12:01 AM
@TedShifrin Indeed. It is the first case of that that I have come across.
In the associated diagram, I got to use my text along a path code.

@robjohn You are an utter show-off!!
Munchkin would probably say udder show-uff.

$\ddot\smile$

hola

ted: her preferred epithet is "poo-poo butt." no doubt a gift from her classmates at day care.

Close enough to udder.

12:06 AM
I don't understand close votes from the review queue

@leslietownes my wife and I refer to select individuals as "poo-poo heads"

most of them are along the lines of "not enough context" even though the questions are well posed, and explain in at least half a sentence where the problem is coming from

Are they not obviously someone's homework, with little or zero effort? That's about the only time I vote to close.

I mean, there's often no way to tell
but then I encounter such questions math.stackexchange.com/questions/4341960/…

astyx this is a grey area. simply identifying the source of a problem is probably not enough to avoid a vote to close. identfying context can be, depending. it is an ultimately subjective evaluation of, how much effort has been put into this.

12:09 AM
that have zero context and get 3 upvotes and no close votes
This one has -1 upvote even though it shows effort from the OP math.stackexchange.com/questions/4346341/…

if people find a question interesting they will upvote it perhaps despite general site guidelines. i wouldn't assume that the same level of review is applied to each post. some people flag topics of interest and only visit those things. and if a question is of some kind of general interest it might be seen by more people quickly.
sometimes people confuse downvoting with "i don't think this is an interesting question" and upvoting with "i do think this is an interesting question."

Yeah, I consider that zero effort. On the other hand, I just solved one with zero effort, but I'm pretty sure it's a non-mathematician asking this.

the trouble i have is that often a restated version of a textbook hint (not expressly described as such) can sound like effort, when it isn't. i can catch those pretty quickly in my field of expertise but am not so good outside of it.

@Astyx perhaps the users that scour the site for PSQs don't frequent those tags. There are many poorly posed questions (note that I did not say bad questions) that have escaped scrutiny.
@Astyx If you think the question is worthy (I am not giving my judgement), you could upvote it.

Yes, I confess that I get annoyed when people who are not remotely experts in differential geometry or manifold theory pass judgment on questions in that sort of field. I know enough algebra, for example, to judge up to graduate level algebra just fine, but I don't believe the police here are well-rounded in every field.

12:21 AM
@robjohn Yes that's what my thoughts were as well, finite automata/CFG questions are not that common on this site

@robjohn The one I just linked just above I finally did get my graphic inserted. No finesse, but it'll do.

@robjohn I don't think it's a particularly good or interesting question. But I think the OP is honest, has given this problem some thought before posting it online, and is just stuck and need some hint to carry his on work on

How the hell can you tell honesty and having given thought? It doesn't gave any such signs to me. "However, I couldn't do that." That's it.

@TedShifrin it is infinitely better than the one in the question

I think you're attributing your own personal attributes to the OP, @Astyx.

12:23 AM
I think we're not talking about the same question, Ted

Yes, @robjohn. His sucked and misled me.

Yes that's my point

13 mins ago, by Astyx
This one has -1 upvote even though it shows effort from the OP https://math.stackexchange.com/questions/4346341/prove-inequality-for-vectors-in-mathbbrn/4346479#4346479

The first question is effortless and has 3 upvotes, the second one shows more involvement yet has a downvote

Yeah, I just looked at the second. I consider that effort. You dismissed the alternating signs. I wonder if they caused the OP an issue.

12:25 AM
I explained that to him in a first comment

Then I gave him a hint about CS, and a few hours later gave a complete solution (had I the time I would have worked harder so he would have come up to the solution on his own)
anyway I'm just rambling, time to go to bed
bye

Bonne nuit.

I'm taking the dog to the park. BBL

does anyone know what are some advanced enumerative identities?(e.g. hook-length formula)

8 hours later…
8:58 AM
Since we are told that we are approximating $f(x)$ with the linear function $\lambda x$. Why can't we just substitute that into $3\int_0^1 xf(x) dx$ and show that it equals $\lambda$?

