto (0,2,4,6,8,...)$

likewise for the odd numbers.

However is {odd} U {even} expressiable in terms of ordinal arithmetic?

It seems this union is not just in terms of what elements were there, but also the ordering information. Does that mean ordering is destroyed in set unions?

Liad

7:17 AM

@MatheinBoulomenos Thank you! nice solution :)

Akiva Weinberger

@BalarkaSen I felt like you should know this:

The original quote is actually "It's over 8000"

The dubs are wrong

Daminark

My whole life is a lie @Akiva

Even the part of my life that came before hearing that quote has been retroactively made a lie

Aldon

8:28 AM

Hi i have a quick question about a trig function

wwwf.imperial.ac.uk/metric/metric_public/functions_and_graphs/… here at the last part it says the positive root was used for convention

is it really by convention tho, like what if i wanted to use the negative root?

Balarka Sen

9:16 AM

@LeakyNun That G acts freely and properly discontinuously on $X$

You want $X \to X/G$ to be a covering space, that's all

@AkivaWeinberger No, this can't be

I am in denial

Kaustabha Ray

If we have a group of 10 people who are either mutual friends or mutual strangers, can we say that if we select one person out of these ten people then this single person will be friends with at least 5 people or starngers with at least 5 people? This is what my reasoning goes. The text says atleast 6 friends or 4 strangers. I think this is a tighter bound than my reasoning?

abenthy

9:33 AM

This might be a stupid question, but geometrically, why does rotation on $S^1$ preserve the orientation? If I chose one orientation, i can always rotate in the "opposite direction", why does this give me an orientation preserving map $T_pS^1 \to T_pS^1$ in each tangent space? It seems to me that this map should be negation.

Balarka Sen

11:04 AM

@abenthy That's not what orientation means. Each tangent space of $S^1$ has a consistent "right-hand rule". That is preserved under rotation

Rotation by 180 degree apparently "flips" the direction of a tangent vector (eg at the points (-1, 0) it was pointing upwards and it's pointing downwards at (1, 0)), but that does not imply orientation-reversation.

Indeed, the orientation on $S^1$ is imposed so that those are exactly the right directions of orientation

Perturbative

12:17 PM

Hey everyone!

Leaky Nun

@BalarkaSen thanks

abenthy

12:38 PM

If $G$ is a linear algebraic group defined over a subfield $K \subseteq \mathbb{C}$, it has a functor of points given by $A \mapsto hom(K[G],A)$ for a $K$-algebra $A$. If $G$ is embedded in $GL(n,\mathbb{C})$ and I choose $A=k$, do I have an identification $\hom(K[G],K) \cong G \cap GL(n,K)$?

Here $K[G] = K[X_{11},\ldots,X_{nn}]/I_K(G)$ is the coordinate ring of the $K$-group which is embedded in $GL(n,\mathbb{C})$.

Simply Beautiful Art

2:08 PM

Hey @LeakyNun

user193319

Problem: Determine the complete ring of quotients of the ring $\Bbb{Z}_n$. Okay. So I trying to determine what $S^{-1}\Bbb{Z}_n$ is, where $S \subseteq \Bbb{Z}_n$ is the set that doesn't contain any zero divisor. I determined that $S=\{\overline{m} \mid gcd(m,n)=1\}$. At this point I am unsure how to proceed. What exactly is $S^{-1}\Bbb{Z}_n$, besides being equal to $\{\frac{\overline{r}}{\overline{s}} \mid gcd(s,n)=1, \overline{r} \in \Bbb{Z}_n \}$? Is it isomorphic to some familiar space?

Leaky Nun

@SimplyBeautifulArt hi

Secret

Hmm.... It is easy to embed $\omega$ into some subset of $\Bbb{Q}$ (e.g. $(0,1,2,3,4,...) \mapsto $(0,1/2,2/3,3/4,4/5,...)$

Alessandro Codenotti

I don't think there's a simpler way to describe that localisation @user193319

user193319

@AlessandroCodenotti That's what I suspected. Thanks!

