@jak

@Jakobian

it appears they dont have the book :(

Could complex manifold be described by analytic geometry? I know that there is non-algebraic complex manifold, so algebraic geometry is not enough, but then how about analytic geometry?

Okay it seems yes : math.stackexchange.com/questions/2211188/…

@nickbros123 same here

I tried searching on ndl, it's even showing open access but couldn't find the download link

Do you guys know what a difference set is?

I'm wondering if $\Delta A = A - A = I \leqslant R$ an ideal of a ring, and you take $\Delta(A \setminus \{a\}) = ?$ would you still have $I$ the original ideal. Intuitively you would, but requires proof.

$A - A = \{ a - a' : a,a' \in A\}$.

Also $R = \Bbb{Z}$ so this is an infinite setting

@Debug $\Delta A$ need not be an ideal, for an infinite set $A$ of integers. For instance, $A$ is the set of all squares of integers, $$A = \{n^{2} - m^{2} \mid m,n\in\mathbb{Z}\},$$ then $\Delta A$ contains $1 = 1^{2} - 0^{2}$. If $\Delta A$ were an ideal of $\mathbb{Z}$, then $\Delta A$ would have to equal $\mathbb{Z}$ and thus also contain $2$. However, $2$ is not a difference of two squares of integers.

Hi 👋👋

Hi

I feel that this theorem is missing the hypothesis that $V$ and $W$ are finite-dimensional over $K$. As far as I can tell, (i) and (ii) are both fine, but (iii) is false in infinite dimensions...

Hi. How would you go about solving the following problem that I made. I have a solution but I want to find the most elegant or efficient one. In both problems each letter represents a different digit. NINE - FIVE = FOUR

What are the values of ONION that are multiples of 3

2. WATER + WAVES = OCEAN and ACORN+ TERROR = COCOON

Find each digit?

is the solution even unique?

Yes I believe so

For the second question at least (there is also only one solution to number 1)

E can be almost any digit

Yes but the value of ONION is the same

And there is only one when it must be a multiple of 3

Should I also give the solutions?

Actually, I'm not interested in these kind of problems

What kinds of problems do you work on

If onion is a mupltiple of $3$, then $2N+2O+I=3k$

We can also immediately seduce R and O

*deduce

Hi

Hi

@nickbros123 I see. Thanks for checking

@Joe yes, of course

but in this case the vertices would be all these points?

context?

calculation of maximums and minimums

as I said the last time, you parametrize the curve and you have to check the function at the endpoints

A vertex of a curve is something else

@Pizza how is the study of Stokes theorem proceeding?

So on the frontier to find the maxima and minima I just have to parameterize the curves?

In this case

yes

In Folland's text when he proves the well ordering theorem (any set $X$ can be well ordered), he states that we define a partial order on $\mathcal W$, the collection of well orderings of subsets of $X$, as follows:

If $\leq_1$ and $\leq_2$ are well orderings on the subsets $E_1$ and $E_2$, then $\leq_1$ precedes $\leq_2$ in the partial ordering if

(i) $\leq_2$ extends $\leq_1$, i.e. $E_1\subset E_2$ and $\leq_1$ and $\leq_2$ agree on $E_1$, and

(ii) if $x\in E_2\setminus E_1$, then $y\leq_2 x$ for all $y\in E_1$.

I'm trying to verify that this is indeed a partial order.

Reflexivity; it holds that $(E_1,\leq_1)$ precedes itself ((ii) is vacuously satisfied).

I really struggle with antisymmetry and transitivity. How do you show $(E_1,\leq_1)$ equals $(E_2,\leq_2)$? This confuses me.

$(E_1,\leq_1)$ is not just a set, so I'm not sure what equality between $(E_1,\leq_1)$ and $(E_2,\leq_2)$ means.

The thing is, maybe I'm confusing myself by introducing the notation $(E_1,\leq_1)$. After all, Folland only writes that $\leq_1$ should precede $\leq_2$, but I feel like these well orderings are tied to a set...