9:27 AM
@LearningCHelpMeV2 one needs to show that the error in the approximation is small enough that it does not interfere with the substitution you mention.
Oh, wait. That passage is showing how to compute what $\lambda$ is to minimize that square integral. We are not given that $f(x)$ is close to $\lambda x$. We are simply trying to find the best $\lambda$ for the approximation.

@MadSpaces Hi, that $\zeta$ outside didn't look right to me.

10:22 AM
Bonjour, I'm trying to study the following DE: $\displaystyle \frac{{\rm dy}}{{\rm d}x}=\frac{x^2+3y^4(x)}{x-4y(x)}$. I think using the existence and uniqueness theorem we can conclude that solution there exists. But can we find the solution in closed form? Using change o variable, of techniques in homogeneous equation all seems does not work. How can I work in this problem?

1 hour later…
11:48 AM
If I have two Bernoulli R.Vs $X$ and $Y$ with $X \sim Bern(\Theta)$ and $Y \sim Bern(p_0)$. Then I have $Z = XY$, what's the joint pmf of $Z$ and $Y$?

12:11 PM
Hello, does anybody know: We have equation 3x+5 = 2 in mod 10. I applied the extended10 euclidian algo on $3x+10k = 3$ and got to the equation: $10\cdot 1 - 3\cdot 3 = 1 \iff 10 \cdot 3 + 3\cdot (-3) = 3$.

However we written that $x = -3-10m$ and $k = 3+3m$, however I do not understand where we got x and k to be in that form (OK for -3-10m it is because of mod 10, but what about 3+3m ??)

12:35 PM
How do you react when you answer someone's question, and they respond "can you prove this using this?"? It feels like it's not my problem.
@VLC First, note that $10\cdot 3 + 3\cdot(-3)$ is not $3$

@Jakobian Sorry I meant to multiply the equation with $3$, so we would get: $10\cdot 3 + 9\cdot (-3) = 3 \mod 10$
However $3$ and $-9$ are only one of the solutions, we would need to get it into the form :
a_0 = a/ gcd(a, b) in b_0 = b/ gcd(a, b)
I do not really understand how it workd

I don't really get why use extended Euclidean algorithm in the first place
$3x+5 \equiv 2\pmod{10} \iff 3x \equiv -3\pmod{10}$ and we can use that $\gcd(3, 10) = 1$ to obtain, theoretically (you don't need to know what it is), some $y$ with $3y \equiv 1 \pmod{10}$ from which multiplying both sides by $y$ we get $x \equiv -1 \equiv 9\pmod{10}$

1:05 PM
@Jakobian Does gcd(3,10) work only because there is $3$ at both sides of the equation ?

You want to divide both sides by $3$ basically
if you have $ax \equiv ay \pmod{n}$, then you can kind of divide by $a$, but the equivalent equation won't be mod $n$ anymore
It will be $x\equiv y \pmod{m}$ where $m = \frac{n}{\gcd(a, b)}$
$\gcd(3, 10) = 1$ allows you to stay in the realm of modulo $10$

1:25 PM
@Jakobian I see, I get it now

1:39 PM
I came across this definition of a smooth mpa. A map f: U \subset R^m \to R^n is called smooth if it agrees with another smooth g an open set O, so g|(U\capO) = f|(U\cap O). So then how do we define smooth on g....? I thought smooth map meant it has a derivative everywhere.

Yes, but the point here is to define smoothness on domains which aren't necessarily open sets
We can't talk about the smoothness directly, so instead, we say that it extends to some smooth map in some small neighbourhood of every point of that set
And this generalizes smooth maps

Does anybody know the best way to get 3x = 4y mod 13
To find x and y

So to define smoothness on g, we use open sets first and talk about g has a derivative everywhere and to find out if a function f on an arbitrary set U is smooth, we compare it to known constructions locally (with g)? And if this local comparison is true everywhere in U, then f is just smooth on U.

@Hawk Yes, pretty much
@VLC You an do $3x\equiv 4y \equiv -9y \pmod{13}$, divide by $3$, and then express the whole thing with respect to $x$ for example. Idk if this is what you're looking for.

