Mathematics - 2023-05-30 (page 1 of 2)

2:18 AM

Proving something with sequences and dense subsets is a tough job...

2:37 AM

@noballpointpen Why? Example?

robjohn

@冥王Hades I put the work here

🤷‍♂️🙄🙄👀

4:10 AM

here is my solution to that euclidean geometry problem

4:30 AM

Yup, Make Way for Ducklings!

5:11 AM

Do I need to know field theory seriously to study Algebraic geometry?

I took a class covering field theory about 2 or 3 years ago and forgot a lot.

5:44 AM

i am 20+ years away from my last algebra class (which did a decent coverage of field theory) but the impression i got at the time was that the commutative algebra needed for AG was at least somewhat a different kettle of fish.

@onepotatotwopotato dont think so, I did fine on the first part of a two part course which covered chapters 1 and 2 in entirety of hartshorne, and assessed all exercises for each section as homework every week, with just atiyah-macdonalds commutative algebra textbook, as well as notes by gathmann

i.e. maybe necessary or at least good to know the basics, but OK to not be up on the finer points of kummer theory or whatever.

@leslietownes i think you end up using pretty much all of atiyah-macdonald if you use hartshorne

@onepotatotwopotato i would recommend getting acquainted with sheaves using a better source than hartshorne or AG in general

is there a ton of field theory in atiyah-macdonald? i have never opened that book. or eisenbud.

no there isnt i was just referring to the different kettle of fish comment

its a very thin book though

so your comment is probably still accurate

its pretty much entirely idiosyncratic after a point

the commutative algebra needed, that is

5:48 AM

I'm trying to read 'moduli of curves' but its preface says one needs to know basic algebraic curve priori.

i deliberately peaced out of algebra right around the point where i would have had to learn this.

same, although i was always in analysis, and just took two AG courses so that i feel like a proper mathematician /s

@onepotatotwopotato i would highly recommend you read raymond o.wells jr's book 'Differential Analysis on Complex Manifolds' for sheaves

even better if you read it from the start

i think you will get sheaves morally and build a strong intuition by using complex analysis as motivation

rather than jumping in with kind of unmotivated definitions if you were to use hartshorne chapter 2 section 1

its also a very short book

but yeah, if you know you are doing hartshorne, you need some commutative algebra

and honestly motivation wont help you too much in hartshorne until you get to chapter 2 (chapter 1 is just gymnastics with commutative algebra)

@porridgemathematics I am also learning AG from Hartshorne recently, about your recommendation for sheaves, do I need to learn Varieties first or can I directly go for sheaves?

you can learn sheaves independently of varieties

although knowing about varieties will give you some concrete examples of sheave

*s

for instance the sheaf of rational functions ..

but if you use something like well's book, you can learn about sheaves, and also see a nice proof of de-rhams theorem on the isomorphism between singular and de-rham cohomology in the abstract

which uses sheaves

and if you keep reading, you can learn about the hodge theorem.. now it seems like im advertising the book

but it really is very good

oh yeah, and of course, the kodaira embedding theorem too

which if youre interested in AG should be something you care about

6:06 AM

Thank you for the recommendation, I am going to try the book.

np

Silly Goose

6:45 AM

hey all. so is this notion of closure i) just happens to coincide with topological closure (contains every limit point), or ii) a completely different definition of what it means to be closed?

because to ym understanding a limit points and limits of sequences are distinct concepts

yes, it's the topological notion. closed with reference to the topology that GL(n,C) inherits from M(n.C). a topology where convergence is indeed fully captured by sequential limits.

7:39 AM

Can we have a Peanno curve fill out S^2 (2-sphere)?

there is a continuous onto map f: [0,1]--> D (the closed unit disk)

Half of disk is homeo. to northern hemisphere, and the balance half to the southern hemisphere.

Silly Goose

ah okay thanks :D

specifically, let $D_L$ be the left half of the disk (its boundary also contains diameter). Let $g_L: D_L\to S^2 $ be a homeo. to the northern hemisphere. Pasting these, define $g(x)= g_L(x)$ if $x\in D_L$ and $g(x)= g_R(x)$ if $x\in D_R$. $g$ is continuous and onto.

