1:12 PM
@AkivaWeinberger Hey, man! Do you have time to clarify me a doubt?

1:26 PM
Hey everyone!
Yo @BalarkaSen

eyy

@BalarkaSen You know of any good place to learn about Pseudo-Riemannian manifolds?

@nbro Incidentally, where are you from?

@Perturbative Nope. I only ever learnt a little of the Riemannian theory

Ahh okay cool, no prob

1:35 PM
If the orthogonal projection of acceleration vector $\gamma’’(t)$ of a curve onto the tangential space is zero, why can one conclude it is a straight line in Euclidean geometry?

@Balarka Have you gone ahead with the covering space stuff? (sorry I've been a bit busy trying to learn enough DG to read this paper on pseudo-riemannian stuff)

I mean I can still have a circle as the acceleration can be orthogonal to the tangent?

@Perturbative i already know a chunk of the covering space story so, no, not really

I just don’t understand why it is sufficient to only look at the tangential part of the second derivative to define a geodesic.

@AkivaWeinberger Why do you want to know that? I don't want people to judge all my compatriots based on their experience and relationship with me...
Unfortunately, many people make up an idea about a group of people based on their experience with a member of that group
But that idea isn't always completely true.

1:45 PM
I was just wondering if you were from a Spanish-speaking country, 'cause the phrase "clarify me a doubt" seems like a mistranslation of the Spanish word duda
@nbro
Second guess would be India

No, both guesses are wrong :)

I have an apple, I have a doubt, uh... apple doubt

Ah, well

Anyway, how would you say "clarify me a doubt" otherwise?
"clarity to me"?
What?
Erase a doubt that I have by clarifying to me a few statements?
How would you express such a desire/need that I have?

I generally try to avoid using the word "doubt" when I really mean a "question"

1:49 PM
Provided you're a native speaker

I'm not actually sure, to be honest

I would probably just say "Do you have time to clarify a problem that I have?"
Though none of these things are communicationally superior to the other
Everyone understands what you mean

A doubt is by definition something which needs to be clarified, no?
and you need some clarifications

@BalarkaSen Ahh okay I see, I'll get back to the covering space stuff in a couple of days if that's cool with you, need to sort out what I need to learn for the Riemannian stuff first

@nbro Yes, but it's generally not used to indicate that you have a question. It indicates uncertainty as an emotion, not an intellectual dilemma, in my view
A lot of people use it though
There's nothing per se wrong with it

1:53 PM
I've heard that Indian English uses the word "doubt" more similarly to how Spanish uses duda than how other varieties of English use "doubt"

@AkivaWeinberger I suppose we Indians confront a lot of doubt

But I think this specific example was probably fine, I was a bit overzealous

@AkivaWeinberger But are you a native speaker or not?

I am

@Perturbative Totally fine
@AkivaWeinberger Epic failure

1:56 PM
Oh, so if that sounds strange to you, then I shouldn't probably use that expression anymore, even though it doesn't really seem wrong and, in any case, I think, it is understandable.

“Can you clear up a doubt for me” seems closer in English
The difference being, I guess, that a doubt is not a question but rather the grounds for a question

ok, suppose we have the set $U_1=[a,\frac{a+b}{2}) \cup (\frac{a+2}{2},b]$ where $a,b$ are rational. It is easy to see that there exists a countable cover which consists of intervals that converges towards, a,b and $\frac{a+b}{2}$. Therefore $U_1$ is not compact. Now we can construct $U_2$ by taking the midpoint of each half open interval of $U_1$ and we can similarly construct a countable cover that has no finite subcover.

By induction on the naturals, we eventually end up with the set $\Bbb{I} \cap [a,b]$. Thus this set is not compact

@BalarkaSen I had to go take a call

I know it was a typo, I was pulling ze legs

My glowbrick stopped being an internet browsing device and started being a cellphone

2:00 PM
@Semiclassical Unforgiving: Northern Hymn is a fantastic horror game
Check it out if you haven't

So that means the set $\Bbb{I} \cap [a,b]$ has at least one countable cover that has no finite subcover. Now to think about how to compute the infimum of the length of intervals involved...

$\Bbb I$ is $\Bbb R\setminus\Bbb Q$?