2:33 PM
Hey, I have a probability question: If you are rolling two die and you want to find the probability that none of the rolls will yield two faces with the same number, it's just $1-P(S)$, where $S$ is the probability that the faces will have the same number, correct?

2:48 PM
yes. i'm a little unsure of the language here, 'none of the rolls' when there's just two of them, and 'the faces' when there's just two of them. if you mean just the two rolls, yes.

Yeah, whoops. I mean when you roll once and are looking at the faces of the two die
*roll them both once

so there's a single roll of two dice. yeah. then yes.

Ok, good, thanks.

or alternatively two consecutive rolls of one die. oddly enough.
it gets subtler if you roll a die three, four, five, or six times because the event that 'at some point two values match' is different from simple equality between two fixed outcomes.
if you roll more than six times, you can be pretty sure that two of them will be the same. :)
unless the die falls off the table or vanishes in some kind of quantum incident

That last thing you said reminded me of something that's stumping me as well. I'm shaky on determining the total number of outcomes.
If you roll once, the total number of outcomes is 6*6, correct?
So, if you roll six times, is it (36)^6?

2:57 PM
you sort of get to choose. i would say yes to that, with a caveat like, say one die is red and one is blue and you distinguish them. if the dice are indistinguishable you could choose to model it with a smaller number of outcomes because you don't have a way of seeing "this one is 3, and this one is 4" you just have "we rolled 3 and 4". with distinguishable dice the probability of each 'outcome' is the same. with indistinguishable dice, different outcomes may have different probabilities.
i find it's usually helpful to model these things with distinguishable dice. this results in a uniform distribution on the set of outcomes, and you can care about whether or not the distinguishability matters in whatever your event is in how you count them.

Aah, ok. Yeah, I would assume distinguishability then.

Let $(X,\Omega, \mu)$ is a $\sigma$-finite measure space $x$ is a bounded linear operator on $L^2(X,\mu)$. There exists an ascending sequence $\{E_n\}$ of measurable subsets of $X$ of finite measure whose union is $X$. If $1_{E_n}$ denotes the indicator function on, why does there exists a function $f : X \to \Bbb{C}$ such that $f 1_{E_n} = x 1_{E_n}$ for every $n \in \Bbb{N}$?
In fact, $f$ is supposedly unique.

this doesn't seem right. consider X = R (which is sigma finite), you could take E_n = [-n, n], and x to be translation by 1 (say to the right). x 1_{E_n} is nonzero on the interval [n,n+1] but for any multiplication operator f, f 1_{E_n} is supported on E_n and cannot be.

3:13 PM
Hmm...that's no good. I'm reading this math.berkeley.edu/~brent/files/209_notes.pdf
The claim is made on page 11 under section 2.4.2
We do know that $x$ is actually in the double commutant of $L^{\infty}(X,\mu)$, when viewed as left multiplication operators...Not sure if that makes a difference.

okay, here x is in the bicommutant of the multiplication algebra.
yeah.
i took math 209 once. :)
this does make a difference.

Oh, cool!
von Neumann algebras are cool.

Can someone recommend a very good probability textbook at the undergrad level? The ones we're using in class kind of suck.
Eh, I probably shouldn't say they suck. It's most likely just me. Still...recommendations?

if you let g denote the function x 1_{E_n} then it is certainly an L^2 function. it is also supported on E_n because 1_{E_n} g = 1_{E_n} x 1_{E_n} = x 1_{E_n} 1_{E_n} = x 1_{E_n} = g. x commutes with multiplication by 1_{E_n} because it is assumed to commute with things that commute with multiplication operators, which include multiplication operators.
a priori i guess it's not clear why g should be in L^infty and not just L^2 but maybe he gets to that later.
long story short, the function here is just x applied to the constant function 1, which a priori is in L^2 and under further reasoning winds up in L^infty.
in my postdoc i helped a student who was stuck on his thesis. he had all of this pretty cool theory worked out, but it depended on something being in L^infty and not just in L^2, and he couldn't show that. it was not a von neumann algebras setting. he had been stuck on it for months. one day it just hit me and he copied my proof verbatim into his thesis without attribution.
which doesn't bother me, i think it was ignorance of scholarly norms and not hostility. and i am more famous than he is anyway.
under: i liked feller for self study although it is not the kind of book that you could really base a course around.