This should work.

that stuff about closure in the relative topology not necessarily meaning closure in M(n,C) has illustration even in the case n = 1, where GL(1,C) = nonzero complex numbers is not a closed subset of C but is a closed subset of itself and hence a matrix lie group.

I should say that $g_L$ is an embedding.

The meat of the matter is though how I created $f$.

but the f is valid because one can create such f for a square.

8:21 AM

@leslietownes There is a video on 'how to pick up a duck' and it has unbelievable no. of views on it.

:)

marmots are so cute. They are also found in India. I'll visit them one day. I never saw a marmot in real life before.

Identity generators of the group $P_f$ where $f(z) =|\sin z| +|\cos z| $ and $P_f$ is the set of all periods of $f$ i.e $z_0\in P_f \iff f(z_0+z) =f(z) \forall z\in\Bbb C$

8:41 AM

sourav: a few thoughts, not exactly hints. 2pi is certainly a period of f (e.g. thanks to the identity theorem and facts about the analytic sin and cos functions on the real axis). the question may be asking you to think about whether pi might also be a period of f, given that it is a period of the restriction of f to the real axis, although |sin| and |cos| are not analytic and so the identity theorem doesn't immediately imply that pi is a period of the complex funciton.

also, there may be some fairly close-to-the-surface reasons why f is not going to be doubly periodic (i.e. it is not going to have a period p with Im(p) nonzero). unless i'm missing something.

Ofek

hey, i found a "fractal" in collatz conjecture, and im wondering if it had been discovered before. if you look whether an odd number reaches 1 or not, you can scale it by 4 around $5/3$. in other words if $n$ is an odd number, and $m=4(n-5/3)+5/3$ is an odd number. then $n$ reaches 1 iff $m$ reaches 1.

ofek: it wouldn't surprise me if a lot of results of that general flavor are known, although i don't know of a reference or any obvious place to start looking for where things like that might be written down.

Ofek

yea, same, that's why i was asking here

9:17 AM

@SouravGhosh the fundamental period should be pi/2

as leslie said, you may prove that the only thing worthy of consideration are real periods

after that draw a circle

you should be able to see that pi/2 is a period of the manhattan distance

and there is no smaller period

So $P_f$ is a cyclic group generated by π/2

i believe so

(infinite cyclic)

Then only two generators π/2 and -π/2

isnt that just one?

Any infinite cyclic group is isomorphic to (\Bbb Z, +)

9:33 AM

oh sorry, it wont be a circle, it should be an ellipse

after you fix the imaginary argument

but it should still be pi/2 for an ellipse

9:48 AM

scratch that again, its always a circle

but the $y$ in $(sin(x+iy),cos(x+iy))$ only changes the radius

Q: Can $\mathbb{R}$ be written as an uncountable union of disjoint uncountable subsets?

0

Can $\mathbb{R}$ be written as an uncountable union of disjoint uncountable subsets? I was thinking of the following: Consider an uncountable proper subfield $F$ of $\mathbb{R}$, then consider $\mathbb{R}$ as a vector space over $F$. If we accept the axiom of choice, this vector space has a basis...

so you may as well fix it to zero

@porridgemathematics Clearly π/2 is a period of $f$ . I need to show that π/2 is a fundamental period. Then $P_f$ will be a cyclic group generated by π/2 . Hence -π/2 will also be a generator and $\{π/2, -π/2\}$ is the complete list of generators of $P_f$.

then youre just asking what real period is smallest for the manhattan distance on a circle, which is going to be pi/2

so pi/2 is the fundamental period

@SouravGhosh yeah, I just spelled out how you can do that

just notice that $|(sin(x+iy),cos(x+iy))| = g(y)$

as in it only depends on $y$, and is monotonic increasing in $y$

so that you cannot have a period $n_1 + in_2$ with $n_2$ nonzero

now just show that pi/2 is the smallest real period of $|sin(x)| + |cos(x)|$ by geometric consideration

In this question above, can we combine the arguments that Cantor set is uncountable and between any two real numbers, there are uncountably many real numbers?

