12:45 AM
@leslietownes used functional analysis for a topology question :-)
1:02 AM
hello I was thinking about $M=\Bbb R \times \Bbb C - \lbrace 1~\mathrm{point} \rbrace$
@zetaspace I was thinking about going outside to get the mail.
@leslietownes Believes in nothing!
I was thinking about $M$ and I thought the following thought: What do we know about $M$?
Well, it has two pointy bits at the top, and a pointy bit at the bottom.
And some sexy serifs if you are using the right font.
Yes Xander, you are right actually
Making a sketch may help!
1:14 AM
It's totally obvious that we have a biholomorphism from $M$ to a complex cylinder
correction: not just one complex cylinder, many of them
Infinitely many?
yeah cylinders within cylinders
@XanderHenderson I wasn't thinking
@zetaspace that its homotopic to $S^2$?
How is that @Jakobian ?
because $S^2$ is homeomorphic to a deformation retract of $M$
1:27 AM
I don't see what happened with the punctures. $S^2$ is closed (compact w/out boundary)
punctures? There is just one
yes but the punctures extend across the real line
what do you mean by that
I mean the best way I can describe it is infinitely many copies of $\Bbb C - \lbrace 1~\mathrm {point}\rbrace$ stacked on top of one another
1:42 AM
so you mean $\mathbb{R}\times (\mathbb{C}\setminus\{0\})$?
brackets are important you know...
in this case it's homotopic to $S^1$
It's also homeomorphic to $\mathbb{R}^2\times S^1$
2:13 AM
when you solve for spherical harmonics using separation of variables in spherical coordinates, why do we end up gettinng only a solution that seems to "orient" around the z axis? of course since you have spherical symmetry these could be rotated around the sphere in any manner but the m=0 harmonics specifically seem to be around the z axis. is it the separation of variables assumption (along with the specific choice of coordinates) that ends up locking you into this one specific orientation?
like, does assuming the spherical harmonic breaks down into indepndent functions of r, theta, and phi lock you into this orientation?

3 hours later…
5:10 AM
Given a square, Alice chooses a random point on its boundary and Bob chooses a random point in its interior. The probability that Alice's point is closer to the center than Bob's is, is (1-ln2)/3 ≈ 0.102284273147.
What happens for a cube rather than a square? (I do not know the answer, other than that it should be larger)
PS, for a triangle it's (3-2ln2)/9 ≈ 0.179300626542.
For a pentagon it involves the logarithm of something golden ratio-y.
(Puzzle: Find the exact value for a regular n-gon. This is doable with pure pencil and paper and not too much pain if you do it cleverly, but there's no shame in using Wolfram Alpha to bash integrals.)
5:24 AM
hi akiva
Hi Allie
do you remember me
Not sure...
meow mix
last time i remember speaking with you i thin k iwas an insufferable teenager
Who among us
5:27 AM
true
6:02 AM
wb
@Jakobian I've just started off in either topic, but all the time spent studying Euler totient function n its properties, fermats little theorem etc, We also see these things come again through Lagrange or Euler theorem. I haven't studied galois or ring theory yet

2 hours later…
8:02 AM
I am at the point where I am trying to define "=", so pretty high confidence! :)

3 hours later…
11:13 AM
(=_=)
Just er... a small question, but if $A$ is a commutative ring, and $S\subseteq A$ is multiplicatively closed, then how would you pronounce $S^{-1}A$? (In [AM], it's called the "ring of fractions of $A$ with respect to $S$", but presumably people have a shorter way of pronouncing it.)
11:49 AM
@psie that's where I draw the line (or two, one over the other)
12:37 PM
Hey @AkivaWeinberger long time no see! What are you up nowadays?
12:59 PM
I think I will just surrender to the idea that by definition equality is reflexive, symmetric and transitive (and hence an equivalence relation). After many internal negotiations, I think this is the agreement I have reached.
@psie Equality is a pretty primitive notion in mathematics. Indeed, thinking of it as an equivalence relation (which it is, but...) is kind of backwards. Equality is the ur-equivalence relation---all other equivalence relations are just a shadow of equality.
that's a nice way of thinking about it too
But unless you are actually studying mathematical foundations, there is really no need to think too hard about equality. It means what you think it means.
