Mathematics

user193319

1:22 AM

Hmm...If $\iota : X \to X \times X$ is given by $\iota (x) = (x,x)$, then I agree that $(id_X \times f) \circ \iota : X \to X \times Y$ is measurable...But how does that show that $graph(f) = \{(x,f(x)) : x \in X \}$ is a measurable subset of $X \times Y$? All I can see is that $graph(f)$ is the image of $X$ under $(id_X \times f) \circ \iota$...but the image of a measurable set under a measurable function is not necessarily measurable...hmmm...

D Left Adjoint to U

0

Q: Astonishing pattern adding two prime numbers, at least for the first few examples; if it doesn't continue forever, then why?

$$ 3 + 5 = 8 \\ 3 + 7 = 10\\ 5 + 7 = 12 \\ 5 + 11 = 16 \\ 7 + 11 = 18 \\ 7 + 13 = 20 \\ 11 + 13 = 24 \\ 11 + 17 = 18 \\ 13 + 17 = 30 \\ 13 + 19 = 32\\ 17 + 19 = 36\\ 17 + 23 = 40 \\ 19 + 23 = 42 \\ ? = 44 \\ 19 + 29 = 48\\ ? = 50 \\ 23 + 29 = 52 \\ 23 + 31 = 54 \\ ? = 56 \\ 29 + 31 = 60 \\ ? = ...

Ask me any questions you might have

D.C. the III

2:36 AM

3

Q: Describe the equivalence classes generated by T

Suppose $S = \{(x,y) \in \mathbb{R}^2\mid y = x + 1\text{ and } 0 < x < 2\}$. Question Describe the equivalence relation T on the real line that is the intersection of all equivalence relations on the real line that contain S. Suppose that $$A_1 = S$$ $$A_2 = \{(x,y) \in \mathbb{R}^2\mid y = ...

elementary-set-theory proof-verification equivalence-relations

Two questions: How were all of the $A_i$ determined? and from that how were the equivalence classes determined?

leslie townes

without even checking the definitions of those A_i, you might generally look up the concepts of 'symmetric closure' and 'transitive closure' of a relation. not because they are theoretically deep but (aside from adding in the diagonal if it happens to be missing) they capture the spirit of "toss in the stuff that any symmetric/transitive relation containing this stuff would have to include."

i'm guessing they are doing something like that.

computing the transitive closure of an arbitrary relation can be computationally, i don't know if difficult is the right word, but more interesting than the other kinds of closing up, in that you might think more about how to organize an algorithm to do it to make sure you get everything, than you do for symmetry and reflexivity.

D.C. the III

This is the first chapter of Munkres's Topology. So though I could decipher what symmetric closure and transitive closure mean based on experience, I'm wondering if this is something I should have known prior

leslie townes

i want to be very clear, there is nothing about 'symmetric closure' or 'transitive closure' that would require a separate section of a textbook or special definitions.

they're just the words for the general thing that this kind of exercise is asking you to do.

D.C. the III

Ok fair enough.

leslie townes

so for purposes of googling, that's what i might google. definitely not saying "oh if you haven't gotten there in the textbook, forget it, they must want you to do it some other way."

however you do it, you will be confronting those concepts in this example.

but maybe someone somewhere has some cute way of organizing the work, or speeding it up, when the relation is a subset of R^2, or something. or someone has written out a lot of examples.

if you haven't guessed from the form of the exercise, the specific example in the problem is pretty random and made up. you will not be confronting a lot of this later in the book :)

D.C. the III

2:44 AM

I'll take a look at the concepts you suggested and try to work it out a but

No I agree I won't be confronting it. It is more so for me to just get more comfortable with getting dirty with elements and understand what is going on beyond the usual "abstract" idea

user193319

@TedShifrin Hmm...If $\iota : X \to X \times X$ is given by $\iota (x) = (x,x)$, then I agree that $(id_X \times f) \circ \iota : X \to X \times Y$ is measurable...But how does that show that $graph(f) = \{(x,f(x)) : x \in X \}$ is a measurable subset of $X \times Y$? All I can see is that $graph(f)$ is the image of $X$ under $(id_X \times f) \circ \iota$...but the image of a measurable set under a measurable function is not necessarily measurable...hmmm...

leslie townes

3:02 AM

user: what are your hypotheses on the sigma algebras on X, Y, and X x Y? they matter. e.g. if {emptyset, X} and {emptyset, Y} are the sigma algebras on X and Y then every function from X to Y will be measurable, but graphs of functions will not be measurable in X x Y with the product sigma algebra (outside of trivial cases e.g. where X and Y are empty, or Y is a singleton)

D Left Adjoint to U

I don't think we can prove Twin Primes without proving Goldbach and vise versa - they're intricately linked

You need to prove both of them simultaneously for an induction argument to work

Ted Shifrin

3:19 AM

@leslietownes what about all that blsh completion stuff blah done for Fubini?