I'd like to extend the proof from ysharifi.wordpress.com/2022/08/31/… from C to any field, will I have to make any major changes? The only problem seems to be that they use the fact that C is algebraically closed

hi

@ILikeMathematics You'll have trouble extending it to other fields because it's not true over other fields, in general :)

you need to assume that the field contains all the eigenvalues of the matrix you're looking at

if you want to this to hold for all matrices you can think of you need the base field to be algebraically closed

@RyderRude hi :)

@BenSteffan do you like formal logic

@RyderRude nope, not at all

really not my shtick

can we say that "a+b=b+a" is a statement independent of group theory axioms, in the Godel sense

@BenSteffan oh

@RyderRude well it doesn't even hold under the group theory axioms

@BenSteffan yeah. But it doesnt not hold either. So it may be called an unprovable statement

it's not a statement

unless you tell us what $+$ is

oh. I meant + defined by the axioms of group theory. So it obeys those four things

also you need to quantify $a$ or $b$, and tell us what they are

@BenSteffan yeah...

@BenSteffan but first order logic leaves these as simply variables

that need not belong to a set

but i need to write $\forall ab (a+b=b+a)$

yes, that's looking better :)

it's about Arrow's theorem

this theorem proves that rank-based voting systems are not good

but score based ones may still be good

this video proves the theorem

@SineoftheTime Good enough :)

@BenSteffan Yeah, I'll require the charpoly to split, then we should not need algebraic closure for the deg p_i(x) = 1 corollary anymore

And other than that, I think everything can stay the same

yeah, that should be the only thing

Thanks!

@RyderRude nice!

@RyderRude That's not really what Arrow says. What he says is that such election systems cannot satisfy all of a set of properties which determined were necessary for a "fair" election.

@BenSteffan US elections are coming up, and many folk on YouTube are American (or interested in American politics).

@XanderHenderson ah, of course

that makes perfect sense

@psie equality of pairs. So $E_1=E_2$ and $\leq_1=\leq_2$

@Jakobian ok, do you know why condition (ii) is needed at all? I feel like I can conclude reflexivity, antisymmetry and transitivity without it.

@psie if both are less or equal than each other then from $E_1\subseteq E_2$ and $E_2\subseteq E_1$ you have $E_1=E_2$. Moreover you know the well-orderings agree on this set

@Jakobian yeah, that all follows from (i)

@psie you can. But for transitivity you need to check for (ii) also. And adding (ii) just means its harder to compare two elements

Why do we need (ii)? It surely makes sense with the rest of the proof

For example suppose we have an order $\{n\in\mathbb{Z} : n\leq 0\}$ with standard Euclidean ordering

You have a trivial maximal chain of well-orders in the partial ordering satisfying only (i)

So you need (ii) to be able to say the well-orders don't shift weirdly, so that all maximal chains have an upper bound

yeah, I believe (ii) is needed for a subtle detail in showing that a chain $\mathcal C\subset\mathcal W$ has an upper bound $\bigcup \mathcal C$ that belongs to $\mathcal W$, see e.g. ProofWiki's proof, which is taken from Folland's book (but his first edition I believe, I only have the second one). Maybe this is what you're saying.

@psie it is what I'm saying. Don't you understand the example?

I'm pretty fried :) haven't had breakfast yet

anyway, time to have breakfast! thanks for helping :)

If you take $\{\{n\in\mathbb{Z}: m\leq n\leq 0\} : m\in\mathbb{Z}, m\leq 0\}$ this is a maximal chain on $\{n\in\mathbb{Z}: n\leq 0\}$ with respect to the partial order satisfying only (i) but it doesn't have an upper bound

@psie haha, I experience the same. Also my grain fries if I do too much math

@XanderHenderson yes. I'm just defining those properties as "good"

Hi

What is the term for something that is both an equalizer and coequalizer?

@SoumikMukherjee Is there a good reason to think there is such a term?

If you really need one I'd go with biequalizer

biproduct is an established term, after all

@BenSteffan just curious, as there is a term for injective+surjective

@BenSteffan oh okay, thanks

@SoumikMukherjee in what precise sense?

