hi everybody, I have a question related to a compact-supported function I just learned is only continuous: since the 1D wave equation u_xx= 1/c^2 u_tt has traveling solution like F(z) with z=x-ct... then any function F(z) is solution? Will F(z)={0, |z|>=1; z ln(z) e^(z^2/(z^2-1)), |z|<1} works as a traveling wave?

@Joako maybe in some generalized sense?

just a guess

While PDE's is totally not my field, I know they consider weak solutions and similar there

(With reference to the picture above) Can we also claim that, P(E_{i_1}E_{i_2})=P(E_{i_1})P(E_{i_2}) ?

@ThomasFinley yes, for $n = 2$

or do you mean if its equivalent

either way they forgot to say $i_1, ..., i_n$ must be distinct

@Jakobian No I meant. Can we say that $P(E_{i_1}E_{i_2})=P(E_{i_1})P(E_{i_2})$ or say, $P(E_1E_2...E_r)=P(E_1)P(E_2)...P(E_r)$ where, $E_1,...,E_r\in\{E_{i_1},...,E_{i_n}\}$ and $r\leq n$ just from the fact, that $P(E_{i_1}...E_{i_n})=P(E_{i_1})...P(E_{i_n})$ holds true?

I am not sure about the exact term for this, but when we say that a function is $e^{O(\frac{1}{q^m})}$ do we not mean that it is bounded above by some $e^{c\cdot\frac{1}{q^m}}$? And so such a function would converge to 1?

@ThomasFinley If $E_1, ..., E_r$ are independent then $E_1, ..., E_{r-1}$ and $E_r^c$ are independent. So $P(E_1...E_{r-1}) = P(E_1...E_r)+P(E_1...E_r^c) = P(E_1)...P(E_r)+P(E_1)...P(E_r^c) = P(E_1)...P(E_{r-1})$ and so on

that is you can use induction to eliminate $E_{i_k}$ for any $k$ and the equality will still hold

so, if I interpreted your question correctly, from $P(E_{i_1}...E_{i_n}) = P(E_{i_1})...P(E_{i_n})$ for some fixed $i_1, ..., i_k$ it follows that for any distinct $E_1, ..., E_r\in \{E_{i_1}, ..., E_{i_k}\}$ the same equality holds

@Jakobian ok

@ephe I would understand it as a function of the form $e^{f(q)}$ where $f(q) = O(1/q^m)$

I don't know what you mean by converge to $1$.

@Jakobian This means, that we say that $P(E_{i_1}E_{i_2})=P(E_{i_1})P(E_{i_2})$ or say, $P(E_1E_2...E_r)=P(E_1)P(E_2)...P(E_r)$ where, $E_1,...,E_r\in\{E_{i_1},...,E_{i_n}\}$ and $r\leq n$ just from the fact, that $P(E_{i_1}...E_{i_n})=P(E_{i_1})...P(E_{i_n})$ holds. Did I get you correctly?

@ThomasFinley you have copy-pasted what you wrote above

but, yes, I was answering your question in an affirmative

if that wasn't clear from what I wrote

@Jakobian Ok, I get it. Thanks for the help!

@Jakobian Sorry I think I should have been more clear. $q$ is the constant here not $m$. I meant as $m\to\infty$ an $O(1/q^m)$ function would converge to $0$ and so a function of the form $e^{O(\frac{1}{q^m})}$ would converge to $1$.

Then a function of the form $e^{f(m)}$ where $f(m) = O(1/q^m)$.

If $q > 1$ then of course it will converge to $1$ then

This is the context for my question. If what I say is true then that probability is close to 0 for large $m$.

as for if it means the same as being bounded by $e^{C/q^m}$, I'm not so sure

this definitely depends on what exactly do we mean by the symbol $O(...)$, if we mean it generally or for non-negative functions

@Jakobian I really hoped you'd say no

but we can certainly conclude it implies that it is bounded by $e^{C/q^m}$

@Jakobian Here it's meant as the time complexity so the absolute value of the function $f(m)$ would eventually be bounded by a constant multiple of $\frac{1}{q^m}$.

