Mathematics - 2023-05-23

Ted Shifrin

00:09

She probably wants to explode the debt ceiling, too.

Akiva Weinberger

00:23

@SoumikMukherjee Nice. So now the values of the maximums?

Ted Shifrin

00:41

Ugh @ mums

shintuku

01:22

can i find a list of prime ideals of Z[x] anywhere?

Ted Shifrin

There can be no “list.”

shintuku

i mean a couple of examples

there's $\langle x^2 +1 \rangle$ and $\langle x \rangle$

Ted Shifrin

This is what happens when you skip a year of algebra.

shintuku

there's nothing on prime ideals in any first year of algebra books :(

Ted Shifrin

Of course there is stuff on prime ideals. Look at a good book. There are all prime ideals of $\Bbb Z$ and all the irreducible polynomials. What about combining?

shintuku

01:36

seems likely that those might generate prime ideals

Ted Shifrin

So go make your list.

shintuku

ah

that's only true in integral domains

and i needed: in an integral domain, every prime is an irreducible

thanks ted

Ted Shifrin

But there are lots of those!

leslie townes

02:14

hahaha. thanks ted. still waiting on the list.

ted deliberately withheld prime ideals from the regular version of algebra book. he released it as DLC that you had to buy through his website. it was mostly just an inducement to get your credit card number.

冥王 Hades

Didn't know Ted was a gamer like me

shintuku

the discussion on prime ideals is secretly disguised under titles resembling 'factorization of polynomials' and you have to secretly know prime ideals correspond to irreducible polynomials

leslie townes

shin in all seriousness i think ted is right, most algebra books give at least examples of prime ideals in polynomial rings. at least any book that discusses ideals at all and not just group theory.

冥王 Hades

Unrelated but, since when can a professor tell me not to eat chocolate because "your teeth will fall out"?

leslie townes

fraleigh's intro to abstract algebra does, at least. its got like a whole section on it.

my daughter's dentist said something similar about candy. i don't understand the incentives. you'd think a dentist would profit from that happening.

shintuku

02:18

actually most books deal with it like fraleigh

they have a section called prime and maximal ideals

and speak of maximal ideals the whole way through and only look at finite rings

冥王 Hades

@leslietownes that's why they give you the lollipop at the end

But I can't quit chocolate. I'm addicted

shintuku

but if you realize the secret code indicated at :63629608 you will once again find the plenitude of prime ideals everywhere

and i've just now found the gem in Artin: f is irreducible in Z[x] if and only if f is a primitive polynomial irreducible in Q[x]

leslie townes

02:43

that's in fraleigh too, although i had forgotten how he sprinkles the stuff about polynomial rings across a bunch of different sections

shintuku

fraleigh is cool because there's a solution manual. instant 10 stars

leslie townes

he also added some crap about groebner bases in some edition since the one i used in college. that's as close as you can get to trying to chase a trend while also writing a first semester algebra book.

Sourav Ghosh

02:56

@shintuku Gauss lemma :)

shintuku

yes :sunglasses:

weren't there a couple of emoticons available on chat?

is there a sunglasses emoticon

leslie townes

😎

shintuku

!!

how??

leslie townes

🤐

shintuku

i have bookmarked your reply

now i can use my most used emoji 😎

leslie townes

03:06

i dunno what device you're on, on my version of windows, the windows key + the period key brings up an emoji menu that generates the input

dunno how to just type it in a la discord

maybe it's just looking into some subset of unicode?

shintuku

i've come across unicode codes through my entire lifespan on the internet, and each and every time it has been a headache to deal with

copy pasting is the solution 😎

D.C. the III

03:35

Working through the first chapter of Munkres's Topology. Currently doing the section on relations. An example that was given was "Define two points in the plane to be equivalent if they lie at the same distance from the origin."