The course is basically self-study at this point, so that's fine. -_-
I don't mind doing things this way too much because the concepts tend to stick better.

3:27 PM
Wow, really? That's crazy!
What field do you work in?

@UnderMathUate Try Resnick. For something more advanced there's Billingsley

it's also somewhat more 'pure mathy' in its approach. not that its contents are useless and cannot be applied, but some books take an applied approach from the beginning and if that's why you're into probability you might not like feller.
my wife used resnick. it is a good book.
user: i am 'retired' from math now. the problem involved specific C* algebras.

I prefer the pure approach more than applications. The course is more applied, but I can still read it alongside the required textbooks. Either way, I just need to better solidify my understanding of what's actually happening.
@Jakobian Thanks, I'll check this out too.

@UnderMathUate Oh, I actually never read Billingsley, sorry. The book I was trying to remember is Probability with martingales by Williams. It has a small amount of exercises, but they're challenging and fun to do.

Is there a faster way than to loop oneself to find: $$x^3 \equiv 5 \mod 13$$ ?

3:42 PM
@Jakobian Ah, ok. Thanks!

@VLC You could note that $5 \equiv -8 \pmod{13}$ and use that $x^3+8 = (x+2)(x^2-2x+4)$, then solve for $x^2-2x+4 \equiv 0 \pmod{13}$. Not sure if that's any faster

@Jakobian But still I would have to brute force until I loop right ?

to solve this one?

Yes
I would have to go from 1 to 13 until a loop
in x

You'd have $(x-1)^2 \equiv -3\equiv 36 = 6^2 \pmod{13}$
and you can do $(x-1)^2-6^2 = ...$ now
It's not like, an algorithm or anything, but if you can see such things then it's maybe a bit easier
or maybe not

3:55 PM
@Jakobian So I would have to find the roots ?

@Jakobian hang on kinda elementary, but we can't define smooth functions on non-open sets? What about something like f(x) = x on [0,1)?

@VLC Yes, roots of $x^3-5$ in $\mathbb{Z}_{13}$

Rate my integral

So basically we can say $x^3 = x'$ and then we use $x' -5 = 13k$, and $x'-13k = 5$, now we can use extended euclidian to find $x'$ and then we know that $x' = x^3$
@Jakobian

@Hawk This kind of smoothness at the edges of the interval is seemingly different from the smoothness of functions in the theory of smooth manifolds, but you can show the following math.stackexchange.com/a/2162283/476484
You have $f:D\to \mathbb{R}^n$, and for all $x\in D$ there should exist some open $U_x$ and a smooth function $g_x:U_x\to \mathbb{R}^n$ such that $g_x$ is an extension of $f\restriction U_x\cap D$
what "smooth" means varies from author to author, sometimes it means $C^\infty$, sometimes it means $C^1$ and sometimes just differentiable

4:07 PM
I was asking how "g" is smooth here. but apparently in book we only define maps from R^n to R^m on open sets where their components are calculus smooth.

We can take one universal $g:\mathbb{R}\to\mathbb{R}$ defined as $g(x) = x$ here and $U_x = \mathbb{R}$
I don't think there's any doubt a polynomial would be smooth

brb. bus

@VLC Idk why would you do that or what would that achieve
What you could do, I guess, is note that $(-x)^3 = -x^3$ so you can just check $x = 1, 2, ..., 6$ to see if you get either $5$ or $-5 \equiv 8 \pmod{13}$