@nickbros123 I found a scan of the book I was searching for on Anna's archives

Why do we build infinity categories out of compactly generated Hausdorff spaces instead of plain old topological spaces? I always thought that this was so that we can get an equivalence to the construction using simplicial sets but that doesn't seem to be the case. Lurie says this is so that we can facilitate the comparison with more combinatorial approaches to homotopy theory but I don't understand what else he could have meant.

@ephe Do you mean compactly generated weakly Hausdorff spaces?

I can't comment for why anything about infinity-categories, but the standard category of topological spaces is bad in the sense that some laws don't hold for it. For example the exponential law

@Jakobian Oh yes sorry, total typo

ncatlab.org/nlab/show/exponential+law+for+spaces

they were introduced by Brown iirc

ncatlab.org/nlab/show/nice+category+of+spaces

the category of all topological spaces is bad from the standpoint of category theory

@Jakobian I am mainly confused because I had based my understanding of this fact on thinking that we couldnt get a quillen equivalence between Top and the category of simplicial sets. What kind of an issue does the lack of cartesian closure cause wrt homotopy theory?

@ephe since I'm not an algebraic topologist, I can't really answer you, but perhaps @Thorgott or @BenSteffan will

@Jakobian Thanks anyways! I feel like there would be a problem with the Map(X,Y)s but I can't really formalize it.

well, for example the exponential law says that Map(X, Map(Y, Z)) is the same as Map(X x Y, Z) so perhaps it is about that

this fails for Top, but works for your category

To address the original question, I don't think we build infinity-categories out of either CGWH spaces or plain old spaces. The $\infty$-category of spaces is either built out of Kan complexes or CW-complexes. Further generality is not really needed there.

What I meant was directly defining an $(\infty, 1)$-category to be a category enriched in CGWH

To address the second question, I believe there are Quillen equivalences $CG_{Quillen}\leftrightarrows\mathbf{Top}_{Quillen}$ and $CGWH_{Quillen}\leftrightarrows CG_{Quillen}$, so all three are Quillen equivalent to $\mathbf{sSet}_{Kan}$.

@Thorgott I was thinking something like if f,g are two morphisms from a to b and h:b->a is both an equalizer and coequalizer of the pair f,g then what would h be called. But now I see that this would imply f=g and a=b.

@ephe ah ok, then it's just that you want cartesian closure

for example, $\mathbf{Top}$ is not enriched over itself, but $CGWH$ (or any cartesian closed category) is enriched over itself

@Thorgott Sorry if this is very obvious, but I can't see how the usual cartesian closedness property guarantees this. How does that work?

If $\mathcal{C}$ is a cartesian closed category, then the functor $-\times Y\colon\mathcal{C}\rightarrow\mathcal{C}$ has a right adjoint $\mathcal{C}(Y,-)\colon\mathcal{C}\rightarrow\mathcal{C}$. For any two objects $X,Y\in\mathcal{C}$, you choose $\mathcal{C}(X,Y)$ to be the corresponding Hom-object. The unit and composition maps are then constructed using that adjunction.

If $\mathcal{C}$ is a convenient category of topological spaces like CGWH, then this Hom-object is given by the space of continuous maps together with the compactly generated topology induced by the compact-open topology.

In plain old $\mathbf{Top}$, you still have a candidate for the Hom-object, namely the space of continuous maps with the compact-open topology, but the fact that this is not generally adjoint to the product functor means the unit and composition maps you want to write down are not necessarily continuous.

The answer here also explains this and gives a counter-example for why this doesn't work in $\mathbf{Top}$

I feel like I should just put your name on the presentation at this point. Seriously, thank you so much.

no problem, I like this topic

actually, what I said about Quillen equivalences above might be slightly inaccurate, I'm not sure the obvious adjunction between CGWH and CG induces a Quillen equivalence, but it should be true in any case that these categories are all Quillen equivalent to sSet

Hi 👋

@Thorgott I'd like to clarify one thing about this equivalence. The way I have understood it is as follows: There is a Quillen equivalence between the category of CGWH-enriched categories and simplicial categories and Quillen equivalences are the correct notion of equivalence for homotopical data which is what we care about. So we say that the two constructions of infinity categories are equivalent. Or is it the equivalence between sSet and CGWH that makes us say this?