9:52 AM

then youre done

Q: Can $\mathbb{R}$ be written as an uncountable union of disjoint uncountable subsets?

3 mins ago, by Soumik Mukherjee

0

Can $\mathbb{R}$ be written as an uncountable union of disjoint uncountable subsets? I was thinking of the following: Consider an uncountable proper subfield $F$ of $\mathbb{R}$, then consider $\mathbb{R}$ as a vector space over $F$. If we accept the axiom of choice, this vector space has a basis...

set-theory real-numbers

@SouravGhosh I dont understand, {pi/2} is already a complete list of generators

why do you keep insisting on its negative as also being another generator?

i mean, i thought you are asking for a basis

over Z

so thats just {pi/2}

any recommendation on functional analysis textbook @porridgemathematics?

I want to find all possible ( two in that case as $P_f$ infinite cyclic group) generators.

@onepotatotwopotato kreyszig is good

very long though

9:57 AM

@onepotatotwopotato Conway

@porridgemathematics @porridgemathematics This was an exam question where they specifically asked to write down all the generators.

oh.. i mean, surely there is only one generator though?

unless im missing something very big

If $G=<a> $ then $G=<a^{-1}>$

oh, sure

okay, thats a bit weird to me

but you are not wrong

so then the full list of generators is indeed as you say

so if P_f was a lattice

I only wrote $\frac{\pi}{2}$ and they didn't consider it correct:/

10:02 AM

wow, so you got 0 marks for that?

and full marks would be awarded for {pi/2, -pi/2}?

lol

i would have thought most of the marks would be for your steps showing pi/2 is the smallest possible period, and there are no periods with nonzero imaginary part

and maybe one point for giving the full list of generators

rather than a single possible generator

(of the whole period module)

functional analysis course, which will be held next semester, uses Rudin's book which I empirically know it's not introductory. So I asked for a recommendation btw.

@onepotatotwopotato how much FA do you already know?

@SoumikMukherjee What is your D-score?

@porridgemathematics Nothing. I only know RA in Stein book level. I'm planning to read RCA this summer.

@onepotatotwopotato take a look at the 'elements of functional analysis' chapter in Folland (its short) and do the exercises, then if youre not satisfied with your preparation for your course, look at the references folland cites in that chapter (he always goes through good references for further study on the last page of each chapter) and choose according to his descriptions and your course syllabus

since you dont know any FA, but know some real analysis

it would not be a waste of time to get a taste of FA from a good real analysis book

and he does go through a lot in that chapter

10:08 AM

I'll note that. thanks.

maybe after that you could try reading rudin if you want

Correct! A linear functional on a normed space is bounded iff kernel is closed.

iff kernel is not dense.

@onepotatotwopotato Do you know metric spaces?

in general topology? yes

Functional analysis is very interesting:)

@porridgemathematics yes, and full marks for $\frac{\pi}{2}, \frac{-\pi}{2}$

@SouravGhosh $47$, this question cost me the interview call :/

10:15 AM

@SoumikMukherjee Bad luck :(

yes:(

Anyway, all the best for your interview @SouravGhosh

@SoumikMukherjee Thank you bro :)

:)

@porridgemathematics Actually one had to only write down the answers, they may do rough works but no points were given based on rough works.

damn, thats harsh @SoumikMukherjee

also kinda... silly

what if you just use trial and error for real z, and conclude pi/2, with 0 other justification

you would be right, but you ought to be given a 0 in my book

oh, and you remember to include -pi/2 too.. lol

functional analysis isn't just metric spaces of course, but that might depend on how deep you want to go into it

@SouravGhosh non-zero linear functional

this also holds for continuity in arbitrary (Hausdorff) topological vector space

iff $f$ is continuous at some point iff $f$ is bounded at neighbourhood of some point

10:40 AM

@porridgemathematics Yes, the exam pattern is kind of silly in the sense that one can write the correct answer without having any concrete reasonings.

@SoumikMukherjee False. It is possible in case of True/False questions but you get $-2$ marks for incorrect response and $+2$ for correct response.