@ephe no actually I was still thinking more generally about real functions. But given this is about probabilities it does mean a function $f$ such that $|f(m)|\leq C\cdot 1/q^m$ and the function needs to be negative pretty much

because there can't be negative probability

No sorry if we were talking about non-negative functions then $0\leq f(m)\leq C\cdot 1/q^m$ so it doesn't change anything

I was hoping to miss something pretty stupid :/

I think that's the opposite of what the theorem should say. The entire point is to show that we can attain that code with a high probability using those linearly independent elements.

The proof only shows that the probability is greater than $(1-\frac{1}{q^m})^{\frac{q^m}{q^m-1}}$ which does converge to $1$ (I hope).

it does

from my understanding they've shown that $f(m)\leq C\cdot 1/q^m$ for some $C < 0$ or something

you know this is the confusion that follows because people use $o$ and $O$ notation

that's why I don't like it

maybe it would be better if they wrote something like $P(...) = 1-e^{-O(1/q^m)}$

@Jakobian Honestly, I don't remember a time when I wasn't confused about the Big-Oh notation

@Jakobian but such a function would still converge to 0, right?

$1-e^{-\omega(1/q^m)}$ I suppose

Or it seems to me like $e^{-O(\frac{1}{q^m})}$ would also do the job on its own

its already between 0 and 1

$\omega(x,y) = \left(xy - \frac{1}{x}\right) dx + x^2 dy$

@SineoftheTime But since the set is not simply connected, to calculate a primitive do I have to divide the set into simply connected components?

To calculate a primitive, must the form exact? If I'm not mistaken, it was enough for me to know that there were components that were simply connected even without specifying which ones.

Right ? Or im wrong?

@Pizza do you need simply connectedness to talk about a primitive?

@Pizza yes, you talk about primitive when the form is exact

which is the same as closed, on a simply connected domain

@SineoftheTime Yes but it is correct that in this case $D = \{(x,y) \in \Bbb R^2 : x \neq 0\}$

So the set is not simply connected

it's not even connected

So the partial derivatives match

so it's closed

Yes

the problem when the domain of the differential form is not connected is that it's no longer true that 2 primitives differ by a constant

what are you required to do in the exercise?

I calculated the primitive and found $x^2y - \ln|x| + k$

But then I had this doubt

@SineoftheTime Study the differential form

So theoretically domain, closure and exact, calculate a primitive

it's a bit tricky, but if it's required to find only one primitive you don't have problems. Just take $k=0$

in fact, $\int \frac 1x dx=\log |x|+c$ is imprecise

No, but I mean, can I calculate it even if the form is not exact?

how would you do that?

Shouldn't the domain be simply connected?

a form is said exact if it's the differential of a scalar function, which is called primitive

@Pizza no.

Hi, guys. Can anyone confirm this? In the Lorentzian space $\mathbb{R}^{2,1}$, a vector $v$ is called spacelike if $\langle v,v\rangle <0$, timelike if $\langle v,v\rangle >0$ and lightlike if $\langle v,v\rangle =0$ (?)

you need simply connectedness to say exact $\iff$ closed

Where $\langle \cdot,\cdot \rangle$ is the pseudometric, of course.

@SineoftheTime But are there cases where it is not possible to calculate a primitive?

to talk about a primitive, you work with open sets

@Pizza when the form is not exact

@SineoftheTime I don't understand, so why could I do it here?

@SineoftheTime In my case this is not valid

you can work on the two connected components

what definition of primitive did you see?

@SineoftheTime So $x > 0$ , $x<0$

@SineoftheTime $dF = \omega$

yes, but what are the domains of F and omega?

open sets or open connected sets?