@Pizza hi

Maybe it is the latter because we just ask there to be an induced isomorphism between the homotopy categories (for weak homotopy equivalence) and the CGWH - sSet adjunction basically provides this with the geometric realization and singular complex functors

I have a question, but I think the answer is no. $\int \sqrt{2^x - 1} dx$ , I set $u = 2^x , du = \log(2) \cdot 2^x dx$ , so the integral becomes $\frac{1}{\log(2)} \cdot \int \frac{\sqrt{u-1}}{u} du$

Now

I know there is a formula for forms like $\int \frac{\sqrt{u^2 - a^2}}{u}$

In this case there is no way to apply It , i think

Otherwise I have to proceed with $v = \sqrt{u-1}$

@Jakobian Did you manage to download it?

@Pizza procede with this substitution

@SineoftheTime Yes I'm doing that, it's just that I found a site with many formulas for immediate integrals, some seem interesting

I have a book (1200+ pages) with tables for a lot of integrals, so there may be a formula for integrals of the kind $\int \frac{\sqrt{x-a}}x dx$, but it probably takes more time searching for the result intead of solving the integral

I found this site, maybe you have better ones? Check it out

studenti.it/matematica/formulari/…

This

Although honestly, only a few of these can be useful to me, also because some have formulas that are difficult to remember

I was taking a look at them, maybe they can be useful to save me time

@SoumikMukherjee yes. I have both Rings of continuous function by Aull and Normal topological spaces by Shapiro now

@SineoftheTime Maybe you can advise me which form would be best to remember?

@Pizza I don't know if it's a good idea. If your 100% sure you can solve all the integrals in the exam and you're fast enough, then I'd use this formulas, otherwise it's not a good idea

personally I know my limits and strength, so if I'm evauluating an integral and I know the strategy, I won't go on with all the steps. For example when computing line, surface and double integrals, once I find the bounds I rarely complete the whole exercise

@Pizza I only know the basic ones, like $\int f(x)dx$ with $f(x) \in \{\sin x, e^x, \cos x, \dots\}$ and one formula, i.e. $\int \frac{dx}{a^2+x^2}$ which often appears in the computation I do

@SineoftheTime but for example with green gauss, I don't have to compare if I find the same results, you say if I'm sure I did well?

you can compare, but if it's not asked I would not do it if I don't have enough time

Because its asked to evaluate the integral directly and with the Green Gauss formulas

then you have to do it

or at least this request almost always comes up

@Jakobian Do you have the membership? Can you kindly tell me how to download?

@SoumikMukherjee Never Illegally Download

@ephe I think it should be true that there is a Quillen equivalence between appropriate model structures on CGWH-enriched categories and sSet-enriched categories (note that even just constructing these model structures already is difficult), but that's a much stronger claim than just CGWH and sSet being Quillen-equivalent.

@SineoftheTime This is legal( according to my laws:P)

@Pizza yes, when doing double integrals a lot of time you have $\int_0^{\pi/2} \sin^2 x dx$ of $\int_0^{\pi/2} \cos^2 x dx$, maybe you can see if these appeared a lot of time and try to remember

@Thorgott Which one of these equivalences is the one that makes us say "ah yes these two different constructions are equivalent"

@SoumikMukherjee just go for slow download and wait a bit for the link to pop

Ah okay

or go for the extremely slow download

check the box to refresh the page

Do I have to log in or register?

no

Okay it's saying I have to wait 537 seconds before I can download

I was not expecting this much slow

yeah. But what is 600 seconds. It's not like you need the book right now

yeah

I'm fine with this

Also I never knew about this website

My go to website was libgen

libgen doesn't get updated anymore from what I know

@ephe That is, I think, a moral question. In HTT, 1.1.4. Lurie uses only a consequence of the Quillen equivalence between CGWH and sSet to motivate these theories as equivalent.