If I understand this correctly: The equivalence relation is the distance between two points from the origin. The equivalence relation is acting on the set $\mathbb{R}^2$ (i.e the plane). The collection of equivalence classes would be all circles centred at the origin. And each equivalence class is defined by the distance the points in that set are from the origin. ?

shintuku

the equivalence relation is defined $a \sim b \iff |a| = |b|$

yeah, an equivalence class is a circle

D.C. the III

Yea that's what I meant but I was trying to describe in words.

So the relation is being applied to the plane?

shintuku

i don't know what that means

i.e., is it partitioning the plane? yeah

D.C. the III

Just definition unravelling: A relation on a set $A$ is the subset $C$ of the cartesian product $A \times A$.

then of course equivalence relation has all of its necessary properties as well

@shintuku reading things I think this is an equivalent way of saying it

shintuku

hm i think partitioning is something more than what you've just said, you've given the set-theoretic definition of relation

D.C. the III

03:43

The definition of partion is a collection of disjoint nonempty subsets of A whose union is A

shintuku

i wasn't sure what you were saying by 'applied on the set'

one potato two potato

If $F$ is holomorphic on $\Bbb C$ and $F\in L^p(\Bbb C)$ then $F\equiv 0$? i.e., $F$ is bounded?

shintuku

@D.C.theIII yeah

D.C. the III

just trying to unravel things..........everything is set this set that....

so each circle would be their own equivalence class then?

shintuku

if you take the points in an equivalence class defined with that relation they draw a circle

D.C. the III

03:49

so the distinguishing feature that makes the equivalence classes disjoint is the distance, i.e radius of the circle.

shintuku

the distance of an arbitrary element from the origin yeah

D.C. the III

ok. cool. thnx

shintuku

np

Thomas Finley

04:46

Hello everyone! Can anyone take a look at this:

0

Q: Prove that $\Bbb R$ satisfies Dedekind Theorem $\implies$ $\Bbb R$ satisfies Cantor's property of Nested intervals.

I was recently looking at the proof of the fact that " The assertion that the real number satisfies Dedekind's characterisation of cuts is same as saying that it satisfies Cantor's Theorem on nested intervals. To elaborate on this, I am trying to understand the validity of this: $i.$ Let $A,B\in...

Soumik Mukherjee

05:59

@AkivaWeinberger $1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}$. Wow, I wasn't expecting a pattern!

shintuku

any faster way to solve $x^3 \equiv 6 \mod 7$ than trying $x=1, \dots, x=6$?

leslie townes

shin: is there anything faster than evaluating seven expressions mod 7? no

looks like you used your noggin to skip x = 0, but even just not skipping it might have been as fast

shintuku

but what if we can gain speed, even more speed

ok i'll compute them manually

leslie townes

there are different schools of thought on this, but if an algebra exercise can be solved in six lines without using essentially anything, my first instinct would be to do that first, and look for cleverer stuff later if at all

there are few things in life simpler than evaluating this polynomial at six or seven values, even without the "mod 7"

copper.hat

use brute force where possible

shintuku

06:06

words to live by

leslie townes

less facetiously i think (1) sometimes people use "but i don't know some complicated way of doing it" as a weird kind of procrastination for avoiding just attacking something that they definitely do know how to solve, (2) if there's one thing everyone should have learned from calculus class, it's that a few lines of hand calculation is not a big deal, and six is not even "a few lines"

consider the mountains of ink people will spill over a multivariable integral, plus drawing a picture

what crap

copper.hat

there is that story of Von Neumann involving trains and a fly...