@SineoftheTime $\Omega = \Bbb R^2 : \{x=0\}$

I mean, in the definition not in this case

an open, but unconnected set

so in your definition of diff form, the domain is a generic open set

no hypothesis on connectedness

To ensure that a closed form admits a primitive, the domain must be open and connected (or simply connected to ensure global uniqueness of the primitive)?

what do you mean by uniqueness of the primitive?

it is unique up to an additive constant

So in this case for $x > 0$ , $x^2y - \ln(x)$, for $x<0$ , $x^2y - \ln(-x)$

if you have two primitives and the domain is no longer connected, like in your case, you can't say that two primitives differ by a constant

@SineoftheTime ah ok

@Pizza this is one primitive, so if you're asked to find just one let the constants equal to 0

@SineoftheTime aren't they 2 primitives?

so $F=x^2y-\log |x|$ is defined on $\Bbb R^2 \setminus \{ x=0\}$ and $dF=\omega$

@Pizza no?

@SineoftheTime So I found a primitive that is valid for x>0 and x<0

Yes, since the primitive have the same domain of the diff form

Oh ok, I'll do some more exercises to see if I understand better

this is a pleasant to read

@Pizza I don't think it's that important in your case tho, since you can work on the connected components

Oh, I guess I was wrong. Spacelike vectors are the ones with $\langle v,v\rangle>0$ instead of timelike, right? That makes more sense.

but calculating the work or circulation of a differential form is the same thing, i.e. do I have to do the curvilinear integral of the differential form along the given curve?

@Gian'sPizzeria $\int_{\gamma} \omega=\int_a^b \sum_i \omega_i(\gamma(t)) \gamma_i'(t)dt=\int_a^b F(\gamma(t))\cdot \gamma'(t)dt$ where $F(x)=(\omega_1(x),\dots, \omega_n(x))$

Dooes anyone know learning material for segmenting nurbses to B-splines? I would like to learn how to make a Python program to do the job but I'm not sure what to do if knot vector has wrong length.

Can one help me in my question $222^{333}+333^222$ when divided by 2,3,7 the remainder will be? Here are my effort I know I can solve it using modular arithmetic so I rewrote the equation as $11^{333}.2^{333}+11^{222}.3^{222}$

From here it is simple I could re write them as $ (9+2) ^{333}.2^{333}+(9+2). 3^{222}$

One way to start is to note that if $a>n$, then $a^m\pmod n\equiv ((a\pmod n)^m\pmod n)$.

@आर्यभट्ट did you mean $333^{222}$

Yes

can you tell what the remainder would be mod 2?

is the number you are deaing with, even or odd?

It could be 1

it has to be 1

since the first term is even, the next odd, so the sum is odd, so it leaves remainder 1

checking by 3 is also trivial

recall the common divisibility rule for 3

It is same 1

no, 0

222 is divisible by 3

and so is 333

Oh shit yes 0

Can u tell me is it divisible 37

what do you think?

hint: $37\mid 111$

I could not read syntax here because I m on phone

37 divides 111

@आर्यभट्ट so the answer is?

Yes

indeed

Remain 0

checking disiviblity by 7 might be a little more difficult

One thing how u predict something so divisible or not? Ima very bad at number theory

honestly, just comes with experience

checking divisibility by 2 is very trivial

Hey wait one minute I think my first question if incomplete

$222^{333}+333^222$

This was the original question

It is not divisible by 2 and 3

222^{333} mod2+333^222 mod 2$

I don't understand

@आर्यभट्ट This IS divisible by 3, as we discussed

and leaves remainder 1 on being divided 2, as discussed

I tried to solve it using modular arithmetic

You want to check divisibility by 2, 3, and 7 right?

and also, your mathjax is slightly off point, you are not using the {} on the exponent with 333

Oops sorry for that it been a long time since I used the site

@robjohn can solve it using modular arithmetic?

For $\bmod 2$, you can note $222\equiv 0$ and $333\equiv 1$. Can you then conclude your expression mod 2?