The Quillen equivalence between CGWH-enriched cats and sSet-enriched cats with appropriate model structures is not even in HTT, I think

cause I don't think Lurie actually cares that much about topological categories at the end of the day

Google points me to this paper, however, so at least the claim should be true: arxiv.org/pdf/1110.2695

@Jakobian Why? what happened to?

@SineoftheTime Yes, I have to see what comes out more often

@SoumikMukherjee I don't know, I'm not interested to know

same for sci-hub

ooh

with the VPN, libgen should work

@Pizza yes, expecially because you're not allowed to have notes

@Thorgott Oh, okay. That's actually pretty good that you didn't answer with "yes here is 500 pages of a totally new math subject that shows precisely why this equivalence shows the constructions are equivalent." And yes in HTT, Lurie is sort of like "yeah here is how you could define it but just dont". I have not seen that equivalence in HTT but again I haven't gone far with it. Thanks for the answers and the reference.

it's best to start and stick with one definition of $\infty$-categories before you start worrying about different models and what it even means for models to be equivalent (especially cause you can very easily make circular statements when you get there)

@Thorgott That is very sound advice that I wish I had thought about before writing 28 pages of my report :\

I tried to basically center the entire thing around these two constructions - their equivalence - infty cat of spaces

@SineoftheTime that's not what I was talking about.

@Thorgott it's very hard to really do that though, no?

at least if you do quasicats

since for some things you need to pass through simplicial categories

but in double integrals with polar coordinates, when does one integrate first for d$\theta$ and when first for d$r$?

@BenSteffan The way I've read it in HTT is that since the associative law holds strictly in Cat_Top this causes problems later on and you have "straighten" your morphisms to stay in the category, whatever that means. My takeaway is to always just define infinity categories using simplicial sets.

sure, you quasicats is the best model

but topological categories and simplicial categories aren't in HTT just for sport

some things are hard to do directly, so you pass through one of the other settings

like defining spaces

definitely, but there's a difference between knowing what simplicial cats are and how to use the coherent nerve functor to construct quasicats (or even its adjoint) and knowing that there is a Bergner model structure on simplicial categories for which this adjunction exhibits a Quillen equivalence to the Joyal model structure on sSet

the latter is worth having heard about, but not worth worrying about

at least I'm not yet at the point where I've worried about it :P

it ought not be too relevant to topology per se :)

there are apparently some people at the MPI here who care about stuff like that

or so I hear

but yeah

do what lurie does

@SineoftheTime Can you by any chance take a look at the question I asked above?

never mind, found it

@ephe typical math paper learning experience :')

what a headache

if the domain of integration is "rectangular", i.e. $(r,\theta) \in [r_1,r_2] \times [\theta_1,\theta_2]$, then if $f(r\cos \theta, r\sin \theta)$ is "nice enough" (for example is positive or is in $L^1$), then the order does not matter

if for example you fix $r$ and find $\theta$ as a function of $r$, the you integrate first $\theta$ then $r$

@Thorgott this definitely had to be stated as a single theorem

no way to make this any more readable :^)

I'm getting green book appendix vibes

@SineoftheTime What is $L^1$ in your notation?

@BenSteffan the corresponding theorem for models of $(\infty,2)$-categories is even longer!

@not_a_chatbot that's true about $\Delta A$ not necessarily being an ideal, but I'm talking about when it is an ideal.

@Gian'sPizzeria $f\in L^1(E)$ if $\int_E |f|<+\infty$

?

I don't understand the notation

what is not clear?