Yes
because Irrationals

I haven't been following what you've been writing, but it's instructive to see why we can't define a measure $\mu$ on $\Bbb Q$ satisfying $\mu\big((a,b)\cap\Bbb Q\big)=b-a$

I am currently working under the Lebesgue outer measure, though I did not know we cannot define any measure where subsets of rationals have nonzero measure
The above workings is basically trying to compute $\lambda^*(\Bbb{I}\cap[a,b])$ more directly without using the fact $(\Bbb{I}\cap[a,b]) \cup (\Bbb{I}\cap[a,b]) = [a,b]$ where $\lambda^*$ is the Lebesgue outer measure
that is, trying to compute the Lebesgue outer measure of the irrationals using only the notions of covers, topology and the definition of the measure

2:08 PM
@Secret You mean $(\Bbb I\cap[a,b])\cup(\Bbb Q\cap[a,b])$

yup, typo
What I hope from such more direct computation is to get deeper rigorous and intuitve insight on what exactly controls the value of the measure of some given uncountable set, because MSE and Asaf taught me it has nothing to do with connectedness or the topology of the set

Hm. Are there uncountable dense $X\subset\Bbb R$ such that $\mu(X\cap[a,b])=\frac12(b-a)$, I wonder?
Like splitting the reals in half
Clearly it can't be cocountable

Problem: Let $X$ be some measurable space and $f,g : X \to [-\infty, \infty]$ measurable functions. Prove that the set $\{x \mid f(x) < g(x) \}$ is a measurable set. Question: In a solution I am reading, the author just asserts that $g-f$ is measurable and the rest of the proof essentially follows from that. My problem is, how can $g-f$ make sense if either function could possibly take on an infinite value?

Hi everyone
is it true $\sin\theta\sin(120-\theta)\sin(120+\theta)=\sin 3\theta$ ?

I don't know Fawad, but here's how to check:

2:16 PM
Also $\tan\theta\tan(60+\theta)\tan(60-\theta)=\tan 3\theta$

$\sin (a \pm b) = \sin a \cos b \pm \cos a \sin b$
and $3 \theta = 2\theta + \theta$
so expand both sides and see if they're equal

Ok checking

proofwiki.org/wiki/Tangent_of_Sum here's one for tangent but it's less commonly memorized :)

@AkivaWeinberger For $\lambda^*$ I can think of simple examples like: If $\frac{a}{2} < \frac{b}{2} < a, b$, then I can always add some $\frac{c}{2}$ to $\frac{a}{2},\frac{b}{2}$ to generate the interval $[\frac{a+c}{2},\frac{b+c}{2}]$ which will fullfill the criteria. But if you are interested in some $X$ that are not intervals, I am not very sure

the $\sin$ one I mentioned is common enough to be worth memorizing, if you can
also $\cos(a \pm b) = \cos a \cos b \mp \sin a \sin b$

2:20 PM
@Secret I want $X$ not to depend on $a$ and $b$

notice that the $+$ on the LHS becomes $-$ on the RHS and vice versa, for the $\cos$ one

@GFauxPas I memorises many formulae I guess then :D sin3x=3sinx-4sin^3 x
@GFauxPas I know those formulae :) I am revising trigonometry

that one I didn't memorize, but if I wanted to derive it, I'd write $\sin (3x) = \sin (2x + x)$ and use the one I posted above!
nice
most mathematicians memorize the $\sin$ and $\cos$ one I typed because it's so useful

Ah right, use the complement of a fat cantor set which has measure $\frac{a+b}{2}$

also, $\sin 2x = \sin (x + x)$ and you can use the identity again!

2:22 PM
@GFauxPas you are mathematician?

I'm not a mathematician but I play one on TV

I substituted $\theta =90^{\circ}$ I got lhs=-1/4 and rhs=-1

@Fawad Write $\sin(\pi/3 + \theta)\sin(\pi/3 - \theta) = \frac{1}{2}\left (\cos(2\theta) - \cos(2\pi/3) \right )$. That makes things a lot easier, I believe.

@AkivaWeinberger So, my doubt is, again, about the consistency of the proof we were talking about yesterday.
More specifically, we derive the Fourier series for $g$ w.r.t. $c_n$. We then need to show that $\frac{c_n}{(-1)^n}$ is $d_n$.

(cont.) Let a countable cover $C = \lim_{i \to \aleph_0} C_i$ of $\Bbb{I}\cap [a,b]$ be built in stages $C_i$. Let:
$$C_0 = (a,q_1)\cup(q_1,b)$$ .Then $\lambda^*(S_0)=(b-q_1)+(q_1-a)=b-a$

Let $C_1 = (a,q_2)\cup (q_3,q_1) \cup (q_1, q_3) \cup (q_3,b)$, where $q_1=\frac{a+b}{2}, q_2 = q_1 -1, q_3 = q_2+1$