At any level of mathematics, you eventually have to accept that what you are doing is built on a foundation which you just don't have the time to fully explore (again, unless you are really planning on focusing your work on those foundations). There are certain things which you just have to treat as axiomatic.
ok, good idea. I was thinking, maybe somehow one could define equality through set inclusion; you could say two objects are equal if they are members of the same singleton sets. But then the issue arises; how do you determine a set is a singleton?
But I will stop. That was just a thought.
1:15 PM
@psie What is a "singleton set"? It has been a long time since I looked at basic set theory (and even then, I've never gotten really deep into the nitty-gritty), but "counting" comes along much later than basic notions like set inclusion.
At that level, everything is a set, and $A = B$ means that $x \in A \iff x \in B$ (the two sets contain exactly the same elements).
1:28 PM
right, hmm. By singleton set I was simply thinking of a set with a single element. $A=B$ in set theory would be equivalent to $A,B$ containing the same elements, but also be contained in the same sets. Logically, $$A=B\iff \forall C((C\in A\iff C\in B)\wedge (A\in C\iff B\in C)),$$ $\forall A\forall B$.
That makes no sense.
:)
1:41 PM
The properties of equality are hardcoded into first-order logic, they happen at a lower level than anything set theory says (this is kind of a modern convention, you will see older texts talking about "first-order logic with equality" since it used to be treated as a binary relation symbol in older books)
Among the properties of equality that are just part of first-order logic is symmetry, the fact that for all formulas $\varphi(x_1,\ldots,x_n)$ if $a_1=b_1,\ldots,a_n=b_n$ and $\varphi(a_1,\ldots,a_n)$ holds, then $\varphi(b_1,\ldots,b_n)$ holds etc.
In ZFC there is an axiom (extensionality) stating that two sets are equal iff they contain the same elements, because that makes life easier
Extensionality is not a property of equality though, it is a property of ZFC. For example in ZFC we have that $V\setminus V_\omega$ is a model of all ZFC axioms except extensionality
interesting 👍
1:57 PM
Guys, could I ask about the following statement in the wikipedia article $Under any reflection or rotation ρ, the delta function is invariant$. Is there a "not too technical" proof of this statement?
To be honest, I'm more interested in how to deal with something like $$\delta[\mathcal{R}^{-1}\mathbf{r}-\mathbf{r_0}]$$ where $\mathcal{R}$ is a rotation
@Joe S inverse A
If you are developing axiomatic set theory within the framework of first-order logic (with equality), then equality is a primitive notion, and so $A=B$ cannot be treated as an abbreviation of $\forall x(x\in A\iff x\in B)$. See here.
On the other hand, it is of course true that $A=B$ iff $\forall x(x\in A\iff x\in B)$. The forward direction follows from the logical axioms of first-order logic, not from the axioms of ZFC; the backwards direction comes from the axiom of extensionality.
@SoumikMukherjee: Thank you.
2:20 PM
interesting, @Joe
@psie so the way I defined it here is a way of defining it if one does not have equality as a primitive notion, i.e. FOL without equality
2:44 PM
Why do you not want to take it as a primative notion?
well, I'll accept it as a primitive notion as well :) just curious
ok
Just stay away from using it in stuff like 1/0 = ∞ and ∞ = ∞.
good idea
If a is a real number, then a = a.
💯% ✅
Btw, inequality symbols are also taken as primative notions.
3:02 PM
Hello
<, >, ≠ and ≈
hi
≤, ≥
I wonder if anyone has done work on spaces that parametrize isomorphism classes of zeta functions. i.e. moving in different regions of the space you get zeta functions all isomorphic but tuned differently
3:30 PM
@Joe localization of A with respect to S
@user20458579510081670432 are they
3:53 PM
Yes.
@user20458579510081670432: The sense in which the inequality symbols are primitive is quite different to the sense in which equality is primitive in set theory. Set theory doesn't even have a symbol for inequalities: the symbol $\le$ would be a defined notion that is added to the language of ZFC.
The only sense in which inequalities are primitive is that some people simply state the axioms of the real numbers (i.e. that of a complete ordered field) without worrying too much about what the real numbers actually "are". If they do that, rather than, say, defining $\mathbb R$ using Dedekind cuts, then $\le$ is primitive in the sense that you are never thinking about what set-theoretic object it is – you only care about its properties, specifically how it behaves as a relation on $\mathbb R$.