@robjohn Sorry this is late (fell asleep). But the approximation integral is the area between the graphs. Wouldn't the best approximation occur when $f(x) = \lambda x$. I.e. approaching zero area between the graphs

so you'd get $1, 8, 1, 12, 8, 8$ for $x = 1, 2, ..., 6$ respectively (note that to calculate $a^3$ you can always try calculating $a^2$, reducing mod $13$ and multiplying by $a$ again to make your job easier for bigger $a$)
so solutions are $x = -2, -5, -6$
Also, this hints at a convenient number theory trick, checking something mod some number. Here for example, we see that a cube needs to be $0, 1, 5, 8$ or $12$ mod $13$. So if in some equation you get a cube, and the other side is some other number mod $13$ than the ones listed, you know it's not solvable. And this trick usually works modulo other numbers, for example, a square needs to be $0$ or $1$ modulo $4$

4:30 PM
@LearningCHelpMeV2 Yes, but we don't know that $f$ is the best approximation, so we compute the best $\lambda$ from the observed (or computed) values

Oh, I see. Thanks Rob

Ok I will try to do some more problems and apply this trick
Does anybody know a good pdf online for stuff like this, so discrete math?

which topic is better to learn first? counting and combinatorics or number theory?

I don't know if you should learn those unless you really want to indulge yourself into a topic or it's for a course

i agree with jakobian. number theory might have slightly more structure to it and more obvious interactions with other fields as a beginner (i do not mean to criticize combinatorics, which is everywhere, but i think the connections come later on)

4:55 PM
@leslietownes Wait, so for each $n \in \Bbb{N}$, we define $g = x 1_{E_n}$. But then $g$ depends on $n$. Are we supposed to take some sort of limit to get $f$?

user i was just thinking about it at the individual level for purposes of thinking it through. let g = x 1. its restriction to E_n will be x 1 E_n.

Ah, okay. Got it! Thanks!

@VLC $x\equiv4t\pmod{13}$ and $y\equiv3t\pmod{13}$?

the vibe of this is that if x is a multiplication operator (which it turns out to be) then its symbol has to be x 1. although it is not a priori clear that x 1 (belonging to L^2) also belongs to other function spaces in which one considers multiplication operators.
this comes up a lot in physics treatments of operator theory which ignore this. and they may be right to do so, there's nothing about it that's wrong in common settings, that i know of. it's just, if you want to pencil in the details, you do have to show by some route or another that multiplication by x 1 induces a bounded operator. that's not going to be true of general L^2 functions.
and if it does induce an unbounded operator that can be fine too, because we can work with unbounded operators. but that's not on page 11 of your notes. :) that might be page 90 of your notes.
90 chosen at random, not a pin cite. just, it comes later. often in assessing commutants of general von neumann algebras.
there's no route i know to understanding general von neumann algebras that does not involve unbounded operators, which is weird because they are by definition algebras of bounded operators.
tomita had a very deep understanding of this although some of his results were phrased oddly and confusingly. takesaki is the standard reference point although it would be in any book by now.

5:29 PM
Hi, I know that if $z=f(x,y)$ then we can consider the parametrization for the surface as $x(u,v)=(u,v,f(u,v))$. Similarly if $x=f(y,z)$ then we can consider the parametrization for the surface $x(u,v)=(f(u,v),u,v)$. Then if the surface $S$ is given by $x=y+z^{3}$ then $x(u,v)=(u+v^{3},u,v)$ is a good parametrization for $S$? I know that the parametrization it not only, but my parametrization is correct?

looks good to me.

Thank you leslie, my doubt was because when I try to calculate the Gaussian curvature, the principal curvatures become hard to calculate. So I had my doubts.

in general, calculation of curvatures is difficult. the expressions work out simply only in special cases.

Oh, so I will continue working. Perhaps I can simplify things

5:48 PM
I am working on this problem, I have some nice topological spaces (metrizable & separable) $X_i$ which are Baire spaces, and I'm trying to prove that their product $X$ is also Baire
So I've tried taking $G_m\subseteq X$ which are open and dense, and $p_N:X\to X_1\times ...\times X_N$ be the projection, then consider $G_m^* = \bigcap_{N=1}^\infty p_N^{-1}p_N(G_m)$ but it doesn't get me anywhere tbh.
I already have this result for finite product of them
So I'm trying to somehow prove that $\bigcap_{m=1}^\infty G_m^*$ is dense, and I expect it to somehow magically imply that $\bigcap_{m=1}^\infty G_m$ is dense as well
idk
I could assume that $G_{m+1}\subseteq G_m$, that usually seems like a good property to have

6:30 PM
@Alex At the origin only, or everywhere?