Similarly for $\bmod 3$, it is obvious that $333\equiv 0$ and $222\equiv 0$, and thus conclude

$\bmod 7$ it appears to me you need to use fermat's little theorem; although it could be simpler noting that exponentiation is cyclic mod n etc etc

Got it 👍

@आर्यभट्ट with addition to those comments, you should look at Fermat's little theorem as well as its generalizations Euler's totient theorem and Carmichael's totient theorem

(in increasing generality)

Verify $sin(45^{\circ})^2-cos(45^{\circ})^2-tan(45^{\circ})^2=\frac{2sin(45^{\circ}^2-2sin(45^{\circ})^4-1}{1-sin(45^{\circ})^2}$ for $x=45^{\circ}$. You can use a calculator & solve it very easily but how would you go about solving it without one?

$sin(45)^2-cos(45)^2-tan(45)^2$=$\frac{2sin(45)^2-2sin(45)^4-1}{1-sin(45)^2}

@VulpesInculta your LaTeX is broken and unreadable

maybe now's a good point to point out that you should write \sin instead of sin and similar for the other trig functions, too

@SineoftheTime so are they the same thing?

@VulpesInculta sin, cos, and tan of 45° are very simple. You should know them by heart!

me when i try writing \cosec :((

@Derso Both conventions for the sign of the spacetime interval are in use. But that's probably a better discussion for the Physics chat room chat.stackexchange.com/rooms/71/the-h-bar

@PM2Ring I remember them, should I convert them all to their values & try solving then?

@VulpesInculta Yes.

some more note on your LaTeX: you don't need to write ^{\circ} you can write ^\circ

(but it also doesn't hurt to write the curlies)

@Jakobian don't tell Don though :)

also, you do need to write the degrees in something like $\sin(45^\circ)$. There is a difference between $\sin(45)$ and $\sin(45^\circ)$

@Jakobian okay thank you I'll take note of that

I can't be bothered writing ^\circ on the phone, especially since ° is so easy to type, and looks nicer. ;)

maybe the world would be a better place if instead of $\sin$ we used $\sin^*(x) = \sin(2\pi x)$

and of course, no degrees either

No one wants to use $e^{i t}$ either, you want to use $e(t) = e^{2\pi i t}$ and forget about the connection to $\pi$

@PM2Ring I'm on my laptop & don't know how to do that lol

so $sin^2(45^\circ)=1/2$ , $cos^2(45^\circ)=1/2$, & $tan^2(45^\circ)=1$? does this seem right?

Yes

like Ben said above, you should use \sin, \cos and \tan instead of sin, cos and tan

@Jakobian Yes, cycle notation is good.

@Jakobian sorry I didn't notice that message, I'll do that from now on.

The world would also be a better place if - was not symmetric, and if the obelus was eliminated from the collective memory of humankind.

I'll write out the whole question in a moment here hopefully my LaTeX is alright for this one

@anak are you trying to sell us that $-$ is symmetric?

@BenSteffan Reflect it across the y-axis.

oh, as a symbol

gotcha

I am referring to the symbol itself, not the operation it stands for.

Symmetric symbols should not be used for non-commutative operations.

$\sin^2(45^\circ)-\cos^2(45^\circ)-\tan^2(45^\circ)$=$\frac{2\sin^2(45^\circ)-2\sin^4(45^\circ)-1}{1-\sin^2(45^\circ)}$

anti-symmetric is almost as good as symmetric :^)

@BenSteffan Within $\epsilon$.

within algebraic topology

okay that's the correct question now. So I should use the trig values & sub them in, then I can easily evaluate whether they are equal or not

@BenSteffan I have not seen you around these parts before (not that you have seen me around here before, or that I even frequent here much). What math are you a fan of?

algebraic topology, (stable) homotopy theory in particular :)

and you?

I do geometry stuff, but I am not picky.

I see, that's nice :)

If it features an infinite dimensional Lie group, I am probably game.