@Jakobian Yes :)

History of Banach Spaces
and Linear Operator by Albrecht Pietsch

jay

11:20 AM

hi folks does anyone see how the first equation on top of page 3 (page 1237) comes about? from this paper sciencedirect.com/science/article/pii/S0021782405000528 the authors say a straight forward calculation

1:11 PM

@SouravGhosh Yes, but in the other part, one can try guessing without any risk of negative marks.

1:28 PM

@SoumikMukherjee 🐼

I've got to number 2, brb

That's math, right? The "natural" numbers. Nature calls

1:49 PM

Hi @AlessandroCodenotti

PlaceReporter99

2:01 PM

Guess what $f(x,y)=x^{\sqrt{\log_xy}}$ does.

2:40 PM

how would you show $\mathbb Z[1/2]$ is not a field? seems clear that $3$ is not a unit, so I'm trying the approach of showing there is no polynomial $p(1/2)$ that equals 1/3

PlaceReporter99

@shintuku there is the constant polynomial $\frac13$.

@PlaceReporter99 how so?

if you have a concrete description of Z[1/2] as {a/2^k: a integer, k nonnegative integer} it would suffice to show that a reduced version of one of those things can't be an inverse for 3

i'll try that approach, thanks leslie

PlaceReporter99

@shintuku I don't understand fields, but I know that polynomials are of the form $c_mx^m+c_{m-1}x^{m-1}+...+c_1x^1+c_0x^0$. Note that $x^0=1$ for all $x$. The constant polynomial is the polynomial when $m=0$, which is essentially $c_0$.

2:47 PM

@PlaceReporter99 $c_0=1/3$ is not in $\mathbb Z$

or even an unreduced one of those things

hm, here we have $Z[1/2]$ as the set of polynomials in $Z[x]$ evaluated at $1/2$, so i think your approach would be the same as mine, showing no such polynomial can be equal to 1/3

if you wanna be galaxy brain about it i guess you could look for ideals in Z[x] that strictly contain (2x-1)

if you are fine with what "1/2" is i think my characterization of Z[1/2] is slightly easier to work with than yours, although it is equivalent to it

Show there's a nonzero proper ideal

oh, right because a field only has the trivial ones

at leslie: noted

3:17 PM

@shin Write it out explicitly and show that $3$ must divide a power of $2$.

Mats Granvik

4:08 PM

Suppose $p(1/2) \in \mathbb Z[1/2]$. Let $\mathbf 2=2^n2^{n-1}\cdots2$ Then $p(1/2) = z_n\frac{1}{2^n}+\cdots + z_0=\frac{z_n\mathbf 2/2^n+z_{n-1}\mathbf 2/2^{n-1}+\cdots+z_0\mathbf 2}{\mathbf 2}$. Suppose we have $p(1/2)3 = 1$. Clearly, the numerator $\mathbf z$ of $p(1/2)$ is an integer, so $\mathbf 2 = 3\cdot \mathbf z$, and therefore $\mathbf 2$ is divisible by $3$. But $\mathbf 2$ is a power of $2$ so it is even, so we have a contradiction.

therefore there is a nonzero element of $\mathbb Z [1/2]$ that is a nonunit, so it cannot be a field

Your denominator is ridiculous. $2^n$ is all you need.

leslie pointed out that $\mathbb Z[1/2]$ is isomorphic to the ring with elements $z/2^n$, but I didn't see how I could show this without anyway having to use the above denominator

You're being absurd.

Simplify $1/2 + 3/4$.

oh

.. yeah i see what you mean

4:20 PM

@SouravGhosh have you ever proved continuity of a continuous extension from dense subset? I have imagined how it would go, but to construct a proper sequence and steps which will prove the thing is tough.

thanks ted

LOL, you're welcome.

@noballpoint Tough? Come on. How did you define the extension?

btw is this the best way of showing $\mathbb Z[1/2]$ is not a field? the business with ideals seems to me to require anyway showing that the generator of an ideal is not a unit

A limit of a Cauchy sequence in complete metric space.