We then manipulate the $c_n$ for the Fourier series of $h$ to obtain a new $c_n$, but expressed w.r.t. $g$.
Now, I am still not understanding why by doing what we have done we're logically showing that this new $c_n$ is the $d_n$ which we need. Why would this $c_n$ be the $d_n$ associated with the Fourier series of $g$?

morning chat

Morning

so, I got a book from the library yesterday

2:37 PM
Then $\lambda^*(C_1)=(b-q_3)+(q_3-q_1)+(q_1-q_2)+(q_2-a)=b-a$

and today I found the following theorem in it

We can proceed indefinitely by induction and picking rationals using the cantor pairing function. So in the end we obtain:

@fawad that's using the $\tan$ identity i linked to (but didn't memorize)

test: $\arccos$

given the convention $\sinh$ is sinch, $\cosh$ is cosh, $\tanh$ is tanch, how would you pronounce $\operatorname{sech, csch, coth}$?

2:42 PM
"The following statements are equivalent for $\alpha,\beta,\gamma\in[0,\pi]$:

1) The matrix \begin{pmatrix} 1 & \cos \alpha & \cos \beta \\ \cos\alpha & 1 & \cos \gamma \\ \cos \beta& \cos \gamma & 1\end{pmatrix} is positive semidefinite.

2) There exist unit vectors $v_1,v_2,v_3\in\mathbb{R}^3$ such that $\alpha=\arccos(v_1^T v_2), \beta=\arccos(v_2^T v_3),$ and $\gamma=\arccos(v_3^R v_1)$.

3) $\alpha\leq \beta+\gamma$, $\beta\leq \alpha+\gamma$, $\gamma\leq \alpha+\beta$ and $\alpha+\beta+\gamma\leq 2\pi$."
If I replace $\alpha,\beta,\gamma$ with their doubles, that's exactly the matrix I had the other day

$\lambda^*(\Bbb{I}\cap [a,b]) = \lambda^*(C) = \lim_{i\to \aleph_0}\lambda^*(C_i) = \lim_{i\to \aleph_0} (b-q_i) + \sum_{k=1}^i (q_{n(i)}-q_{m(i)}) + (q_{i+1}-a)$. Therefore, computing the Lebesgue outer measure of the irrationals directly amounts to computing the value of this series. Therefore, we first need to check it is convergent, and then compute its value

but with the advantage now of there being a direct geometric interpretation

If you know the measure of $\Bbb Q$ and the measure of $\Bbb R$ it's easy to get the measure of the irrationals

What I'm trying to confirm right now is the boundary case of these theorems, e.g. when $\alpha+\beta+\gamma=2\pi$.
And I think it should correspond to all three vectors lying in a common plane in R^3.

39 mins ago, by Secret
The above workings is basically trying to compute $\lambda^*(\Bbb{I}\cap[a,b])$ more directly without using the fact $(\Bbb{I}\cap[a,b]) \cup (\Bbb{I}\cap[a,b]) = [a,b]$ where $\lambda^*$ is the Lebesgue outer measure
35 mins ago, by Secret
What I hope from such more direct computation is to get deeper rigorous and intuitve insight on what exactly controls the value of the measure of some given uncountable set, because MSE and Asaf taught me it has nothing to do with connectedness or the topology of the set
Alessandro: and typo for the third $\Bbb{I}$ in the quote, which should be $\Bbb{Q}$
(cont.) We first observed that the above countable sum is an alternating series. Therefore, we can use some machinery in checking the convergence of an alternating series
Next, we observed the terms in the alternating series is monotonically increasing and bounded from above and below by b and a respectively
Each term in brackets are also nonegative by the Lebesgue outer measure of open intervals, and together, let the differences be $c_i = q_{n(i)-q_{m(i)}}$. These form a series that is bounded from above and below
Hence (also typo in the subscript just above): $$\lambda^*(\Bbb{I}\cap [a,b])=\sum_{i=1}^{\aleph_0}c_i$$

2:53 PM
@BalarkaSen oops, I meant to direct the messages above in your direction

can you rigorously prove something by drawing a commutative diagram that shows such and such a property?

Consider the partial sums of the above series. Note every partial sum is telescoping since in finite series, addition associates and thus we are free to cancel out. By the construction of the cover $C$ every rational $q_i$ that is enumerated is ordered such that they form expressions $-q_i+q_i$. Hence for any partial sum by moving through the stages of the constructions of $C$ i.e. $C_0,C_1,C_2,...$, the only surviving term is $b-a$. Therefore, the countable sequence is also telescoping and:
Finally
$$\lambda^*(\Bbb{I}\cap [a,b])=\sum_{i=1}^{\aleph_0}c_i = b-a$$
QED

@GFauxPas why not?