I like a good lie group, but only as a topological object :P

I do wish I knew more geometry

Well you sound like you are interested in things that are geometry-adjacent so you will pick it up passively anyway.

maybe

although stable homotopy theory is not that close to geometry

what with being very high up in the air in terms of abstraction

Out of curiosity I was just browsing the nlab page on stable homotopy theory and they have a note "The reduction of geometric phenomena to solvable problems in stable homotopy theory has remained an important mathematical theme, the most recent major success being Stolz’s use of Spin cobordism to study the classication of manifolds with positive scalar curvature."

I am familiar with Stolz's paper, but I don't remember it being relevant to stable homotopy theory...

it's the other way around

you use stable homotopy theory to study geometric things

cobordism being perhaps the most notable instance of this phenomenon

in essence, computing cobordism groups turns out to be a problem of stable homotopy theory after a transformation, and surprisingly this problem is generally feasible (which is somewhat rare for problems like this in homotopy theory)

you identify the cobordism groups with the homotopy groups of a certain spectrum, or "stable homotopy type"

@Jakobian what does the * mean?

oh wait

Okay, so here they are just saying because it was with the spin cobordism, it's stable homotopy theory.

Interesting... do we have any good examples of people who purposefully used the language of stable homotopy theory to solve a geometric problem like Stolz?

@anak maybe. I don't know Stolz's paper, but this sounds to me like they imply that Stolz used some fact about spin cobordism derived from the AT study of spin cobordism, like knowledge of the respective cobordism groups

@anak I really don't know too much about the geometric side and am probably the wrong person to ask, but Rokhlin's theorem comes to mind (sort of)

@Sahaj I just wanted to distinguish my new sine from the old one

Rokhlin's proof of his theorem makes use of the fact that the third stable homotopy groups of sphere is $\mathbb{Z}/24$

I will have to check that out. Sounds more explicitly homotopy theory than Stolz's paper.

how this fact translates into a geometric statement is closely related to the way the cobordism correspondence I mentioned above works

Stolz's paper is a great geometric read though, maybe you can read it at some point and identify the more general theory being used.

but in fact this was early in the history of AT (40s I think), so there was absolutely nothing for tools to tackle that computational problem, except to translate it into the geometric side and use geometry

so in effect Rokhlin did not end up doing much homotopy theory per se, even though we would classify the homotopy group input to the proof as homotopy theory

it's a very impressive feat of strength, and pretty much exhausts the method

it is prohibitively difficult to use this geometric method to compute any stable homotopy groups of spheres beyond that, and we didn't push beyond $\pi_3 \mathbb{S}$ until the advent of the Serre spectral sequence a few years later

You sound like you'd be a pal to Balarka.

who's Balarka?

A guy who used to (?) frequent here. Prodigious in all things topological and geometric.

Don't know 'em

but I haven't been around this chat for that long

Chat him up if he ever pops in. Anyway, I gotta go. Nice chatting. :D

Nice talking to ya

see you around :)

If you're verifying a trig equation "algebraically" is it still appropriate to use identities such as $\frac{\sin(x)}{1+\cos(x)}$=$\tan(\frac{x}{2})$ in your verification?

that's a question to your teacher

@Jakobian okay thank you

@Jakobian where did you learn math, did you do a degree or learn it on your own?

yes

That seems to be a common answer here. What resources so you mostly use when learning on your own?

its just that you asked me this twice

@VulpesInculta various books and math.se

rarely wikipedia

Sorry I can't remember who have asked lol

$\iiint_T z \ dxdydz , T = \{(x,y,z) \in R^3 : x^2+y^2 \leq z^2 , x^2+y^2+z^2 \leq 2y , z \geq 0\}$

I also use articles and monographs

I don't really use notes often

any good monographs for someone around my skill level, precalculus with perhaps introductory calculus?

I'm having trouble finding the extremes of integration, I'm trying with $x = r \cos(\theta) , y = r \sin(\theta), z = z, dV = r dr d\theta dz$

@VulpesInculta I honestly have no idea what to recommend to someone on a precalculus level

but you might check out those notes tutorial.math.lamar.edu

@Jakobian I'll bookmark this, seems very helpful. Thank you.