Howdy, Ted.

Howdy.

4:23 PM

@shintuku did you figure it out ?

So the first issue is why it's well-defined.

@DLeftAdjointtoU yeah, i used the approach of showing 3 is not a unit, but I'm wondering if there are some other methods. showing $\mathbb z [1/2]$ has a nontrivial ideal seems to me to require showing the generator of that ideal is not a unit

robjohn

@shintuku That was the approach I was thinking about

Does every ideal have a single generator?

robjohn

there's a name for that ;)

4:25 PM

@shintuku maybe $\Bbb{Z}[X] / (2X - 1) \approx \Bbb{Z}[1/2]$ might work because $X = 1/2 \iff 2X -1 = 0$ so that would give you $\Bbb{Z}$ coefficients. Then you need to show that $(2X - 1)$ is not maximal? IDK lol

Yeah. That's just rephrasing the same thing in fancy language.

at ted: if the ideal is also in Z, since Z is a principal ideal domain, we know it is principal. Otherwise, it is generated by 1/2, which is clearly a unit

But it does make it a computation with ideals in $\Bbb Z[x]$.

Otherwise? So why are there no other ideals?

hm

If $(2X - 1)$ were maximal then there exists $a \in R, \notin (2X - 1)$ such that $\Bbb{Z}(2X - 1) + \Bbb{Z}a = 1$ or $\alpha (2X - 1) + \beta a = 1$ in which case:

4:31 PM

BTW, @noballpoint; your statement omitted an important hypothesis, I guess.

@DLeftAdjointtoU There exists? How 'bout for all?

@TedShifrin yes that's right!

Uniform continuity of the first function?

Right :)

Not sure how to prove that $(2X - 1)$ is contained in a proper ideal...

Well, to go back to our earlier example. You need to show, for example, that $(2X-1,3)\ne (2X-1)$, i.e., that $3\notin (2X-1)$.

Q: If $A_n$ is a countable set for each $n ∈ \Bbb N,$ then $\cup_{n=1}^{\infty}A_n,$ is countable.

4:35 PM

Hi guys! It happened so, that I was going through a proof of the theorem : "An infinite union of countable set is countable" , but there was a portion which appeared to me to have lacked clarity.

1

I was recently, going through a proof of the following theorem, which states that: If $A_n$ is a countable set for each $n ∈ \Bbb N,$ then $\cup_{n=1}^{\infty}A_n,$ is countable. [Definitions that are used in the proof: Countable Set: A set A is countable iff $\exists $ a bijective mapping $f:...

real-analysis proof-explanation

reading chat then suddenly WALL

Oh ic now. :D When you multiply a formal univariate poly by $2X - 1$ the result is always non-constant unless $0$.

fight fight fight

@DLeftAdjointtoU but that doesn't guarantee it is proper yet, i think, since it could be a subset of only the unit ideal

@Shintuku that's right

new upvote/downvote design is ugly to me

6

4:38 PM

@Jakobian what about it

It's bolded or something

it's an arrow inside a circle

illuminati

I sorta like that.

@ThomasFinley which portion?

I like my buttons round and there's are as round as they can be

4:40 PM

They haven't changed any design from 2008.

I like big buttons and I can not lie, those other algebrothers can't deny...

I believe that "union of countable amount of countable sets is countable" uses axiom of choice of some kind

Why?

Because in ZF it's consistent that R is a countable union of countable sets

I wouldn't know that.

4:43 PM

what's the definition of countable

bijection with N

Countable sets are small ( sigma ideal) in sense of cardinality :)

I was once told that most logicians don't consider it AOC unless it's in the uncountable situation.

Is the bijection taken as given or must one choose one?

oh I see the issue

but there's people that distinguish finite countable from infinite countable, finite countable is injection with N

4:45 PM

Howdy @Thor. Long time no see!

if we suppose countable the bijection is given

I don't want to bother with finite countable, personally.

@Jakobian Countable sets have outer measure 0 and countable union of countable sets also have measure 0.Hence outer measure of R is 0.

hey @Ted, doing alright?