Balarka Sen

I think everything in user's question are Borel measurable

X, Y top spaces, X x Y product top, all equipped with the sigma algebra of Borel sets

I think this is fine as long as Y has a countable basis, or something

Jakobian

definitely don't want to mix topology with measure theory unless I have some assumption like that, that's true

ending up in set-theory madness resulting from not assuming that is not something that seems productive

Balarka Sen

most of measure theory is "Stuff is true if everything is nice"

which is good enough for those of us who couldn't care less about geometric measure theory

D Left Adjoint to U

4:00 AM

I think I just proved Goldbach -_-

This is getting ridiculous mon

Still Corners - The Trip [LYRIC VIDEO Spanish/English] Subtitulado Español

►

Who wants to see the 2 paragraph proof of Goldbach?

Check my profile. It's my latest post

I argue that if not Goldbach, then this monoid doesn't generate the odd numberse

I mean this generating set doesn't generate the odd number monoid

It's solid as a rock, so far

It's actually more solid than that

I would say it's the center of a black hole

?

Take the odd numbers

it forms a monoid

:D

It's all finite products of odd primes and 1

In other words both $2\Bbb{N} + 1$ and $2 \Bbb{Z} + 1$ take your choice are both monoids, this is well known

Soumik Mukherjee

Yeah sorry

D Left Adjoint to U

@SoumikMukherjee did you check my post?

Want a link?

I need someone to check it, since I was kind enough to make it 2 paragraphs, surely someone could glance at it and find the flaw.

0

Q: Extremely simple monoid generation principle for Goldbach's conjecture.

Let $M = 2\Bbb{N} + 1$ be the monoid of odd naturals. It contains all odd primes. If for some $2n \in \Bbb{N}$ we have that there exists no prime solution to $2n - p = q$, then that's the same thing as saying the submonoid $N$ generated by all $2n - q, n \in \Bbb{N}, q \in \Bbb{P}$ such that $q...

elementary-number-theory solution-verification prime-numbers open-problem goldbachs-conjecture

Thx to whoever on mse upvoted that

Soumik Mukherjee

4:34 AM

@DLeftAdjointtoU yes

D Left Adjoint to U

What are your thoughts or questions

Soumik Mukherjee

When constructing the subset, you are considering all $2n-p$ as generators, in the later argument you are saying that for each odd integer ( or better for each prime) $p$ you will have a prime $q$ such that $p+q$ is even. But this does not guarantee that $p+q$ will be the same $2n$ as you started with.

D Left Adjoint to U

The $2n - q: n \in \Bbb{N}, q \in \Bbb{P}$ are the generates of sum submonoid, right?

Since you can generate one from any subset

It doesn't have to be the same $n$ that we started with - I'm not sure what you're asking

Those are just dummy / internal variables

*some

@SoumikMukherjee

Look at the third paragraph which expounds upon the first paragraph's argument that it glanced over

Any $q \in N$ must be of the form $2n - p$ since you can't form a prime via generation ops

You can't write $q = xy$, for nontrivial x,y, so...

Soumik Mukherjee

@DLeftAdjointtoU Yes, your post precise shows that any $q \in \Bbb{N}$ will be of the form $2n-p$, so in particular any prime $q$ will be of the form $2n-p$ for some other prime $p$, but that does not imply that any even natural number will be of the sum of two primes.

you've shown that $\forall p\in \Bbb{P} \exists q \in \Bbb{P}$ such that $p+q=2n$, what you need to show is that $\forall 2n\in 2\Bbb{N} \exists p, q \in \Bbb{P}$ such that $p+q=2n$

Thomas Finley

5:00 AM

Hi guys! I was recently going through a whole new proof outline of the Schröder-Bernstein, and I never seen a proof like this. However, I am stuch at a portion in the proof, which halts me from proceeding with the remaining lines of the reasoning in the proof. Any help regarding this will be much appreciated.

I made a post. The link is:

1

Q: A confusing proof-outline for the Schröder-Bernstein Theorem.