@Gian'sPizzeria $L^1$ is the space of absolutely integrable functions, up to equivalence when they're equal a.e.

that is, it consists of equivalence classes of functions $f:E\to\mathbb{R}$ (or $\mathbb{C}$) such that $\int_E |f| < \infty$, where two such functions $f_0, f_1$ are equivalent when $f_0 = f_1$ except possibly on a set of measure zero

I wasn't aware of this, thank you very much.

by the way the $1$ stands for integrable with the power of $1$, where something like $L^p(E)$ means the same thing but for functions such that $\int_E |f|^p < \infty$

👍

I don't know if studying double integrals without Fubini's theorem is a good idea

remember that the result of a double integral is a number, so if for example $r_0\le r \le g(\theta)$ and $\theta_0\le \theta \le \theta_1$, if you integrate $\int_{r_0}^{g(\theta)} \int_{\theta_0}^{\theta_1} f d\theta dr$ you get a function of $\theta$, and not a number

Fubini's theorem says that the order of integration does not change if the domain is a rectangle, right?

@Gian'sPizzeria $$\int_{R} f(x, y) \, dA = \int_{a}^{b} \left( \int_{c}^{d} f(x, y) \, dy \right) dx = \int_{c}^{d} \left( \int_{a}^{b} f(x, y) \, dx \right) dy$$

I need to look at these topics better.

@Gian'sPizzeria here you can find example and exercises along with a general explanation

I have a basic question.

> Proposition: $\operatorname{card}(X)\leq\operatorname{card}(Y)$ iff $\operatorname{card}(Y)\geq\operatorname{card}(X)$.

> Proof of $\implies$: If $f:X\to Y$ is injective, pick $x_0\in X$ and define $g:Y\to X$ by $g(y)=f^{-1}(y)$ if $y\in f(X)$, $g(y)=x_0$ otherwise. Then $g$ is surjective.

To show $g$ is a surjection, we need to show $\forall x\in X$, there's a $y\in Y$ such that $g(y)=x$. Now, if $x=x_0$, is it true that there is a $y\in Y\setminus f(X)$ such that $g(y)=x_0$ and a $y\in f(X)$ such that $g(y)=x_0$?

@psie how do you define both symbols?

@psie no

for a start $Y \setminus f(X)$ may be empty

I can see what you mean, there definitely always exists $y\in f(X)$ such that $g(y) = x$, no matter the $x$

ok

@Jakobian the first one means there's an injection from $X\to Y$, the second one means there is a surjection $Y\to X$

sometimes $Y\setminus f(X)$ is also non-empty, and it gets sent to $x_0$

ok

@psie then the proposition is false

it fails precisely when $X$ is empty and $Y$ is non-empty

ok, you have a point :) we need to assume $X$ and $Y$ are both nonempty

I suggest defining $\text{card}(Y)\geq \text{card}(X)$ as there existing a surjection $Y\to X$ or $X = \emptyset$

that equivalence definitely requires AC btw

I suggest defining $\operatorname{card}(X)$ as the cardinal number of $X$ ;)

not this direction at least

so if both $X$ and $Y$ are nonempty, the answer to my question is yes, right?

it's still no

it's still no?

the proposition is true then, but the answer to your question is still in the negative

..ah

ok, I believe my question is a 'no' for all injective functions $f$ which are also surjective, i.e. bijections. Then $Y\setminus f(X)$ is indeed empty.

yes, it's equivalent

I definitely do not suggest defining $\ge$ to mean anything but $\le$ read the other way round

I mean this is no different than defining what $\text{card}(X)\leq\text{card}(Y)$ means, it's a set up for something that should be equivalent to inequality of cardinals, if those cardinals were to be defined

@Thorgott: I agree, and I also think it is bizarre to say that the definition of $\operatorname{Card}(X)\le\operatorname{Card}{Y}$ is that there exists an injection $X\to Y$, when the natural reading of the statement is that the cardinal number of $X$ is less than or equal to the cardinal number of $Y$. Unfortunately, this seems to be rather common in some textbooks. If you want to say "there exists an injection $X\to Y$", then you can just use prose.

have you considered that not every textbook wants to set up cardinals

also note that here it's not just $\leq$ or $\geq$ that is subject to the definition, it's whole expression altogether

yeah, I think either definition works

this definition is good enough for government work, if you fix the empty set thing