For context: $L$ is a vector space over the field $K$, and $a_i \in K, l_i \in L$ for each $i$. I would like to ask, in your opinion, what the author means when writing "is uniquely defined".

My guess is that, since the author defined a product $(a,l) \mapsto al$ as a function $K \times L \to L$ (the usual product of a vector for a scalar element of the field), each $a_i l_i$ has a unique value (because of the definition of function) and so $\sum_i a_i l_i$ as well has a unique value for the same reason. Is this correct?
@ZaWarudo that no matter in which way you do the operations, the result is the same
@ZaWarudo: This is a slightly awkward and imprecise way of saying that the symbol $\sum a_il_i$ makes sense, because of associativity of addition. If $+$ is an operation on a set $S$ which is not associative, then it's unclear what something like $\sum_{i=1}^3 c_i$ would mean, since it could be either $(c_1+c_2)+c_3=c_1+(c_2+c_3)$, and those are not in general equal.
@ZaWarudo In a vector space expressions like $u+v+w$ are "uniquely defined". A priori, we "can't" sum three vectors. As you mentioned for the scalar product, the sum is an operation that takes TWO vectors and gives us a new one
4:01 PM
it's basically the "$x(yz) = (xy)z$ implies that $x_1\cdot ...\cdot x_n$ is the same no matter how we put the brackets in"
But we can do $(u+v)$ first and then add $w$. Or also $(v+w)$, and then add $u$ with this one. All of them gives us the same final result
3

Consider the following o.d.e. system on $p,v:I\to \mathbb{R}^3$ given by \begin{align*} p' &= (\cos (at+b))(p\times v)\\ v' &=(\sin (at+b)) (p\times v) \end{align*} with orthonormal initial conditions, i.e. $p(0)=p_0,v(0)=v_0\in \mathbb{S}^2$ with $\langle p_0,v_0\rangle =0$, and $a,b\in \mathbb{... where$\cdot$is any binary operation, in particular it can be addition of vectors Thanks to everyone. I was confused because "uniquely defined": so it doesn't mean that it has a unique output (that is, not a multivalued function) but it means what you all said. in some sense it does mean that it doesn't have a multi-valued output @ZaWarudo: Yeah, I wouldn't have used the phrase "uniquely defined" in that context. It doesn't sound quite right to me. Just saying "unambiguous" would have been better, in my opinion. 4:11 PM Much better indeed I wouldn't say uniquely defined either. That's because I'd talk about this being uniquely defined when discussing semigroups, that I did before :P 4:31 PM I have two functions in a family$f_1$and$f_2$which serve as upper and lower boundaries on a compact connected region$R$in the plane. I know that$f_1$and$f_2$satisfy a functional equation simply differing by a parameter. If I glue$f_1 \sim f_2$then what exactly is the effect on the functional equations? >Is there some way of combining the two functional equations s.t. they agree along the glued functions? Does this look good? I think it passes the eye test The underlying gist is "can the glued functions admit a functional equation perhaps of the same form as the individual ones for$f_1$and$f_2$?" 0 I have two functions in a family$f_1$and$f_2$which serve as upper and lower boundaries on a compact connected region$R$in the plane. I know that$f_1$and$f_2$satisfy a functional equation simply differing by a parameter. If I glue$f_1 \sim f_2$then what exactly is the effect on the fun... In case anyone wants to take a stab at it 4:56 PM yes i got negative 1 > This question does not show any research effort; it is unclear or not useful I want to point out that a downvote could have happened for any of these reasons yepp in fact your question suffers from a severe lack of clarity basically it just asks in general about what happens when you glue functions that satisfy functional equations in general @Jakobian I think that's quite obvious that it does @zetaspace and it's not clear what you mean by that 5:01 PM I'll make the necessary edits I was trying to make it short because then more people will read/comment > Not all questions work well in our format. Avoid questions that are primarily opinion-based, or that are likely to generate discussion rather than answers. very interesting IIRC it's against this site's policies to ask questions which aren't answerable in a somewhat objective way i.e. you need to get to the point of the issue