I have seen a number of proofs of the much popular Schroder–Bernstein Theorem. But a particular proof outline interested me. My version of the Schroder–Bernstein Theorem is : Assume there exists a $1–1$ function $f : X → Y$ and another $1–1$ function $g : Y → X.$ Then there exists a $1–1,$ onto...

real-analysis proof-explanation

D Left Adjoint to U

@SoumikMukherjee thx

Soumik Mukherjee

:)

Thomas Finley

@SoumikMukherjee Hmm...Now, that I take a look, your explanation looks self-contained!

But the attempt of D Left was exciting to see, I admit.

Soumik Mukherjee

@ThomasFinley yes

Thomas Finley

But, now back to my question. Anyone out there who is interested to help me/solve my apparent issue ?

Soumik Mukherjee

5:18 AM

@ThomasFinley I don't think that you should edit so much of your question, you can keep your attempts and write down the progresses in your edit.

Thomas Finley

@SoumikMukherjee oh! I was unaware of that. But I thought it would make it cluttered and made the post fresh as new.

But what do you think about the proof given out there.

porridgemathematics

hi, does anyone know why this is true: if $ f : \mathbb{C} \rightarrow \mathbb{C}$ is lipschitz continuous with lipschitz constant $M $, and $f = 0$ outside $\{ |z| \leq \delta \}$, then $h(r, \theta) = f(r \exp(i \theta))$ is lipschitz continuous with lipschitz constant $M(1 + \delta)$?

I can see that $h$ is lipschitz continuous, but I dont know how to exactly get this constant.

there should be a simple geometric argument here

also $h : \mathbb{R}^2 \equiv \mathbb{C} \rightarrow \mathbb{C}$

thanks for any suggestions in advance

Soumik Mukherjee

5:34 AM

@ThomasFinley Are you asking what the chain $C_y$ is?

Thomas Finley

@SoumikMukherjee To be precise, I am asking for a proof-explanation for the part d.

one potato two potato

rep. theory is hard as hell! Can't understand the lecture

Thomas Finley

@SoumikMukherjee I have edited the post. I hope this makes things clearer.

Soumik Mukherjee

5:50 AM

$C_y$ can be assumed as $C_{g(y)}$, where $g(y)$ is in $X$

So you are comparing two chains $C_x$ and $C_{g(y)}$ where both $x$ and $g(y)$ are in $X$

Thomas Finley

@SoumikMukherjee And as $y\in C$ and $y\in C_{g(x)}$, so, $C_{g(x)} =C$

?

Thomas Finley

6:17 AM

@SoumikMukherjee Oh, I made a typo there. I was talking about $C_{g(y)}$. But thanks, I got it.

Soumik Mukherjee

7:51 AM

:)

PlaceReporter99

8:55 AM

Is $\infty \in \mathbb{N}$?

Shaun

I tested positive for COVID-19 this morning. So that sucks. Maybe some of the regulars here who know me might want to know, hence this post.

The self isolation frees up a lot of my time. I'll probably spend it doing Maths :)

Peter

9:10 AM

@Shaun Not nice , but nothing compared with my situation since middle of February. My 86 year old mother cannot leave her bed since then and moreover we lose much much money because the insurances barely pay anything. The chance that she can at least ever walk again is nearly zero.

Shaun

Oh no! I'm sorry, @Peter. That sounds awful! I hope she recovers soon.

robjohn

9:37 AM

@PlaceReporter99 no

@Shaun I hope it's over soon.

@Peter Sorry to hear that.

user123234

9:59 AM

I have given the torus $\Bbb{R}^2/\Bbb{Z}^2$ and a matrix which has eigenvector $(1,-2)$. I want to identify the torus with the unit square. How can I draw the eigenspace there with my eigenvector? @robjohn can you maybe help me?

robjohn

I am not sure what you mean by "draw the eigenspace". It seems that the torus is already identified with the unit square by its definition. Perhaps I am just not getting what you're asking.

user123234

@robjohn In the lecture we have drawn a line resp two lines in the unit square and our assistant told us that this is the eigenspace given by $(1,-2)$ and I don't get how he got to that lines using the eigenvector

We have drawn this here:

robjohn

Oh, that is one line on the torus

user123234

and he told us that this green lines are the eigenspace.

and I don't get how he could draw this. Could you maybe help me understanding this?

robjohn

the line is all real multiples of the eigenvector

user123234

10:09 AM

But where is (0,0) in this square?

robjohn

any of the corners

all of the corners are identified since you're modding by $\mathbb{Z}^2$

user123234

But I mean if I take the left corner below to be (0,0) then I need to go 1 in x direction and -2 in y direction?

robjohn

but that takes you out of the square

user123234

@robjohn Ah I see because the corners are either(1,0), (0,1), (1,1) and this is all equal to (0,0) mod Z^2

robjohn

yes

user123234

10:11 AM

@robjohn right thats my confusion

ahaa so whait if i start as he did in the left upper corner, then I draw it as I said, but I can only do this up to the boundary of the square so i land at the point (-1,1/2) because then I do mod 1 and start again at the point (0,1/2) and I go to the right lower corner?