@leslietownes bad day?

leslie townes

nah, multivar calculations are crap even on a good day

copper.hat

i am only dealing with single variable problems from now on

unless it involves two variables like driveway completion and total $ paid

leslie townes

solving x^3 = a mod p for p prime and a given, or q(x) = a mod p for p prime, a given, and q(x) some polynomial, is more interesting and maybe time for theory. at some level of generality, "solving polynomial equations mod p" is as generally hard as algebraic geometry or cryptography or whatever though

copper.hat

06:11

discrete problems are always harder. i prefer my convex world...

leslie townes

i was just flashing back to grading multivariable calc homework above, even simple problems often involve a good deal of ink for setup, and that's if you do it correctly

do it incorrectly and maybe you get a full page of crap

copper.hat

indeed.

leslie townes

i also flashed back to one semester, i dont think it was multivar, where i had a grader that i couldn't use because he was incompetent. he would mark correct things wrong and miss obvious mistakes. and for some weird reason the department couldn't replace him with someone else or fire him, and it ended up being more work than not having a homework grader at all

he would have found a way to evaluate 0^3 mod 7 and get 1, or to ding the class for skipping the check of that important case

Ted Shifrin

06:38

@shintuku Of course. Look at $x^3 \equiv -1$.

leslie townes

ted: munchkin somehow went the whole day without getting into trouble.

Sourav Ghosh

07:19

The cardinality of the class of Borel sets in \Bbb R is 2^c

My idea: Borel sigma algebra generated by open intervals.

Hence a Borel set can be obtained by the process of countable union, countable intersection and set complementation.

There is a bijection from the class of all Borel sets and \omega^{\omega}

Countable product of countable sets have cardinality c=\aleph_{1}

Am I right?

porridgemathematics

its hard to say because your proof seems to be a bunch of true statements with no implications in between them

leslie townes

07:35

if the idea is to envision some not-too-big set of labels and then somehow show that every borel set is tagged by one of those labels because of the limitations on the processes that generate all borel sets, OK

but translating that vibe into a theorem seems to be like, 99.99% of the work

porridgemathematics

yeah..

lol i was about to say, you just reworded 'explicit a bijection'

leslie townes

and its not clear to me that you get a bijection out of that, and not just an upper bound on the cardinality (maybe to some extent this is not an issue, if you also have a lower bound and choice making cardinal arithmetic 'nice')

it wouldn't surprise me if the clean answer somehow depends on or needs AC

porridgemathematics

it will depend on choice

that is if by clean answer we mean using what is essentially transfinite induction

leslie townes

by 'clean' i meant by actually identifying the cardinality as 2^c and not just 2^(something bigger than aleph_0, but maybe not much bigger)

but also that

porridgemathematics

ah okay

leslie townes

07:39

of course this has been asked on MSE a hundred times; math.stackexchange.com/questions/1768569/… which seems specifically directed at one of the answers of math.stackexchange.com/questions/70880/… or also math.stackexchange.com/questions/1954108/…

sourav, to make my own personal reaction (a long story) shorter, i definitely think that something like that template works, and that the devil (if any) may only be in the details

some of the answers above seem to be following roughly that program, but maybe in reverse (e.g. constructing the inverse of a bijection that you are looking for)

porridgemathematics

sorry if my reply came off as condescending lol, i was trying to say without the benefit of the doubt of knowing the proof follows those lines (but with way more precision), what you wrote really does read like a bunch of non-connected true statemnets

I guess when you said \omega^{\omega}, a very reasonable grader would award benefit of the doubt points

(because borel codes)

leslie townes

i don't remember enough about cardinal arithmetic to know about what omega^omega is (if that even is cardinal arithmetic) but it seems like any line that says "there is a bijection from the class of all borel sets to X" is basically the guts of the proof

where once you have a non-"borel" description of X it is cardinal arithmetic the rest of the way

porridgemathematics

huh, but saying there is a bijection to \omega^\omega says that the cardinality of the set of all borel sets is |\omega^\omega|

\omega here is just the first infinite cardinal

i.e. |\mathbb{N}|

but then that line is just saying that the cardinality of all borel sets is |\mathbb{R}|

because \omega^\omega is the cardinality of the set N^N

so its the guts of the proof because.. its the statement you are trying to prove??

so how could that be the cornerstone?