@BenSteffan I was once shown a spectral sequence proof along this line, but I never grokked it

for finding the possible values of $\sin(\theta)$ where $0^\circ

sorry ignore that

@Thorgott of $\pi_3 \mathbb{S} \cong \mathbb{Z} / 24$?

from a modern perspective doing that calculation with the Serre specseq is already masochistic

no, I mean of Rokhlin

$3$rd stable homotopy group via Serre SS is OK

I've never bothered to try and understand a proof of that theorem

I looked at what I think would have been close to Rokhlin's proof and it's all things like Seifert surfaces

@Thorgott is it really?

@BenSteffan yeah, that's one of the geometric arguments

(not that I understand that either)

so you're saying there's an AT argument?

@BenSteffan it's in Serre's book, I don't remember it being that bad

in Serre's book?

err

thesis?

Hatcher, dunno why I said Serre

ah ok, that looks relatively fine

think he even computes up to like $\pi_6(S^3)$

though he sadly doesn't identify the generator of the $3$-torsion factor (the answer might surprise you)

@BenSteffan yeah

if you have the ASS and the vanishing line result (I think) you get it for free :^)

I do have the ASS, thank you for asking

sorry I'll see myself in a corner

@BenSteffan get which one for free?

@Thorgott which 3-torsion factor?

@Thorgott all of the first 12 stable stems

not $\pi_6(S^3)$ of course :)

@BenSteffan the group is Z_3 x Z_4

@BenSteffan I suppose, I shamefully never learned the ASS

@Thorgott it's 4 times a generator of the group :)

and Hatcher gives a description of that

@Thorgott you could probably do it in an afternoon, at least setup and first computations

the construction is surprisingly simple

Does the period of $+360^\circ k, k \in \mathbb{Z}$ always give all possible solutions when $\theta$ is between $0^\circ$ & $360^\circ$?

@BenSteffan oh, he does give the cool description, he just doesn't prove it

@BenSteffan probably, just need to actually do it at some point

@Thorgott how do you show it?

it's somehow a consequence of the surjectivity of the J-homomorphism or something

oh, ok

slightly underwhelming

but I guess it makes perfect sense that J should be involved with that map

while you're here, you wouldn't happen to know how to make sense of HTT2.4.1.9?

oh, I guess I was reading that remark wrong

why does Lurie use "final" to mean "final in the homotopy category"

ugh

> Theorem. Let $(X, \mathcal{M}, \mu)$ be a measure space and let's define $\mathcal{N}=\{N \in \mathcal{M}: \mu(N)=0\}$, the set of null-sets, then $\overline{\mathcal{M}}:=\{E \cup F \mid E \in \mathcal{M}, F \subset N \in \mathcal{N}\}$ is a $\sigma$-algebra and there exits a unique complete measure $\overline{\mu}$ on $\overline{\mathcal{M}}$ extending $\mu$.

I want to show that $\overline{\mu}$ is the unique (complete) extension of $\mu$. I'm reading a proof about this in some notes based on Folland's, but have trouble understanding why we require two sets to be disjoint.

> Proof. Suppose $\nu$ is another complete measure on $\overline{\mathcal{M}}$ extending $\mu$. Let $\overline{E} \in \overline{\mathcal{M}}$. Then $\overline{E}=E \cup F \subset E \cup N$ where $N$ is a $\mu$-null set such that $F \subset N$. We may choose $N$ disjoint from $E$. From this, we have$$\mu(E)=\nu(E) \leq \nu(\overline{E}) \leq \nu(E)+\nu(N)=\mu(E)+\mu(N)=\mu(E).$$ Hence, $\nu(\overline{E})=\mu(E)$. But, $\overline{\mu}(\overline{E})=\mu(E)$, so that $\nu=\overline{\mu}$.