Is this correct? @robjohn

robjohn

you go from the upper left corner $(0,1)$ go $\frac12$ in $x$ to $\left(\frac12,0\right)$

that equals $\left(\frac12,1\right)$ and continue from there to $(1,0)$

and you're back where you started

user123234

I think this is the same as I wrote but I wrote (0,0) instead of (0,1) for the upper left corner

robjohn

yeah, I am considering $[0,1]\times[0,1]$ as the unit square

user123234

okey thanks a lot!!

ephe

12:01 PM

Can someone take a quick look at wolframalpha.com/… ? So I've finally figured out a way to solve my problem and got an answer equal to the answer here. What I'm wondering is, is this form okay? There are some terms that I can't put inside the sum like $\ln16$ and $(x-2)[1/2 + \ln4]$. Would it be okay if I just left those as is and stated the answer as it is shown on the page?

冥王 Hades

12:27 PM

They don’t

Mats Granvik

Can someone with Mathematica please help me? What answers do you get for these two one-liners?

Limit[Sum[Log[2*k], {k, 1, n}] - Sum[Log[2*k - 1], {k, 1, n + 1/2}],
n -> Infinity]

Limit[Sum[Log[2*k], {k, 1, n}] - Sum[Log[2*k - 1], {k, 1, n}],
n -> Infinity]

(**)

The reason is that others and I suspect that the answer to the first one-liner is a bug, at least in the current version of Wolfram Alpha and Mathematica 8.0.1.

Is this true:
$$\lim_{n\to \infty } \left(\sum _{k=1}^n \log (2 k)-\sum _{k=1}^{n+\frac{1}{2}} \log (2 k-1)\right)=\frac{1}{2} \log \left(\frac{\pi }{2}\right)$$
?

ephe

@MatsGranvik I'm getting $\frac{1}{2}\log(\frac{\pi}{2})$ for the first one and $\infty$ for the second one.

Mats Granvik

@ephe Great! What Mathematica version are you using?

That is what I got too in Mathematica 8.0.1 and Wolfram Alpha.

ephe

12:44 PM

@MatsGranvik Sorry for the delay, my PC froze for a minute there. $version gives me 13.2.0 for Linux x86 (64-bit) (December 12, 2022)

Mats Granvik

@ephe What do you think, is it a bug answer?

Or a feature.

ephe

@MatsGranvik I think that's a bug

PlaceReporter99

1:00 PM

@MatsGranvik is it $\log_{10}$ or $\log_e$?

Koro

X- connected, A -connected subset of X. Then the complement of any component of X-A is connected.

(the complement is to be taken in X)

PlaceReporter99

@MatsGranvik desmos seems to suggest that the answer to that is $-\infty$

Koro

Let C be a component of X-A. Let f: X-C--> {0,1} be a continuous map. Since A is a connected subset of X-C, $f|_A$ is constant.

How do I prove $f$ to be constant?

Mats Granvik

@PlaceReporter99 $\log_e$

noballpointpen

@Shaun indeed, spend it on math. I will want to know your progress during this time. Wish you to get well.

s.harp

1:15 PM

cool interactive graphics : people.reed.edu/~ormsbyk/projectproject/posts/…

the knots get pretty wild

Soumik Mukherjee

@Koro Continuous function maps connected sets to connected sets. You already wrote that the complement of $C$ is connected.

Koro

It is to be proven that X-C is connected.

Jakobian

@Koro it's not true I think

Koro

I see, I didn't write that there.

@Jakobian NOT true if complement is taken in X-A but true if the complementation is done in X.

Jakobian

ah

Koro

1:22 PM

any ideas on how to show f to be constant. Showing that will prove connectedness of X-C.

Here is a solution math.stackexchange.com/questions/3940866/…

But I want to prove it using cont. function based definition of connectedness.

porridgemathematics

anyone know how to solve this math.stackexchange.com/questions/4712349/… ?