its not even expliciting what the bijection is

its just saying there is one

thats just answering prove x is true by saying x is true to me.. unless im sorely missing something

imo you need to really read things that arent being said to say this is close to a proof, unless 'there is a bijection between class of borels and N^N' was not what Sourav said

leslie townes

uh i think we agree entirely

when i said that the one line was "the guts of the proof" i meant to imply, that is what you need to explain in order to have a proof

porridgemathematics

the idea he has being its natural to think of the set of all sequences (say binary sequences) and borel sets, because of how they are generated, is closer imo to a proof

leslie townes

07:50

that's the vibe that needs to be turned into a theorem

porridgemathematics

ah okay

so uncooked guts

and the cooking still needs to be done

Sourav Ghosh

I thought we can somehow rearrange the indices of the open intervals associated to a Borel set but repeated application of countable union and countable intersection makes me very uncomfortable.

porridgemathematics

@SouravGhosh that is the idea

google transfinite induction and borel hierarchy

leslie townes

more generally, if you say you have a program for proving that the cardinality of Y is X, the line in your program in which you say "there is a bijection between Y and X" is where the substance, whatever it is, will be

porridgemathematics

its precisely that idea formalized

i guess this is just a matter of semantics

for me that line is the thing that the program is proving

because I literally read the cardinality of Y is X as 'there is a bijection between Y and X'

leslie townes

07:53

well in the above, i didn't remember enough about cardinal arithmetic to know about whether 2^c and omega^omega were the same thing

so it's more like "the line in the proof where you say there's a bijeciton between [the thing in your statement] and some other set" is the guts of the proof

whatever that other set is, if that's what you're not sharing with me, that's what you need to share

porridgemathematics

oh i didnt see that

that isnt true anyway

2^c is the cardinality of the power set of the continuum

Sourav Ghosh

Cardinality of N^N

porridgemathematics

and omega^omega is the cardinality of the continuum

leslie townes

OK

for avoidance of doubt, this was a line in the original program

32 mins ago, by Sourav Ghosh

There is a bijection from the class of all Borel sets and \omega^{\omega}

porridgemathematics

yeah that is true

leslie townes

07:55

the action, whatever it is, "guts" you want to call it, seems to be in this line

and the rest seems to be "mere arithmetic" i.e. once you've pinned that down you can forget about what borel sets are

porridgemathematics

but his first line is wrong then, because the cardinality of borel sets should be c, not 2^c

and that is important, because it is clear that omega^omega and c are the same

so that line is the same as what he should be trying to prove

c = 2^omega

so that 2^c is a typo

he meant 2^omega

so its not as delicate and pedantic to point out anymore

Sourav Ghosh

Right. It should be c not 2^c

porridgemathematics

because now the set youre referring to is just N^N

and him making the bijection to N^N is the whole proof

the rest of the arithmetic is ala georg cantor

who showed |N^N| = |R|

one potato two potato

If $f:\hat{\Bbb C}\to\hat{\Bbb C}$ is a local homeomorphism, then $f$ is actually a homeomorphism.

Sourav Ghosh

Sorry for my mistake, it should be c not 2^c in the very first line. I am trying to show that the Cantor set has a non Borel subset.

porridgemathematics

07:59

yeah fair enough

you dont need this then @SouravGhosh

you just need that the cardinality of the borel subsets of the real line is less than or equal to that of the real line

Sourav Ghosh

@porridgemathematics Really?

porridgemathematics

you dont need equality.

yes

Sourav Ghosh

Only surjection is enough

porridgemathematics

yea

Sourav Ghosh

So my task is to show that given any sequence of natural number there is Borel set index by the sequence.

porridgemathematics

08:02

oh wait, im being dumb @SouravGhosh

you do need to have equality :D

sorry!

the argument is just taking the power set of the cantor set yields something with cardinality 2^c

and the borel sets have cardinality at most c

hence there is a non borel subset of the cantor set

oh wait, im being dumb x2, so not dumb, you only need a surjection

sorry again, the 2^c thing made my head spin

@SouravGhosh and yes

thats exactly right

well that there is a borel set indexed by that sequence, and all borel sets are indexed by some sequence

so two things.