@GFauxPas isn't that how you prove the uniqueness up to isomorphism of product and coproducts if they exist in a category?

3:12 PM
@Semiclassical I don't want to see any more trigonometry in my general direction
Go away. Shoo

@AkivaWeinberger Never mind. I think I figured it out alone. Basically, the value of the definite integral for $c_n$ is actually the value of the define integral of $d_n$. So they are the same thing but re-expressed differently.

I had to find the residue of sin tan at pi/2
the answer was to expand it as tan - tan^3/6 + .. and compute each residue
???

Is there a rule for finding the residue of $f\circ g$ at $x$?
I'm guessing it's the residue of $g$ at $x$ assuming $f'(x)\ne0$ or something of that sort

3:20 PM
there's a formula using the n-order derivative of (z - z_0)^nf(z)

That sounds like more work than Leaky's Taylor thingy
So that really is the easiest method then

I have him on ignore so can't see what he's asking

bleh, it's too early for me to already feel this sleepy

Bye, got to go

bis

3:26 PM
yo
:D

3:37 PM
An impression of the irrationals and the rationals

Hard to get a visual impression of a disconnected set

Now, I am thinking something a bit more wild, which goes as follows:
1. Take the axiom of choice, define a bijection $g : \Bbb{I} \to \omega_1$
2. Assume CH is true (to save me the trouble of transfinite induction up too many uncountable ordinals that are successor cardinals)
3. Construct an uncountable cover that has no countable subcover $C$ in stages from $0$ all the way up to $\omega_1$ (this is possible because $\omega_1$ is well ordered) as follows:
Let $\epsilon > 0$ be some irrational if $a$ or $b$ is rational and $\epsilon =0$ if $a$ or $b$ are irrational.
Also let $a-\epsilon=r_0, b+\epsilon = s_0$ and $\forall (i \in \omega_1)[r_i=s_i]$
$C_0 = (a-\epsilon,r_1) \cup (s_1,b+\epsilon)$
$C_1 = (a-\epsilon,r_2) \cup (s_2,r_1) \cup (s_1,r_3) \cup (s_3,b+\epsilon)$
$C_2 = (r_0,r_4)\cup (s_4,r_2)\cup (s_2,r_5) \cup (s_5,r_1) \cup (s_1,r_6) \cup (s_6,r_3)\cup (s_3,r_7)\cup (s_7,s_0)$
...

4:02 PM
What can we say about two topolgies if there exists a function from two topological spaces: f: X, t -> Y, S which is an open map and t / S are topologies on X?say

$C_{\omega} = \bigcup_{k < \omega} (r_{g(k)},s_{g(k)})$
$C_{\alpha} = \bigcup_{k < \alpha} (r_{g(k)},s_{g(k)})$
For the finite cases, the resulting sum is telescoping, therefore their supremum, $C_{\omega}$ is also telescoping.
We can then induct up the successor ordinal and limit ordinal cases all the way to before $C_{\omega_1^{CK}}$
after that I have no idea, because there is no longer a program that exists to check it

@Overflow2341313 Could you state that again?
What do you mean by $t / S$ are topologies on $X$?

@Perturbative My problem asks - at the end after giving the function f defined by f(x) = x is an open map - what we can deduce about "the two topologies t and S on X"

Not sure honestly. Stuck on this problem

If you have a function $f : X \to Y$ between two topological spaces $X$ and $Y$ you can't conclude anything about the topologies, if however the function is continuous, then you can say stuff about the topologies
@Overflow2341313 Could you send a picture or a screenshot of the problem?

Yes, one moment.

4:13 PM
Use the definition of an open map?

@Perturbative I've attached the problem statement and definition of open, as given to me, https://postimg.org/gallery/3e0j3h8ks/

I don't see what more I can say than the "Note"

@Overflow2341313 Ahhh $\mathcal{T} \subseteq \mathcal{S}$
Pick $U \in \mathcal{T}$, then $f[U] = U$ which is also in $\mathcal{S}$ by definition of an open map

Ah, that makes sense now. Thanks overcomplicated that.

Also just notation wise one normally uses the notation $f :(X, \mathcal{T}) \to (X, \mathcal{S})$ to denote a map between topological spaces
Mostly though people just say $X$ is a topological space and $Y$ is a topological space and $g : X \to Y$ and leave out explicitly denoting the topology on $X$ and $Y$
@Overflow2341313 No problem :). Glad to help!

nvm I overlooked something important. Each interval contains a rational, and there are only countably many rationals. This means at the $\omega_1$ limit stage, thre are uncountably many intervals that contains neither rationals nor irrationals, thus they are empty and does not contribute to the sum
So there are only countably many disjoint intervals in the cover $C$

4:24 PM
@Perturbative Okay similar problem if you don't mind guiding me in the right direction. If a function f exists, with the same setup (X, t) -> (Y,S), that is 1-1, open, and continous but not onto construct a topological space which is homeomorphic to the space (X, t).