If you have an injection $\coprod A_n\to\Bbb N$, then you certainly get an injection $\bigcup A_n\to\Bbb N$ by using the well-ordering principle.

4:51 PM

@noballpointpen Uniform continuity is needed in the hypothesis. For an example : f:\Bbb Q \to \Bbb R defined by $f(x) =\frac{1}{x^2-2}$ is continuous but doesn't have any continuous extension on $\Bbb R$ .

I always assume axiom of choice to avoid confusion. Axiom of choice is too natural, and I sometimes just don't know when I'm using it

@Thorgott Yup.

Yes, Sourav, I just didn't bother to mention full statement :)

The numerous times I taught point set topology using Munkres's book, every time I got to a point where, after he discusses AOC, he says "we already slipped an application of AOC by you in a previous section; where was that?" You'd think I would have marked it in my book, but every time I had to stop and search.

glad to hear :)

4:54 PM

Hope you're doing well, @Thor. Almost done with school?

I think the issue is as you suspected that we need to choice to choose countably many bijections with $\mathbb{N}$ in the first place. Everything else workds "by hand".

@ThomasFinley By the way, this was meant for you.

@noballpointpen Then you can use the result: uniform continuous function preserve parallel sequences. ( Two sequences (a_n), (b_n) are parallel iff d(a_n, b_n) \to 0)

That's why I asked what the definition of "countable" is, @Thor. Is the bijection provided or must it be chosen?

@Sourav In fairness to noballpoint, he was worrying, I believe, about proving continuity. I raised the issue of well-definedness.

@TedShifrin Yeah, also doing great. I'm having my first relaxing semester in a while, which is nice. Looking to finish my Master's next semester.

4:56 PM

I figured it would be soon.

I think from how we define it it only requires a brief thought to know that it is well-defined.

Oh, proof by intimidation? Clever.

Ha-ha, no.

Suppose a point is the limit of two different Cauchy sequences.

I didn't just write "let $g$ be a function that is defined to limits blah-blah".

5:03 PM

Let $ (a_n), (b_n) $ both converges to $x$.

Sourav, not you.

Then $d(a_n, b_n) \to ( what? ) $

Hm...

Then $d(f(a_n), f(b_n)) \to (what?)

So, then, what is so tough about the continuity? I assume you're worrying about approaching a fixed element that doesn't belong to the dense subset by others that do not.

5:08 PM

@TedShifrin Ok sir.

The whole thing about "density" that each element is surrounded by all other elements which are either limits or in the dense set somewhat scares me. I usually first imagine proofs related to spaces and then write down ideas rigorously instead of going right down writing rigorous inferences like a machine.

Are you insisting on using a sequential characterization of continuity or are you using $\delta$-$\epsilon$?

First.

Tried second too.

Do you think second is easier?

I think the second is less messy, yes. The only interesting case is approaching by points not in the original subset.

I imagine you're going to want to split $\epsilon$ into pieces either way.

I think that's what I imagined yesterday and haven't written down yet.

5:18 PM

Sketch of the proof: $f:D\subset X \to Y$ is an uniformly continuous function where $D\subset X$ is dense and $Y$ is complete metric space.

So that we can "encompass" each $g(x_m)$ (where $x_m \to x$) in shrinking neighborhoods of $g(x)$ with the help of the original function.

Let $x\in X\setminus D$ then $\exists (a_n) \subset D$ such that $(a_n) \to x$

From Cauchy sequences converging to the various $x_m$ you can use the Cantor diagonal trick to make a single Cauchy sequence converging to $x$.

Define $F(x) =\lim_{n\to\infty} f(a_n) $

A good opportunity to work with his diagonal trick I think, since I haven't ever used it.

5:23 PM

$(a_n) \subset D$ and $(a_n) \to x$ in $X$ implies $(a_n) $ is Cauchy in $D$ . Since uniformly continuous functions are Cauchy continuous, $f(a_n) $ cauchy and by completeness of $Y$ , the sequence $f(a_n) $ converge to some $y\in Y$.

There may be another sequence $(b_n) \subset D$ such that $(b_n) \to x$ .