Alessandro Codenotti

2:18 PM

@Secret Every countable ordinal embeds into $\Bbb Q$, every countable linear order even

Secret

Yes. Meanwhile, $\omega$ has elements being finite ordinals, so if we partition $\omega$ such that one of them forms an initial section of $\omega$, and the other nonempty, then it follows the initial section is always a finite ordinal, while the remaining portion is a countable ordinal of the same order type as $\omega$

$\omega_1$ has very similar properties. By doing an analogous partition on $\omega_1$, the initial section of $\omega_1$ is always a countable ordinal and the remaining portion always an uncountable set with order type $\omega_1$

So then the issue here is that when one tries to do a similar embedding, $\omega_1 \mapsto S \subsetneq \Bbb{R}$, it however does not work. (Proof to be dug up)

Simply Beautiful Art

I recently wrote a function/program which I believe grows as fast as $f_{\psi(x)}(n)$, where $x$ is the first fixed point of $\Omega_{M+x}$, where $M$ is the first Mahlo 'ordinal'.

Secret

I have no idea how can one visualise fixed points at that level since after Bachman howard ordinal, pretty much only the fast growing hierarchy is left to tell you what order the OCF is

as all extensions of veblens will break down way before bachman

Simply Beautiful Art

tio.run/##NY4xDsIwDEV3TtEuCJCDmpEUk4NYHtIkJSgQUShCqPTsJUWwPT37f/…

@Secret x'D

Oh, @Secret you may be interested in a Discord server (Open to anyone) dedicated to the explanation of this nonsense of a nightmare.

@Secret And using Veblens past the Bachmann-Howard ordinal is not the key. Just use more OCF's. :D

(Also, it's Bachmann, two n's)

Secret

Yeah, that's what I am saying, once bachmann is reached, you pretty much just trust the OCF actually continue to churn out bigger objects

Simply Beautiful Art

2:28 PM

@SimplyBeautifulArt g[k] grows really fast :-)

@Secret Well it's not like it's impossible to understand the OCF's.

In fact, a lot of the OCF stuff after the Bachmann-Howard ordinal is just old stuff reapplied.

@Secret What do you think is so special about the BHO? (Bachmann-Howard ordinal)

Large countable ordinal

In the mathematical discipline of set theory, there are many ways of describing specific countable ordinals. The smallest ones can be usefully and non-circularly expressed in terms of their Cantor normal forms. Beyond that, many ordinals of relevance to proof theory still have computable ordinal notations. However, it is not possible to decide effectively whether a given putative ordinal notation is a notation or not (for reasons somewhat analogous to the unsolvability of the halting problem); various more-concrete ways of defining ordinals that definitely have notations are available. Since there...

Lol

Secret

Well, as far I am aware from those exercise, it seems a generic OCF expression looks something like this:
$$\psi (f(\Omega)) [n,f(\Omega)] = [\psi (g(\Omega,]^n\circ \psi (g(\Omega,\omega))$$
where $f$ is some generic monotonic increasing function, and $g$ is a "reduced" version of $f$ (as it happens when taking the $\Omega$ terms apart)

Well, since $\psi (\Omega^{\Omega^n})$ already start to control the number of entries in the transfinite veblen, it follows that $\psi (\Omega^{\Omega^{\Omega^n}})$ will start to control how many "layers of LVO" we can have for the transfinite veblen, and …

Simply Beautiful Art

^ This is the difference between my program and Loader's number.

Secret

Which is why I predict Veblen and its extension will be completely exhausted at the Bachmann howard ordinal

Simply Beautiful Art

@Secret googology.wikia.com/wiki/User_blog:Rgetar/…

It can be done.

But that stuff does get a little crazy.

But on the other side, $$\rm BHO=\psi(\Omega^{ \Omega^{\Omega^{ \dots}}})$$

And so, my question is why do we stop here.

Why not go further? Why is this specific ordinal called the BHO?

Secret

At that point, in theory you exhaust the $\Omega$s since all one can do in OCF is construct ordinal notations from the language {$\Omega$,$\psi$,+,*,^}. So without $\psi'$ we will be permanently stuck at Bacnhamnn

Simply Beautiful Art

2:41 PM

Yay, glad you understand that.

And then there's a really simple extension to this.

$$\psi_1(0)=\Omega^{ \Omega^{ \Omega^{ \dots}}}$$

$$\psi_1(1)=\psi_1(0)^{ \psi_1(0)^{ \psi_1(0)^{ \dots}}}$$

etc.

And then $\Omega_2$ does nestings of $\psi_1$, just as $\Omega(=\Omega_1)$ does nestings of $\psi(=\psi_0)$.

e.g. $\psi_0(\Omega_2)=\psi_0(\psi_1(\psi_1(\psi_1(\dots))))$

$\psi_0(\Omega_2+\Omega_1) = \psi_0(\Omega_2 +\psi_0(\Omega_2 +\psi_0(\Omega_2 +\dots)))$

@Secret I mean, you could almost say the extension is natural.