@Joe the point is that this works in ZF

well it's one point

it's not hurtful as you can introduce cardinals at a later point

or never :^)

or never

I don't mind textbooks not wanting to set up cardinals, or wanting to work in ZF, but then using the notation $\operatorname{Card}(X)$ is confusing in my opinion. (Unless you are doing naive set theory and are writing $\operatorname{Card}(X)$ to denote the size of finite sets.)

you're trading a class for a distinguished representative

I'm not sure it matters usually

you just want to be able to say some set is larger/smaller/same size as another

@Joe I agree that it can be confusing, and I can certainly see a situation where someone would have a problem with this notation. Perhaps it's not always the best notation for educative purposes.

fwiw we often only make a 3-fold case distinction anyway: either the cardinality is finite, in which case you identify it with a natural number, or it's (infinite) countable, or uncountable

we often don't even assign symbols to the latter two

even if the objects aren't fully defined, you can still make sense of the whole expression and that's what's important

In ZF one can always define Card(X) to be the set of all sets of all sets that are in bijection with X and have minimal rank among sets with this property (the so called Scott's trick) to avoid having do deal with proper classes

However it is consistent with ZF that there is no class function C such that C(X) and X are in bijection and C(X) and C(Y) are in bijection iff X and Y are

@AlessandroCodenotti did you write "all sets" twice by mistake?

yes

If you add Grothendieck's universe axiom to ZFC, then isn't it possible to salvage the idea that $\operatorname{Card}(X)$ is the class of all sets in bijection with $X$? I suppose given a set $X$, if $\kappa$ is the least inaccessible cardinal number such that $X\in V_{\kappa}$, you could define $\operatorname{Card}(X)=\{Y\in V_\kappa \mid Y\simeq X\}$. (This still requires you to work with "traditional" cardinals to set up the definition, I guess.)

You don't need universes, look at the least $\alpha$ such that $V_\alpha$ contains a set in bijection with $X$ and define $\mathrm{Card}(X)=\{Y\in V_\alpha\mid Y\text{ is in bijection with }X\}$. This is exactly Scott's trick I mentioned above

Scott's trick is pretty cool

Ah, I see. If you are working in ZFC, are there any advantages to the definition of cardinals that uses Scott trick compared to the canonical representative approach?

No

(as far as I'm aware)

Thanks, that's good to know.

I guess Scott's trick is still useful in ZFC sometimes when defining equivalence relations over a proper class but still wanting to talk about the equivalence classes (for example when taking ultraproducts of the universe)

@Joe the advantage is that it's canonical I suppose

well, not of Scott's trick, the cardinal approach

the Scott's trick for ZFC is definitely worse than just defining it using cardinals

the advantage is to generalize things to ZF while losing the canonical choice of representative

Hi

Hi :)

Riley babysits and works partime at the waterpark/pool as a life guard over the summer. One week, she babysat for 3 hours and worked at the water park 10 hours and made $109. The next week she babysat 8 hours and worked at the waterpark for 12 hours and made $117. How much does Riley make per hour at each job?

If you want people to read that you should fix the formatting

hhuh?

look at your message

what about it

oh, you probably haven't turned MathJax on have you

It's a word problem

it looks like this on my end

ah, I get it

Doesn't look like that for me

no, I know

This seems really weird, the first answer which is the highest voted claims the other two are wrong, yet the other two are also reasonably high-voted math.stackexchange.com/questions/918739/…

since you haven't mathjax turned on

you see, dollar signs delimit math expressions, and your problem contains dollar signs

What is mathja

jax

the system we use to be able to type math online

oh

So who can I trust now, haha

@ILikeMathematics yourself :^)

The two lower-voted answers are made by high-reputation people

@Joey in any case I don't think anybody in here is really out to do your homework for you

cries

@Joey if you've tried a few things and got stuck, and can tell us about it, the situation might be different