7:06 PM
How can a sequential criteria for a nowhere dense set be created?
If a set X is nowhere dense in R then int (cl X) is an emptyset.

A criterion; criteria is plural :P

sorry, yeah. criterion

So how do you give a sequential criterion for a dense subset?

Assuming by dense subset X, you mean dense in R, I would say that for every $x\in R$, there is a sequence of numbers in X that converges to x.

Are we in $\Bbb R$ or a general metric space?

7:13 PM
in $\mathbb R$.

Or a first countable topological space? Or ...
OK, so what does it mean for $X$ not to be dense — at a given point, for starters?

I don't know. I know set being (or not being) dense in some set.

Well, you're still talking about subsets of $\Bbb R$, I presume. That's why I asked.

I know only this definition: Set A is dense in set B, where $A\subset B$ are in $\mathbb R$, if cl (A)=B.

But you gave a sequential criterion for the subset to be dense at $x$. How do you negate that?

7:18 PM
No, I didn't say anything about a set being dense at x. I meant that: if X is dense in $\mathbb R$ then for every $y\in \mathbb R$, there is a sequence in X that converges to $y$.

Excuse me. Look up ten lines.

for example: $\mathbb Q$ being dense in $\mathbb R$

You used the letter $x$.
So I repeated it.
So when you say "for every $x\in\Bbb R$," blah blah blah, that is defining what it means to be "dense at $x$."

yes, I did but not the way you say it.
I used x and X also so that may be a messed up thing (like choosing $\delta$ to denote a negative quantity) but not wrong.

OK. Fine. You defined what it means for $X$ to be dense at $a\in\Bbb R$ in terms of sequences. How do you negate that?

7:22 PM
No. I didn't define X to be dense at $a\in \mathbb R$. I defined $X$ dense in $\mathbb R$.
:(

what does "nowhere dense" mean?

What I said to define "X is dense in R" is a restatement of "cl (X)=R".

You don't listen to what I say. Sometimes you are infuriating.

Astyx: I say X is nowhere dense in R if X contains no open interval in it. And this is same as saying int (cl X) is an emptyset.
Ted, you want me to negate the sentence you said. I can do that but it just doesn't make sense to me.
I never heard of a set being dense at a point $a\in \mathbb R$ and that's what in your last sentence confuses me.

maybe a language issue? what would the consequences of density, per the sequential criterion, be, at a chosen point a in R. i think this is what ted is getting at.

7:28 PM
@Astyx It means $\text{int}(\text{cl}(A)) = \emptyset$

it was rhetorical

is $\mathbb Q$ nowhere dense?

clearly not, if seen as a subset of Euclidean space $\mathbb R$.

Does $\mathbb Q$ contain an open interval?

7:32 PM
I see I messed up the wording in my reply to you earlier.
I should have said if cl X contains no open interval.

I have no idea how to solve the question about products of Baire spaces, going to leave it. Maybe I'll solve it some day

I said that your sentence defined "dense at the point."

What does it mean for $cl(X)$ to contain an open set?

7:47 PM
Ted, I'm afraid I don't understand how that defined dense at the point.
Astyx: I said open interval. cl (X) containing an open interval means that there is an open interval (a,b), a<b such that $(a,b)\subset$ cl(X). If that happens then int (cl X) $\ne \emptyset$

That's true
Do you know another basis for the usual topology on $\mathbb R$ ?

You said "for every $a\in\Bbb R$, there exists a sequence $x_n\to a$ with $x_n\in X$. Thus, "$X$ is dense at $a$" if there exists a sequence $x_n\to a$ with $x_n\in X$.

That might be more closely related to sequences

The set of all algebraic numbers $A$
Or the set of all rationals $\mathbb Q$.

A basis of a topology of $\mathbb R$ is a set of subsets of $\mathbb R$, not a subset of $\mathbb R$

7:51 PM
Oh, I think I don't know that.
Ok Ted, let's define X dense at $a$ like that.
Then X not dense at a would mean that for any sequence $(x_n)$ in X, $\lim x_n\ne a$.