Ofc, you'll run out of $\Omega_2$'s, but that's why we have $\psi_2$.

(after lots of extensions)
And then the limit of this notation would be $\psi(\Omega_{\Omega_{ \Omega_{\ddots}}})$

Secret

> FS #28

φ(1<1,0>)

φ(1<1<1,0>>)

φ(1<1<1<1,0>>>)

φ(1<1<1<1<1,0>>>>)

φ(1<1<1<1<1<1,0>>>>>)

φ(1<1<1<1<1<1<1,0>>>>>>)

φ(1<1<1<1<1<1<1<1,0>>>>>>>)

Hmm, I think I understood what that guy is doing: He is basically nesting a LVO layers of LVOs of LVOS of ... of LVO

that's how he can get to Bachmann so fast and still have room left behind

In our OCF, FS #28 will be the $\Omega$ towers

Simply Beautiful Art

@Secret Nah, his notation doesn't extend past BHO.

Well anyways, I suppose that list of ordinals shall be my christmas present to you.

Enjoy :D

Secret

Anyway, while it is obvious no matter how recursive we can go, we will never even start approaching $\omega_1^{CK}$ (unless the OCF already has $\omega_.^{CK}$ map in it), what surprises me is that even if we have countable length strings by using infinite time turing machine, that will not even bring us close to $\omega_1$

In ZF+large cardinal axioms, $\omega_1$ is not even projective, meaning (roughly) that repeatedly taking projections and then compute the relative complement of it will not reduce its size

It also suggests whatever its structure is, requires a string at least of length $\aleph_1$ to describe

2

Q: The paradox with the first uncountable ordinal

Suppose we have a set $M = (0,1) \subset R$ of reals well-ordered as the first uncountable ordinal. Let $M(a) = \lbrace x \in M : x < a \rbrace$. For every $a \in M$ set $M(a)$ is countable. That's why every increasing sequence is bounded: $$(*) ~~~~~~~~~~ \forall \lbrace a_1,...,a_n,...\rbrac...

$\omega_1$ and uncountable well orderings are very vague without the axiom of choice

Appliqué

3:01 PM

Hello! Could you please tell me, if I modify a pseudodifferential operator on a finite number of its $L^2$ eigenfunctions, will the difference always be a pseudodifferential operator of order $-\infty$?

Secret

I think it is not even clear what self injections of $\omega_1 \mapsto \omega_1$ besides the identity, look like

and for any attempted embeddings $\omega_1 \mapsto S$ while the proof on why $S$ is not an interval in $\Bbb{R}$ is well known in the form of you cannot have uncountably many disjoint intervals in $\Bbb{R}$, I am not sure if there are proofs that showed any generic uncountable sequence formed by picking elements of $\Bbb{R}$ must correspond to an uncomputable function

Perturbative

@BalarkaSen Are you still keen on going through the Covering Space stuff? If you are let me know which book(s) you are using (if any).

Balarka Sen

@Perturbative Very much so! Tomorrow is the last of my exams

user193319

If $I$ is an ideal in $\Bbb{Z}_{p^n}$, is it true that $I = (p^k)$ for some $k \le n$?

Perturbative

@BalarkaSen Great! Good luck for tomorrow! :)

Balarka Sen

3:07 PM

I'll tell you the relevant books I want to use tomorrow

Perturbative

Cool no problem

Liad

I've got a basic approximation question:
is it true that $|\sum_{n=1}^N \dfrac{z \ ^ n - z \ ^ {n+1}}{n+k}| \le |\sum_{n=1}^N z \ ^ n - z \ ^ {n+1}| $ ? $z\in B(0,1)$

$k$ is a fixed natural number

Secret

10

Q: How many ordinals can we cram into $\mathbb{R}_+$, respecting order?

I've been pondering the following question. How can we measure the amount of "space" above an element $p$ in a partially ordered set $P$? One way would be to try to cram the elements of increasingly large ordinals into the space above $p$, respecting order. The least ordinal whose elements ...

order-theory ordinals

> Furthermore if we do embed $\omega_1$, then we can cut $\Bbb R$ into $\aleph_1$ disjoint open intervals, this is impossible because there is an injection from every family of disjoint open intervals into $\Bbb Q$.

ooooooppps

because... (...{x}(a,b){xx}(c,d){xxx}(e,f){xxxx}...) thus we are in trouble

user193319

I think $(6) = \{0,3,6\}$ in $\Bbb{Z}_{3^2}$ is a counterexample. Does that sound right?