Sourav Ghosh

I think in that step I need the AC or more specifically the Axiom of Countable Choice

porridgemathematics

yeah, probably.

well the whole surjection thing implies cardinality is less than or equal to requires AC

and idk anything about countable choice but im guessing its the good enough version for most of math

so yeah

porridgemathematics

08:20

hold on actually , i think I can think of a proof that requires no choice at all.

maybe it is implicitly using choice somewhere though..

Thomas Finley

Hello, can anyone provide me any info about the magazine "Crux Mathematicorum ". To be precise, I came accross this recently, and I want to contribute to it, by solving some problems. But the thing is, I dont know where to start. To be precise, I am new to this and I dont even know where to look for problems whose solutions are needed. I browsed a lot of pages, but wasn't able to get some clarity to it. Is there any deadlines for this submission?

It would be helpful, if someone shares anything about this topic. Thank you!

leslie townes

thomas, forgive me if this is something you have reviewed already, but cms.math.ca/publications/crux/information-for-contributors ?

i don't know this journal specifically, but on skimming this, it looks like this journal might have a feature i have seen in some other places, which is, they publish problems in issue X, and readers submit solutions, and then maybe their editors pick one and publish it in issue X+1 or X+2, perhaps with a list of people who also solved the problem.

and it looks like they also have an article submission process, on top of or separate from whatever problem solving stuff that they do.

Thomas Finley

@leslietownes ah, I get it. The thing is, I have to look at their current issue. After that, the steps are obvious i.e solve it, if done, then send it and wait.

leslie townes

if that's the case, if you want to solve the problems in a way that's interactive with the journal, you might want to focus on the most recent problems, where any deadlines for submitting solutions hasn't passed yet.

yeah.

if you've found solutions to older problems that seem to introduce new ideas or be more interesting than the one they published, maybe they'd be interested in that as a "problem proposal," in which you pitch them a generalized or modified problem that yields to your new method, but does not yield to some modification of whatever solution they published previously.

Thomas Finley

08:36

@leslietownes You are right. But the point is where to look up for the deadlines? I think it will be better if I gaze over the recent issue. I might find something. Who knows?

leslie townes

this kind of thing is sometimes complicated by things being behind paywalls, or not being entirely online, but yes, that's where i'd look.

Thomas Finley

@leslietownes thanks! I found it.

Peter

08:52

Is $F_{285}$ the largest fibonnaci-number which is NOT pandigital (that means that there is at least one missing digit in the decimal expansion) ? Until $F_{10^5}$ , this is the case.

Akiva Weinberger

12:02

6 hours ago, by Soumik Mukherjee

@AkivaWeinberger $1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}$. Wow, I wasn't expecting a pattern!

@SoumikMukherjee Nice! That was exactly what I wanted you to find. Do you think it continues?

Sourav Ghosh

Is the punctured plane \Bbb{C}\setminus\{0\} an annulus? Can outer radius of an annulus be \infty ? I mean r, R\in [0, \infty]

porridgemathematics

12:54

@SouravGhosh what doyou mean by "is an", the punctured plane is homeomorphic to any annulus. But since you talk about \mathbb{C}, you probably want to know if its conformally equivalent to a non-degenerate annulus. The answer is no !

Soumik Mukherjee

@AkivaWeinberger I am thinking about that.

Sourav Ghosh

13:12

@porridgemathematics I want to know the definition of an "annulus".