Jakobian

@Koro it seems to me like you want to extend $f$ to a continuous function on $X$, and to prove it's continuous, use gluing lemma?

applied to $f\restriction_{X\setminus A}$ and $f\restriction_{X\setminus\text{cl}(C)}$

oh sorry those don't cover X

Koro

How would you extend $f$ to all of X?

ephe

1:44 PM

If $\sum_{n=1}^\infty a_n$ converges and $a_n<0$ then $\sum_{n=1}^\infty -a_n$ also converges hence $\sum_{n=1}^\infty |a_n|$ converges. Then since $|(-1)^n a_n|=|a_n|$ we have that $\sum_{n=1}^\infty (-1)^n a_n$ converges absolutely.

Is there an obvious fault in my reasoning here?

Jakobian

@Koro the length of this makes me think the proof with continuous functions is going to be hopeless

at least if we don't want to fill up a whole page

ephe

I feel like I made a mistake here because I didn't even use the fact that $a_n<0$

robjohn

1:58 PM

@ephe If $a_n\lt0$ (for all $n$) and $\sum\limits_{n=1}^\infty a_n$ converges then $\sum\limits_{n=1}^\infty(-1)^na_n$ does converge absolutely.

fantasie

Hello, I have a question about Fourier inversion theorem. In this question, two approaches are mentioned about existence of inverse Fourier transform for $f\in L^2$. It appears to me that using tempered distributions is not only simpler, but also covering a broader situation.

So why does one need the other method, using Plancherel's Theorem and Carleson's Theorem? Am I missing something?

robjohn

@ephe yes, you did. You said that $\sum\limits_{n=1}^\infty(-a_n)=\sum\limits_{n=1}^\infty|a_n|$ converges, which is absolute convergence.

ephe

@robjohn Oh! That made me figure out the gap in my knowledge. I thought $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty -a_n$ both converging implied that $\sum_{n=1}^\infty |a_n|$ converges (for all sequences that is) which is false. Thank you!!

robjohn

@ephe those definitely do NOT imply absolute convergence. They are equivalent and the second tells you nothing more than the first.

ephe

@robjohn Yes those implying each other was what I was thinking when writing the proof which made me question my whole absolute convergence knowledge. Thanks for the verification!

user193319

2:15 PM

@leslietownes I guess we assume that $X$ and $Y$ are Borel spaces.

robjohn

2:27 PM

@MatsGranvik I don't know how you evaluate the $n+\frac12$ in the upper limit any differently than $n$, but the difference of the sums is the log of $2^nn!\frac{2^nn!}{(2n)!}=\frac{4^n}{\binom{2n}{n}}\sim\sqrt{\pi n}$ as shown here. Thus, I don't think the limit is correct.

gotta take the dog to the park. BBL

Hi, bye, @Ted

Jakobian

2:42 PM

@user193319 what's a Borel space?

what's the idea behind the proof that there's $2^{2^{|X|}}$ ultrafilters on infinite set $X$

apparently this is Pospíšil theorem

which I'll write as Pospisil since no such fancy letters on my keyboard

did someone just experiment or is there something deeper going on here

hmm I guess it's not as unintuitive to start with $\mathcal{P}^{<\omega}(\mathcal{P}^{<\omega}(X)) \subseteq \mathcal{P}(\mathcal{P}(X))$

and being a filter does sort of rely on finite subsets

math.stackexchange.com/a/83540/476484 I think this argument by Brian Scott is a more elaborate version of the proof I'm thinking about

Jakobian

3:14 PM

I like how everyone online calls Brian Scott a horrible teacher but his answers were actually amazing many times to me

Koro

He's amazing.

I believe he's excellent as a teacher too.

Jakobian

3:49 PM

I don't know, but there's definitely many factors to it, he may be better as a teacher for mathematicians than engineers etc.

Mats Granvik

4:00 PM

You are right. Another user told me to look at the plot:
(*start*)
ListLinePlot[
Table[Sum[Log[2*k], {k, 1, n}] -
Sum[Log[2*k - 1], {k, 1, n + 1/2}], {n, 1, 120}]]
(*end*)
which diverges, therefore the limit does not exist.

Koro

4:10 PM

Suppose that A is NOT a path connected subspace of B. Then B is also not path connected.

Therefore, I don't see the point of proof of corollary 3.8 here kconrad.math.uconn.edu/blurbs/topology/connnotpathconn.pdf

nvm

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$Mathematics$ Mathematics