Anyone?

mercio

usually, $p$ is used for prime numbers

oh nevermind

your $(6)$ is equal to $(3)$

user193319

3:26 PM

@mercio Oh, you're right! So might be true that if $I$ is an ideal in $\Bbb{Z}_{p^n}$, is it true that $I = (p^k)$ for some $k \le n$?

mercio

yup

user193319

@mercio I've been trying to prove this for quite some time without success. Do you have any hints?

mercio

show that every principal ideal is of this form

then show that if you take any two of them, one is included in the other

user193319

Yeah. I tried that. Given the ideal $I$, $I = (x)$ for some $x$, since $\Bbb{Z}_{p^n}$ is a PIR. But I don't know how to show that $x=p^k$ for some $k$. Showing that $x \mid p^n$ would work, but I don't know how to do that

BAYMAX

4:00 PM

.

Any help on this -

"The variables $X$ and $Y$ are connected by the equation $aX+bY+c = 0$, show that the correlation between them is -1 if the signs of $a$ and $b$ are the same and $+1$ if they are different."

mercio

you're not asked to show that $x=p^k$

but to show that $(x) = (p^k)$

there is a difference

the first one is only true for very few $x$

Secret

Hmm... how about finding a dense (but incomplete) subset $S$ of the irrationals... that way, even if I have $\omega_1$ mapped to $S$, there should not be any issue with countable rationals. It just means $S$ does not embed in the reals even though it is a subset of the irrationals...

Secret

4:47 PM

2

Q: Prove that no uncountable family of subsets of $\mathbb{N}$ is well-ordered by relation of inclusion.

Prove that no uncountable family of subsets of $\mathbb{N}$ is well-ordered by relation of inclusion. I had two ideas how to do it. First was to show that if such family is uncountable, then it is not true that all proper beginning segment (is it correct english name for that?) is of the form $O...

elementary-set-theory order-theory

uh, this proof is much cleaner, and rules out all possible ways of well ordering countable strings

3

Q: Why is $\omega_1$ an accumulation point here, but has no sequence converging to it?

I'm trying to understand the following. I'll denote by $X$ the set of all ordinals that are at most $\omega_1$, the first uncountable ordinal. To avoid confusion, they can be equal to $\omega_1$ as well. Equip $X$ with the order topology. Then how come $\omega_1$ is an accumulation point of $X\...

general-topology

> Take a sequence $(\alpha_n)_{n\in\mathbb{N}}$ in $X\setminus\{\omega_1\}$. Consider the ordinal $\beta=\bigcup_{n\in\mathbb{N}}\alpha_n$. As a countable union of countable sets it's still countable, so $\beta\in X\setminus\{\omega_1\}$. Now $(\beta,\omega_1]$ is a neighbourhood of $\omega_1$ containing no elements of the sequence $(\alpha_n)$, so $(\alpha_n)$ can't converge to $\omega_1$. This shows no sequence of $X\setminus\{\omega_1\}$ converges to $\omega_1$.

So $\omega_1$ is so vast that only $\aleph_1$ nets can converge to it...

Secret

5:09 PM

7

A: An easy to understand definition of $\omega_1$?

While I agree that @Austin Mohr ‘s answer is a good picture of $\omega_1$, it is technically problematic since it is circular. You cannot define an uncountable ordinal in terms of uncountable union since you have not yet defined uncountable or uncountable union. Also, the notation may give the i...

There is however no precise boundary on where uncountability begin

3

A: An easy to understand definition of $\omega_1$?

$\omega_1$ is the first uncountable ordinal, or, equivalently, the set of all countable ordinals. The countable ordinals in turn can be constructed by the following rules: 0 is a countable ordinal If $\alpha$ is a countable ordinal, then so is $\alpha + 1$. If for each $i \in \mathbb{N}$ $\alph...

Unfortunately, we would have got stuck at $\omega_1^{CK}$ before even starting to reach $\omega_1$, and $\omega_{1}^{CK}$ is countable because the set of all computable well orderings can be expressed in finite strings thus there are only countably many possibiliities

7

A: An easy to understand definition of $\omega_1$?

While I agree that @Austin Mohr ‘s answer is a good picture of $\omega_1$, it is technically problematic since it is circular. You cannot define an uncountable ordinal in terms of uncountable union since you have not yet defined uncountable or uncountable union. Also, the notation may give the i...

> 3rd Principle - For every set A of ordinals, there is a least ordinal greater than every member of A, called sup(A).