😪

$A(a;r, R) =\{z\in\Bbb C : r<|z-a|<R\}$

What is $r, R $ here? Non negative real or non negative extended real.

one potato two potato

@AkivaWeinberger is it real?

porridgemathematics

@SouravGhosh it really depends on the context, the degenerate case represent singleton conformal equivalence classes (r=0, and R=infinity respectively), and the third degenerate case when both r=0 and R = infinity is another singleton conformal equivalence class

you should be able to prove that the degenerate cases are all not conformally equivalent to each other, or to any non-degenerate annulus

so the only interesting cases in some sense , is to classify all non-degenerate annuli up to conformal equivalence

Sourav Ghosh

13:35

@porridgemathematics Thanks :)

shintuku

@TedShifrin smart!!

it at least immediately gives us an additional $x = -1$

$x = -2$ also almost immediately

$x = 3$ is the next easiest one

there, that's all i needed for my polynomial factorization

shintuku

14:25

prime ideals that are not maximal in $\mathbb Z[x]$ cycle over either the constants or the polynomials. if they are maximal, they cycle over both. i.e., in the prime nonmaximal case they either partition the polynomials or the constants, and in the maximal case they partition both. that's why $R/I$ is a field if $I$ maximal or just an integral domain if $I$ prime not maximal

I See The Light

i bet this has an interpretation in terms of lattices

*nontrivial partitions

Thomas Finley

15:18

Hi guys! While solving a real analysis problem, I found a strange solution. The thread is math.stackexchange.com/questions/2020407/…. I think the author intended to post a solution applying Cauchy's Convergence Criterion, but what is implied is, "$u_n$ is a Cauchy Sequence, iff $\forall \epsilon \gt 0$, $m\in \Bbb N$ such that $|u_{n+1}-u_n|\lt \epsilon$ , holds true for all $n\geq m$"- which is incorrect.

Am I missing something in here ?

Ted Shifrin

15:54

@ThomasFinley The discussion is incomplete as it stands. Estimate $|x_n-x_m|$ by using the triangle inequality .

Thomas Finley

@TedShifrin I found out the solution, but indeed, Cauchy sequences makes it lengthy, if not convoluted!

I have a better solution though, using MCT's and some elementary algebraic manipulations.

Incomplete discussions like this without apparent extra context is tiring.

shintuku

$A$ maximal ideal of $R$, $b \notin A$. Are there $a \in A, r \in R$ with $1=br + a$?

or: is there $r$ with $br = 1-a$?

almost

one potato two potato

16:18

$R/A$ is a field

shintuku

i'm trying to prove that

the usual proof goes: (b,A) is R, so 1 is in there. i'm trying to see whether there's a more explicit construction of 1

Sourav Ghosh

16:35

A\subsetneq (A, b) and the maximality of A implies (A, b) =R

This is the way we use maximality :)

shintuku

17:07

there has to be some geometrical interpretation of the fact the sum of two disjoint maximal ideals is always the entire ring

seems like a form of bezout's identity too

it's enough for the ideals not to be contained in one another

Ted Shifrin

17:33

@shintuku This is impossible.

shintuku

i will defer to your authority and call it a day, moving on to more algebra

Ted Shifrin

17:50

Two ideals can never be disjoint.

shintuku

hm right, they have 0 in common

the right statemnt is: the sum of two maximal ideals that do not contain each other is the entire ring

Ted Shifrin

They have tons more than $0$ in common. Write down any examples.

That is a silly statement. A maximal ideal is contained in no ideal other than the whole ring.

shintuku

hm right

Ted Shifrin

If you take any maximal ideal and add to it any ideal not contained in it, you must get the whole ring.

shintuku

i hadn't noticed that, thanks

Ted Shifrin

18:02

That's the definition of maximality.

shintuku

19:07

the extreme value theorem (a bounded continuous function has absolute min and max) is equivalent to saying that $\{f(x) : x \in [a,b] \}$ is closed, right?

noballpointpen

19:32

Extreme value theorem requires the domain to be closed, which is therefore compact by Heine-Borel. And therefore the image is closed, by Heine-Borel, again.

Let me check it.

Yeah

Xander Henderson

Hrm...

noballpointpen

19:45

Don't forget that the bounded image of the continuous function doesn't mean that the function has max and min...

Ted Shifrin

@shintuku No, it's not equivalent. You need it to be a closed, bounded interval.