Simply restrict the codomain so that it is onto? Making it bijective and hence invertible.

@Overflow2341313 Yep, that should do it :)
$f[X] \subseteq Y$ would inherit the subspace topology from $S$ and then $f$ would be the required homeomorphism to show $f[X]$ is homeomorphic to $X$

hmm, I don't understand. While I do start with an uncountable cover and using axiom of choice to well order the irrationals, the fact that the rationals are countable means I eventually end up with a countable cover of the rationals. However the telescoping countable sum clearly does not vanish, so this is weird...
In a schematic, we have the following, I will try to figure this out tomorrow before moving on to computing the Lebesgue outer measure of the cantor set:

@Perturbative Okay, kast question. Think I'm starting to get this stuff now.... I want to find a topology t on R such that f: R, U -> R, t defined by f(x) = x^2 is an open map where U is the "usual" topology defined by U = {x in U | x in U implies that x in (a,b) \subseteq U}.

To do this... the smallest t can be is the trivial topology on R - {\emptyset, R}

But, we required that everything in U be in t under f?

4:41 PM
The trivial topology won't work in that case, apply $f$ to the interval $(0, 1) \in U$
The discrete topology on $\mathbb{R}$ which is $\mathcal{P}(\mathbb{R})$ should work though since every possible open set $A \in U$ is an element of $\mathcal{P}(\mathbb{R})$

This is so weird, it should be a different way to prove the same thing, yet I get the wrong conclusion if I start with an uncountable cover (that later becomes countable)?

@Overflow2341313 Also for the previous example, I think it may not be as simple (contrary to what I initially thought), because there do exist functions which are continuous, bijective but do not have continuous inverse
I'm not sure if adding the additional condition that $f$ is an open map will make an difference

For those who are not very familiar about this interest of mine, besides the maths, I am also interested in the notion of a "proof space", that is the set or class of all possible proofs of a given proposition and their relationship

@Overflow2341313 Ahhh previous example is simple because there is the following theorem : *"
A bijective continuous map is a homeomorphism if and only if it is also an open map"*
So because $f$ is an open map we don't need to worry about continuity of $f^{-1}$
Because by that theorem $f^{-1}$ is automatically continuous
So $f : X \to f[X]$ would be a homeomorphism between $X$ and $f[X]$

Elements in a proof space is a proof, which consists of steps and forming a path in this space
For that I have a postulate that given two paths A and B in proof space with the same starting point and a proposition $\phi$. If $A \vdash \phi$ but $B \not\vdash \phi$, then there must exists some condition that make the path $B$ unable to reach $\phi$, or that $B$ is unprovable under the current formal system

1 hour later…
5:55 PM
Hey @orbit-stabilizer

hello
I am scared i'm going to fail my real analysis final today

Chapters 1 - 5 in Rudin

Ah
Don't worry about it. Good luck from my end

Thank you :)
I'm pretty worried haha. I am doing really bad on the practice finals

6:01 PM
Does anyone have a textbook recommendation for complex analysis, say if I have never learned any of it, but I have done plenty of algebra (so I am relatively mathematically sophisticated).

@orbit-stabilizer ugh, I suppose that is worrisome. Don't panic though; that usually ends up being a bad trigger for exams
@Narcissusjewel Stein-Shakarchi

Ahlfors

@BalarkaSen yeah, thanks

6:36 PM
Hi all

6:57 PM
If $(r)$ is a principal ideal in a noncommutative, nonunital ring, what is $(r)^3$? What do it's elements look like?

Hi. I believe I have numerically discovered that $\sum_{n=0}^{K-c}\binom{K}{n}\binom{K}{n+c}z_K^{n+c/2} \sim \sum_{n=0}^K \binom{K}{n}^2 z_K^n$ as $K\to\infty$, where $c=0,\dots,K$ is fixed and $z_K=K^{-\alpha}$ for some $\alpha\in(0,2)$. Any ideas how to prove that?

do you know a standard text book about correlation of functions?

@user193319: What's the definition of the product of two ideals?

I'd like to see a link to signal processing

Heya

6:59 PM
Heya @Krijn

@TedShifrin Ah! I see. So just apply the definition recursively (inductively? not sure which is correct terminology)?