However this fails to be predicative because back at $\omega_1^{CK}$ we are already stuck at uncomputable well orderings

as for $\omega_1$ it (most likely?) fails to be constructive because it is hinted by some really hardcore ZF set theory models that includes large cardinal axioms and other things

6

A: Embedding $\omega_1$ in the hyperreals

What you want is the usual argument that the bounding number $\mathfrak{b}$ is uncountable. Given a countable family $\mathscr{F}=\{f_n:n\in\omega\}\subseteq{}^\omega\omega$, define $$f:\omega\to\omega:n\mapsto 1+\max_{k\le n}f_k(n)\;;$$ then for each $n\in\omega$ we have $f(k)>f_n(k)$ for all ...

Hmm, so it can be done in the hyperreals with some ultrafilter. But since ZF does not necessary have an ultrafilter, it means we cannot really say much about $\omega_1$ in ZF

hmm... looks like the only collection of models I have not test yet, is to replace powerset axiom with something slightly weaker than it, so that it it only strong enough to give a notion of uncountability, but does not result in throwing constructivity out of the window (there is no hope for predictability, as we need to get pass uncomputability of well orderings first)

Now before I end this (.... once again) a big wall of text...

(math.stackexchange.com/questions/173530/…)

[Choiceless ZF, model=default 8 axioms + optional extra axioms]

> The union of countably many pairs may not be countable.

Probably the most intuitive way to understand what an amorphous set looks like, and how linear order will be enough to turn it into a countable set, thus the intuition on why no amorphous sets can be linearly ordered

> The real numbers may be a countable union of countable sets.

Follows from the failure of the countable union of countable sets implies countable, the cardinality of the resulting set can be anything $> \aleph_0$, thus it can include the reals in some models

> We may be able to partition a set into more parts than elements, in particular this set might be the real numbers.

Tug' Tegin

5:34 PM

Hi! I have an easy-peasy question: how do you denote symbolically the statement: "Suppose that the function f is continuous on the set of all real number" in math?

Secret

$\forall x \in \Bbb{R}: \lim_{x\to a} f(a)=f(x)$?

anon

@Tug'Tegin follow-up question: why would you denote that with symbols instead of words?

Tug' Tegin

because i am preparing for exams and am want to write the problem differently (maybe to impress :) ) @anon

@Secret thank you!

anon

when I read people unnecessarily writing that kind of stuff with symbols instead of words, I'm the opposite of impressed

Simply Beautiful Art

@Tug'Tegin Use $\ddot\smile$ to write smiles at the end of a parenthesized statement to avoid confusion.

Tug' Tegin

5:39 PM

H-mm @anon

Simply Beautiful Art

@anon Depends on what kinds of symbols + context

Tug' Tegin

@SimplyBeautifulArt got it!

Semiclassical

just posted this question to the main site:

1

Q: Geometric proof of a trig identity on $\cos t \cos u\cos v$

Consider the following trigonometric identity, valid for any set of angles $u,v,t$: $$\cos t⋅\cos u⋅\cos v =\frac14\left[\cos(t + u + v)+\cos(t + u - v)+\cos(u+v-t)+\cos(v + t - u)\right]$$ This identity and its derivation have previously appeared as a question on this site (though the version...

geometry trigonometry alternative-proof proof-without-words

anon

why not $u,v,w$?

Simply Beautiful Art

@anon Why do you question everything?

Semiclassical

5:40 PM

because it was t,u,v in the original link

anon

I recall the three cycled versions of a+b-c (as an angle) appeared a lot when I was playing around with general cross products

Semiclassical

i'm a little annoyed at how irrelevant the first comment is, though it's nothing i consider worth flagging over

anon

it is indeed confusingly superfluous

Secret

> We may be able to partition a set into more parts than elements, in particular this set might be the real numbers.

Ok I still have not wrap my head around this yet,. Will deal with it later...

> In the real numbers continuity by sequences and by ε-δ are no longer equivalent.

I have not wrap my head around this yet

> Topology breaks down in acute manners, to horrid to begin to describe.

Semiclassical

@anon interesting

Secret

5:44 PM

I like pathologies. The weirder the better

> There might be no free ultrafilters, on any set.

Simply Beautiful Art

@Semiclassical Flagged as no longer needed.

Secret

That could be bad news, given how ultrafilters are useful in forcing (which I don't really understood, but it is used to construct a variety of models with desirable properties)

> In turn some fields might not have an algebraic closure; others could have two non-isomorphic closures, for example the rationals.

I will explore these later...