But $f(b_n) $ doesn't converge to a different limit( due to uniform continuity of $f$).

Claim: 1)$F$ is uniformly continuous.

2) $F$ is the unique uniform extension of $f$

I find that a lot of discussions happened with my post. To clarify, my definition of countable set is a set $A$ such that a bijective mapping exists from $\Bbb N$ to $A.$

The proof I wrote out in the Original post, has a liitle ambiguous statement written when they are considering the case that all of $A_n$ aren't disjoint.

5:43 PM

@Thomas I suggested a better way to avoid that.

$A$ is countable iff there exists an injective map f: A\to \Bbb{N}$

@ThomasFinley According to your definition finite sets are not countable :(

I approve of that, Sourav.

@TedShifrin Yes, to be frank I think I am aware of the line of proof you want to imply. But the thing is, I am much more interested in the reasoning the author wants to suggest to the readers, because I never ever saw this proof before and to be honest it mesmerized me, until I encountered the last case of the proof, which makes me much eager to understand it.

Q: When a function having the Darboux Property represents a function of baire class one?

2

$D(0) =\{f:[a,b]\to\Bbb{R}: f \text{ has the Darboux Property}\}$ $D(1) =\{f:[a,b]\to\Bbb{R}: \exists (f_n)\subset D(0)\text{ and } f_n \overset{\text{pointwise}}\to f\}$ $B(0) =\{f:[a,b]\to\Bbb{R}: f \text{ is continuous}\}$ $B(1) =\{f:[a,b]\to\Bbb{R}: \exists (f_n)\subset B(0)\text{ an...

real-analysis general-topology measure-theory descriptive-set-theory pointwise-convergence

The proof has several sloppy things. First, you don't put set symbols around $A_1\cup\dots \cup A_n$. Second, you should not be looking at $A_n - (A_1\cup\dots\cup A_n)$, but obviously instead $A_n - A_1\cup\dots\cup A_{n-1}$.

I also don't see how you can just ignore the sets $B_n$ that happen to be finite. They are part of the union. Suppose there are countably many of those finite sets.

5:59 PM

@TedShifrin I think you might want to check out my alternative argument(, if it interests you) which somehow coincides with the author's intent. I added it in the original proof.

That's what I'm complaining about. You said to ignore all the finite sets $B_n$. I disagree that that is allowed.

@TedShifrin No, I am not ignoring them. I am just ignoring the empty $B_i's$ that are empty by keeping the corresponding $ith$ row blank. If $B_i$ happens to be finite then my idea is to simply map the $b_{ij}'s$ to the natural number in the $ith$ row and $jth$ column of the arranged array of $\Bbb N.$

I see. "We don't bother with this happening" is not very instructive.

I don't understand. When there is no number in the $j$th column (and past) of $B_i$ what do we do?

@TedShifrin It was when I suggested that there are some natural numbers which are not mapped to any of the $b_{ij}'s$. We then, delete those natural numbers from the array (of $\Bbb N$ ) and conclude, that there exists a bijection from an infinite subset of $\Bbb N$ to $\cup B_i$.

After that, we use the fact that as an infinite subset of $\Bbb N$ is countable, so is, $\cup B_i.$

OK. It is not clearly written.

I think I prefer my approach. :)

6:11 PM

Find the cardinality of the set $\{A\in Gl_2(\Bbb Z_5) : A=LU\}$ where $L, U$ are lower and upper triangular matrices.

|Gl_2(\Bbb Z_5) |= 480

I thought every invertible matrices have LU decomposition.

@TedShifrin But can you judge it's validity ? To be precise, what do you think of my approach ( ignoring how it's presented though, as it seems to be not so appealing) ?

What about $\begin{bmatrix} 0&1\\1&0 \end{bmatrix}$, Sourav ?

But i have to take care of "row exchanges" i.e permutation matrices. It is true that every invertible matrices have PA=LU decomposition.

@TedShifrin Exactly. It doesn't have any LU decomposition.

Isn't every matrix of that form (if you allow $0$'s on the diagonal of $U$)?