Thank you

@ILikeMathematics neither high reputation nor lots of upvotes really mean anything

even people with 100k+ reputation make mistakes and write up answers that are simply false sometimes

@ILikeMathematics you can simply not blindly trust others

it's interesting that none of the answers have any downvotes

you'd expect that if an answer that got 23 upvotes claiming that the other two answers are straight up wrong at least some of the people casting the votes would have bothered to cast a downvote and/or leave a comment on the other answers

I can prove $\Delta A = A - A = \{a-b: a,b\in A\} = 2\Bbb{Z},$ for $A = \Bbb{P}_{\geq 3} \cup \{\pm 1\}$ using inverse-limit language.

In other words, ever even number is either $p - 1$ or $p-q$ for some primes $p,q$.

I advise against writing $A - A$ for that :)

Yes, confusion with set differencing

See my latest question post for the proof. There is an alternative proof as long as inverse-limits commute with $\cap$.

But I didn't use that fact in this proof.

If you replace $\Delta A$ with $A + A$, this might be a method to prove Goldbach's conjecture, but not 100% sure yet

Since if $U$ is a group of units then $U - U = U +U$.

bet ya 5 quid that it isn't

Yes, sure. No new proofs admitted in the mathematical field, unless your last name is Tao

I like your assessment on that

no it's just

it's one of the best known and most studied conjectures, it's old, and it's hard

Don't act as though the ONLY POSSIBLE proof of some prize problem is 300 pages. You'd have to prove that fact to be certain and that's much harder than just proving something with a smaller proof.

it's very difficult to say anything new about it, and these proofs don't generally come in MSE question length

@BenSteffan most studied, but not via my techniques of inverse-limit. I dare you to pull up a reference. You won't because they do'nt exist

are you telling me inverse limits are somehow new???

No, inverse-limits APPLIED TO THIS PROBLEM are new

Directly applied in a big way

The only things you'll pull up on that topic are my own posts from my post history

It's not difficult to say anything new about prize problems. It's difficult to get mathematicians to drop their defenses and actually accept anything new.

If it isn't in last year's graduate level official textbook on the matter, then it isn't true attitude

oh wow, changing name + profile picture always trips me up, but now I know again who you are

well, if you don't want to accept the bet that's fine :)

would have been a chance to earn 5 quid, if you're right

@Debug I don't think it has anything to do with "dropping defenses". It is about heuristics.

Most mathematicians are lucky to be reasonably expert in some small area of mathematics. If you try to talk to us about those areas, we are likely to be able to engage in a useful conversation.

most people who work in mathematics are not in the habit of evaluating alleged proofs of famous open problems. not for their advisors, not for their coauthors, not for their students, not for anybody. it is not like they are putting in 5-10 hours a day at that and just ignoring you

For example, I know a fair amount about geometric and spectral zeta functions, and the relation between them, particularly in more pathological spaces doubling spaces. I can have an intelligent conversation about those things. Because these are related to fractal geometry, there is a very good chance that we could talk about that, too.

But when you start getting out of my comfort zone, my inclination is to say "Who are you, and why do you think that I would know anything about this? Go talk to an actual expert."

Particularly when you start talking about famous open problems.

"if there's something to this, it will be consulted by people who are more expert than i am and come to me via some other route than personal correspondence from the author" is like the default position, even within people who work in mathematics as a day job with titles and responsibilities

So the heuristic is "Oh, this isn't my area, and it is a famous open problem. It is almost certainly wrong, and definitely not worth my time even if it is correct, since I lack the expertise to judge it."

@leslietownes Exactly this. Who are you, and how did you get my email?

@XanderHenderson who are you?

@Jakobian Yup.

Hello Yup

xander: your advisor gave it to me at the last meeting of The Conclave

@leslietownes Oh, shit. How was he doing? Mental faculties all still there? (I worry...)

hard to tell, mostly he was just handing out email addresses and offering prayers to our dark lord

I'm getting the setup of a combinatorics/probability problem: "There's an old professor handing out email addresses at a conference. How many addresses does he have to hand out before..."