instead of creating discussions (or whatever else) @AlessandroCodenotti Moving to Waltham tomorrow, gonna start my master's degree at Brandeis @Jakobian Not quite. There would have to be intent behind the user purposely asking questions that they know aren't answerable 5:17 PM @zetaspace I don't know what you're saying is "not quite" you probably didn't understand what I was saying you said its against the site policies etc. I disagreed that is all disagreed with what that it's against site policies? Well then let moderator explain it to you @XanderHenderson I disagreed that its against site policy to ask questions which aren't answerable in some objective way @XanderHenderson he's probably teaching @AkivaWeinberger Nice, good luck! What kind of maths are you leaning towards lately? 5:49 PM > If your motivation for asking the question is “I would like to participate in a discussion about ______”, then you should not be asking here. However, if your motivation is “I would like others to explain ______ to me”, then you are probably OK. (Discussions are of course welcome in our real time web chat.) Also "...avoid asking subjective questions where ... you are asking an open-ended, hypothetical question: “What if ______ happened?”" 6:05 PM @AlessandroCodenotti Topology probably Btw I did another post 0 A fisher is somewhere on the edge of a rectangle-shaped lake, and a fish is somewhere inside of it. The lake measures$a\times b$. What is the probability that the fisher is closer to the center than the fish is? (Assume that the probability distributions of the fisher and fish are both uniform.)... 6:34 PM @AkivaWeinberger as in general topology or? 6:58 PM Hi ! Anyone want to challenge me to chess? 7:31 PM I am reading a proof of the Baire Category Theorem in Gamelin and Greene's book. There, we construct a nested sequence of open balls such that $$\overline{B(y_n,r_n)}\subset B(y_{n-1},r_{n-1})\subset\ldots\subset B(y_1,r_1)\subset B(x,\epsilon),$$where$x$is a point in a complete metric space$X$, and$y_n\in U_n$, the latter sets all being open and dense in$X$. At the end of the proof, when they've shown that$(y_n)$is Cauchy with limit$y\in X$, they claim that since$y_m\in B(y_n,r_n)$for$m>n$, we obtain$y\in\overline{B(y_n,r_n)}$.I can not understand this last claim. Is this clear to someone? Intuitively it makes sense, but I don't know how to show it with symbols. Forget about the context for a second. What'a going on here is really that if you have a sequence$x_n$in a subset$A$of a space$X$, and$x_n$is convergent (in$X$), then the limit must be in$\overline{A}$Do you see why this answers your question and how to prove it? @AlessandroCodenotti yes, I think so :) the sequence$(y_n)_{n=m+1}^\infty$belongs to$B(y_n,r_n)$. So according to your remark, it has to converge to a limit in$\overline{B(y_n,r_n)}$. Indeed. So why is the remark true? let me think 7:48 PM What is the definition of$\overline{A}$We don't know if$y$is adherent to$B(y_n,r_n)$. If it were, then there must be some sequence in$B(y_n,r_n)$converging to$y$, right? what does "adherent to" mean a point of closure ah. I don't use that terminology$y$is the limit of$(y_m)_{m=n+1}^\infty$as you said a sequence in$B(y_n, r_n)$yeah 7:51 PM so such sequence converging to$y$exists or not? do we know it exists? you seem to have said it yourself, this is such sequence hmm, ok @psie yeah, I seemed to have said it myself, but didn't believe what I said so you agree that$y_m$converges to$y$or not? yes, but I am still doubting if$y$is a point of closure or not and that$y_m\in B(y_n, r_n)$for$m\geq n$@psie I don't really care if you're doubting it, I care about your reasoning why do you doubt it @Jakobian yes @Jakobian I guess by showing that$y_m\in B(y_n, r_n)$for$m\geq n$and that$y_m\to y$we have shown it is a point of closure, or? 7:59 PM @psie I don't know. Why don't you explain it to me Because if there's a sequence in$ B(y_n, r_n)$that converges to$y\in X$, then every open ball centered at$y$contains points of the sequence, so that$y$is a point of closure to$B(y_n, r_n)$. Simple :) 8:39 PM @psie or use the characterization of the closure of$A$as the set of all elements which are limits of sequences from$A\$
you don't need to reprove the statements you already know
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