The whole point is exposition, @Thomas. It's clear that the result is correct.

Your approach is quite equivalent to mine (we throw away every instance of a repeated element after the first appearance).

I am trying to prove that $A\in Gl_n(F) $ is of the form LU iff leading principal minors of $A$ are non zero.

6:18 PM

Sounds believable.

@TedShifrin Diagonals of U are pivots.

Oh, that's not the usual convention. So in that case you need the matrix to have maximal rank.

@TedShifrin Yeah. I agree. I can now safely replace that ambiguous "thing" with my version of proof, and I genuinely feel that it will make the proof look good.

Here is the proof (math.stackexchange.com/a/2955693/977780)

2×2 principal minor is non zero iff $A$ is invertible.

1×1 principal minor is non zero iff $a_{11}\neq 0$

Done!

There are $80$ invertible matrices with $a_11=0$

7:32 PM

is there a finite subring in the integers?

A subring is an additive subgroup with some other conditions.

yeah, i'm testing whether $\{0,2,4\}$ with mod 3 remainder 1 = 2 and mod 3 remainder 2 = 4 is a ring

Is there a finite subgroup of $(\Bbb Z, +) $ ? Is there any non trivial one?

shintuku, how many exercises you do per day during your self-study sessions?

Note: Every additive subgroups of $\Bbb Z $ are cyclic and $ (\Bbb {Z}, +) $ is a torsion free abelian group.

7:42 PM

@SouravGhosh I'd think $\{0,1,2\}$ would be a finite subgroup of $(\mathbb Z,+)$, but I'm not sure it can become a subring if it preserves the operations of $\mathbb Z$

@shintuku Think about my previous comment.

@shintuku Is the set $\{0, 1,2\}$ closed under addition?

no, but modulo 3 yes

Those are not integers.

@shintuku Then you can't call "Subring".

hmm i see

@SouravGhosh thanks for the help

7:48 PM

Equivalence classes of integers are not integers. Very important to write $\bar 2$ or $[2]$ for elements of $\Bbb Z/3$.

hmm right

i thought about changing the operations so that it might be a set of integers, but then it doesn't preserve the operations of $\mathbb Z$ which is a requirement for a subring

Indeed.

so, suppose there are. Then either they're a set of equivalence classes or the operations must be modulated. In the former case, it's not a subset of integers, in the latter, it does not preserve operations. So there are none.

Just take the smallest positive element and look at additive subgroup it generates.

Every additive subgroups of $\Bbb Z $ are cyclic and $ (\Bbb {Z}, +) $ is a torsion free abelian group.

7:52 PM

@noballpointpen oops just saw this. currently rushing to try to prove the nullstellensatz before i need to start my coursework again, so doing as much as i possibly can. when i have econ coursework I manage my daily load with Anki

Like the proof that we have a PID.

Only finite additive subgroup of $\Bbb Z$ is the trivial one,$ \{0\}$

at sourav: I haven't seen the concept of torsion yet

If we estimate progress by finished exercises, how much in a day would be a good estimation, Ted? Considering exercises that are not repetitive "apply the learned technique" which are done in like 5-10 mins depending on technique.

at sourav: but that's an important fact, I'll try to see how come the only finite subgroup of $(\mathbb Z,+)$ is the trivial one

7:55 PM

@shintuku Ok. $G$ is a finite cyclic group of order $n$ iff $\exists a\in G$ such that $|a|=n$

right

$G$ is torsion free means identity is the only element of finite order.

so $(\mathbb Z,+)$ would be torsion free trivially? since there is no element of finite order

Yes, except the identity element.

oh, right right

7:59 PM

1) Every subgroup of $(\Bbb Z, +) $ are cyclic.

right, subgroups of cyclic groups are necessarily cyclic

2) There exists a finite cyclic subgroup of $G$ of order $n$ iff there exists an element $a\in G$ of order $n$

I also would like to point to my question you, guys, shintuku and Sourav.

@SouravGhosh ahhh, i see

thanks, very appreciated

I have discovered a theorem (😁) : A torsion free cyclic group doesn't have any non trivial finite subgroup.