> There could be a tree that every point has a successor, but there is no ω-branch.

Linearly ordered infinite dedekind finite sets

Leaky Nun

@Secret ultrafitlers is how you prove the compactness theorem

Secret

I see

> There could be a vector space which has two bases of different cardinality.

Leaky Nun

@Secret but algebraic closures are isomorphic

or are you ditching AoC now?

Secret

5:47 PM

You know how much I love the wilderness, don't you?

Semiclassical

sign that you may be a math nut: wilderness = ditching axiom of choice

3

Secret

although to be fair, at the final step, I might want to keep dependent choice, because it is important for analysis related theorems

and dependent choice is not powerful enough to blow up $\aleph_1$-dedekind finite sets and $\aleph_1$-amorphous sets, both can hold a lot more structure than their more well known counterparts (e.g. $\aleph_1$-amorphous sets can be linearly ordered)

So in such model, I might get to keep those quasiinfinite sets, while having enough of analysis preserved via dependent choice

> Functional analysis may stop working due to lack of Hahn-Banach, Krein-Milman, Banach-Alaoglu theorems.

This is super bad news

> Cardinal arithmetics can fail for infinite sums and products.

This one I am not sure yet. Sure being able to use addition and multiplication to reach higher cardinalities seemed pretty useful as it provide more pathways to make uncountable sets, but would the cost be too high. Need to check

> In forcing the mixing lemma fails.

Need to check whether that will bring down compactness theorem later

> The partial ordering of cardinalities is not necessarily well-founded.

Wait, are you saying this will blow up axiom of foundation? I am not ready to deal with non well founded sets yet!

> There may be no canonical representative for |A|, namely a function which returns a particular set of the cardinality of A (like the ℵ numbers).

Will found out if that would be a big problem

Simply Beautiful Art

3

A: explicit upper bound of TREE(3)

The underlying rationale will be as follows. The Tree(n) question can be translated into a problem of the length of effectively given bad sequences. I think this goes directly back to Friedman. Such effectively given bad sequences can be mapped into effectively descending sequences of ordinals us...

Doesn't seem super helpful, though it does present an underlying theme on how one would construct bounds... anywho, it's already mentioned by Deedlit:

It was proven by Diana Schmidt in her thesis that the largest order type of any extension of the ordering on trees in TREE(n) is $\theta(\Omega^\omega \omega,0)$. According to Weiermann, we can use this to extract an upper bound of about that level of the fast-growing hierarchy. See math.stackexchange.com/questions/1950116/… — Deedlit Oct 22 '16 at 12:52

Particularly the linked answer

Secret

But one thing that is sure, I will figure out how to replace the powerset axiom with something slightly weaker and more general in order to make $\omega_1$ more constructible, because $\omega_1$ is too important in topology, some analysis and induction to throw away

Simply Beautiful Art

@SimplyBeautifulArt Also, the last few lines of the answer are quite subjective.

Secret

5:57 PM

But my immature comment of set theory is that: We need a model that provide more pathways to construct uncountable sets with a larger variety of structures

Simply Beautiful Art

Deedlit's answer makes it clear that people not being interested in it is not the problem.

Secret

(ok this should get the case $\omega_1$ sufficiently discussed with for now, I really need to spent more time on my chemistry and not procrastinate into foundations of mathematics!)

r9m

8:50 PM

Is there a way to show that if $u \in \mathscr{D}'(\mathbb{R}^n)$, s.t., $\partial_j u \in L^2_{\text{loc}}(\mathbb{R}^n)$ for each $j=1,2, \cdots, n$, then $u \in L^2_{\text{loc}}(\mathbb{R}^n)$ without actually delving into nature of fundamental solution of laplacian?

rudin outlines a neat proof in exercise (pp-224, prob 11) .. but somehow I feel there should be a simpler way of viewing this ..

MatheinBoulomenos

10:09 PM

Hi @Daminark

Daminark

Yo @Mathein, how's it going? How was the talk?

MatheinBoulomenos

Quite well, thanks. I still have not given it yet, it's in 4 weeks, but I still need to prepare it now

And for you?

Daminark

Oh I thought it was yesterday

My talk went quite well, vacuously

MatheinBoulomenos

No, yesterday I listened to a talk, but that was something completely different, not a seminar where student give talks, it was an invited professor from Oxford

What was the topic?

Daminark

There was no talk

:P

MatheinBoulomenos

10:20 PM

Oh

damn

I meant how's it going for you?

Daminark

Things are going pretty well, thanks!

Balarka Sen