I guess we can choose $E$ and $N$ disjoint since we could just replace $F$ with $F\setminus E$ and $N$ with $N\setminus E$ (those are still null sets and $N\setminus E$ is still a member of $\mathcal M$), but where is this disjointness used in the proof?

@BenSteffan I don't think he does

this should just mean "final" in the $\infty$-categorical sense

ah, damn

@psie nowhere, why are you trusting random notes you found online

I glanced over definition 1.2.12.1 too quickly

then I could use some help haha

oh, he phrases it weirdly

@Jakobian ok, well I believe the notes are actually from someone affiliated with a university, at least I found them from a domain from that university, but anyway, thanks for checking

ah, it's cause he wants to talk about finality in simplicial sets that aren't necessarily $\infty$-categories and then the definitions diverge

yeah

@psie Honestly it doesn't matter. You shouldn't trust any kind of proof word for word

but yeah, the lifting problem defining the strong finality for this object curries to a lifting problem that is solvable by the definition of $p$-cartesian edges

@Jakobian other than the disjointness assumption, which I guess we can just safely discard, the proof is alright, right?

@Thorgott ah, ok, so I should work with lifting diagrams instead of working with the trivial kan fibration that defines cartesian edges

@psie you should be answering those questions yourself

1. It will help you develop proof-checking skills. 2. It will benefit your proof-writing skills

yeah, it's all just currying adjunctions

@Jakobian isn't the fact that $E$ and $N$ are disjoint used in going from $ \nu(\overline{E}) \leq \nu(E)+\nu(N)$? We have $\overline{E}\subset E \cup N$, so by monotonicity $\nu(\overline{E}) \leq \nu(E\cup N)=\nu(E)+\nu(N)$, where the equality is due to $E$ and $N$ being disjoint.

@psie no

and you should answer why

the relevant Remark in HTT is 2.4.1.4.

@Thorgott yeah, I figured :)

I'm just really lacking the dexterity with these things right now

or still, rather

in particular the whole slice category shtick

yeah, it took me a lot of grinding to get comfortable with that

If the answer requires you to stare at a piece of paper for an hour then so be it. I am willing to take that sacrifice

its a joke

"an hour" you're being very optimistic :))

I meant psie and why the disjoint assumption is not actually used anywhere

ah

well it does apply in my case too so :)

@Jakobian If you're waiting around for an hour I have a question. Given a value $\sin(\theta)=-0.4321$ could the number of possible solutions be stated to be 2 based solely off the fact that a negative sin value will lie in quadrant 3 & 4? meaning a total number of 2 solutions in quadrant 3 & 4 respectively?

@VulpesInculta I'm not really waiting I am wondering about my own questions right now

someone else can answer you

@Jakobian okay no worries

@Thorgott Rezk's notes are actually pretty decent here

in terms of explaining this sort of argument

@VulpesInculta sine is periodic. So, yes, you have solutions in the third and fourth quadrant, but more than two.

I am wondering what an example of two Hausdorff uniformities $\mathcal{U}_1, \mathcal{U}_2$ in terms of entourages on a topological space would look like such that $\mathcal{U}_1\cap \mathcal{U}_2$ is not an uniformity.

I'm drawing blanks on this

@BenSteffan yeah, I generally like them, especially for stuff like this or saturated class arguments

this isn't a problem with uniformities defined using pseudometrics, any intersection of uniformities using pseudometrics is again a uniformity

they sure came to the rezkue here :))))

entourages break things somehow

@BenSteffan Boo!

ahhhhhh, a ghost

You're lucky that I'm not allowed to kick you for making bad puns!

not so powerful now are you >:)

@BenSteffan Just you wait. Eventually, I will likely want to hang up my diamond. On that day, there'll be a reckoning!

emphasis on eventually

I've browsed Bourbaki for this but they only give uniformities on $X$ given by partitions of $X$ as examples, which are not Hausdorff

useless

@BenSteffan Don't worry. I have list.