Or, better, there is no sequence of elements of $X$ converging to $a$. [If you take an arbitrary sequence, lim makes no sense.]

right, no sequence in X converges to a.
lim symbol won't make sense sometimes so the version I said earlier is problematic.

There's your sequential criterion if you put in appropriate quantifiers.

If for every open interval $(a,b)$, there is a point $c\in (a,b)$ such that $c$ is not in X and no sequence in X converges to c, then we say that X is no where dense in R.
yeah, that looks like a sequential criterion.

No, that doesn't look right.
Why only one point $c$ in the interval?

8:01 PM
by this definition, $\mathbb R\setminus \mathbb Q$ is nowhere dense

there could be more but atleast one such c will be there.

Or even $\mathbb Q$ for that matter

Astyx: why would you think that?

Or maybe you edited since I read, not sure

I didn't edit except possibly adding "if" at the start.

8:02 PM
Why not just say for every $c\notin X$ ...
Why make it complicated with open intervals?

Due to geometrical reasons I thought to keep open intervals.

It's a bit weird. Sequential criteria are meant as a way to avoid dealing with open sets, in my humble experience
I do agree with your def however

Ted, if say that then $X=\{1/n: n\in \mathbb N\}$ will be "dense at $0$".

Yes, it is.
So that set is not nowhere dense. I don't see the issue.

But X is nowhere dense in R.

8:15 PM
You just showed it is not.

$\{1/n:n\in \mathbb N \}$ is nowhere dense in $\mathbb R$.
We can show using the definition I put above, that it's nowhere dense in R.

Oh, I see, the interior of the closure definition says it is.

:)

8:57 PM
The set $K = \{1/n : n\in\mathbb{N}\}$ seems to be an important one

I have been thinking all day about a question I had in exam. I know intuitevely that it is true, but I am not able to show that a X ~ Binomial n=L, p=1/2, and L~Poisson(lambda), is equivalent to say that X~Poisson(lambda/2)

How can you know intuitively that this is true? Doesn't make sense to me. Anyway, how you do this is just calculate probabilities: $P(X = k) = \sum_{n=0}^\infty P(X = k | L = n)P(L = n)$
Anyway, I have a question
We know from descriptive set theory that there are $G_\delta$ subsets of $[0, 1]^2$ whose images under projection are not Borel
can we provide some simple example if we require the image to not be $G_\delta$

Because Bin(1/2) is like halving the Poisson
A prime example of a subset which is not $G_\delta$ I have in mind is $Z\subseteq [0, 1]^2$ defined as $Z =( \mathbb{Q}^2\cup( \mathbb{Q}^c)^2)\cap [0, 1]^2$
wrap it around $so I can use the plugin to read it$P(X = k) = \sum_{l=0}^\infty \frac{e^{-\lambda}}{(l-k)! k!}(\frac{\lambda}{2})^{l}$what's x? Sorry, fixed. x was k 9:20 PM that looks alright one problem though, what is$P(X = k | L = l)$when$l<k$? No way that is possible. L is the n input in binomial. So X=k, is the succeses on Binomial(l, 1/2) yeah, it's just 0 so your sum should begin from l = k But l=0 is a possibility, no? Sorry, Im just a master student 9:24 PM you just said it was impossible as in "probability 0" kind of impossible @JavierMorenoSepena Doesn't that mean you should be able to handle such thing @Jakobian In statistics masters people even barely can do a integral haha Yes, you were right But how can this infinite sum be resolved? change the indexes so your sum starts at l = 0 and then use definition of exp(x) and you're done nothing serious here going on Wow That was so difficult The professor is dumb, is normal none of the 300 students of the master of statistics could solve it 2 hours later… 11:28 PM When we want to find the surface integral of a vector field$\vec A$for a parameterized surface, lets say a sphere of Radius R, while$\vec r(\phi , \theta)$it must be that$\vec A (\rho , \phi , \theta)\$ . Is this assessment correct ?