Note that $[1,\infty)$ is closed.

noballpointpen

Should've added "bounded" before inferring compactness.*

Ted Shifrin

Yes, you should :)

noballpointpen

Actually, shintuku's statement of EVT is completely incorrect.

Ted Shifrin

20:38

Yes, it is. The function $\arctan\colon\Bbb R\to (-\pi/2,\pi/2)$ is an easy counterexample.

schn

21:00

Find a number $N$ such that $|s-s_n|\leq 10^{-3}$ when $n\geq N$ for each of the series (a) $\sum_{k=1}^\infty \frac{(-1)^k}{k^2}$ and (b) $\sum_{k=1}^\infty \frac{1}{k^2}$.

(a) is easy, since the error is at most the absolute value of the first omitted term for alternating series. However, I am stuck on (b). It is not true that for all positive series the first omitted term is bounding the error...

Ted Shifrin

21:11

I would suggest thinking about the integral test and how it works.

schn

Ok.

shintuku

so, correcting: the extreme value theorem (a continuous function $f$ on a closed interval has an absolute min/max) is equivalent to saying that $\{f(x): x \in [a,b] \}$ with $f$ continuous is closed, right?

continuous implies bounded

monoidaltransform

Let $M$ be an n-dimensional topological manifold. Then, a standard fact tells from algebraic topology states that for all $k>n$, $H^k(M,\mathbb{Z})=0$. UCT tells us (since $\mathbb{R}$ has characteristic $0$) that for all $k>n$, $H^k(M,\mathbb{R})\cong Hom(H_k(M),\mathbb{R})=0$.

Now assume a priori we did not know that $H^k(M)=0$ for all $k>n$, but we did know that for all $k>n$ $H^k(M,\mathbb{R})=0$. Could we use the latter fact to prove for all $k>n$ $H^k(M)=0$?

noballpointpen

The hypotheses imply closed image.

Ted Shifrin

@shintuku That’s a separate theorem.

Why are you avoiding the word compact? The continuous image of a compact set is compact.

@monoidaltransform No. There’s no accounting for torsion.

noballpointpen

21:16

Since shintuku is talking about a real-valued function defined on the real-valued set. Those are interchangeable.

Ted Shifrin

What is interchangeable?

monoidaltransform

I see. Thanks

Ted Shifrin

My point is that you need first to prove that a continuous image of a closed interval is bounded. Then you prove the sup/inf are attained.

noballpointpen

The notion of compactness and closed-boundness. Don't get me wrong, they are, of course, in $R^k$.

Ted Shifrin

With the standard metric. But if we’re being elementary, we do as I said. I have no idea if shin has a clue how to prove the EVT.

noballpointpen

21:20

I am curious. Which metric will destroy this?

shintuku

i'm asking because i'm failing to see why we must prove that the supremum is attained if the only possible values of $f$ with $x \in [a,b]$ are those in $\{f(x): x \in[a,b]\}$, and vice versa

Ted Shifrin

Take any bounded metric, for example.

Shin, I have no idea what we actually know. This is completely out of context.

shintuku

We have $f$ is continuous over interval implies the image is bounded over interval, and that's it

so the only way i can understand that there's something to prove would be if we want to get rid of the possibility the image of $f$ is open

Ted Shifrin

So why should we know that the function takes on its sup? For example, that’s false with closed image on an interval $[0,1)$.

noballpointpen

Do you try to communicate that since the image is closed we do not need to prove EVT?

Ted Shifrin

21:30

Or it’s false without continuity at every point, even knowing closed image. This is a real theorem.

Xander Henderson

@TedShifrin Not complex? :P

Ted Shifrin

I imagine not.

Xander Henderson

What is @shintuku trying to do?

I just got here.

Ted Shifrin

I think prove the extreme value theorem.

Xander Henderson

Oh. Doesn't that, like, follow fundamentally from the least upper bound property of the real numbers?

Or is @shintuku looking for something more topology flavored?

@shintuku What do you want?

shintuku

21:33

@TedShifrin at Ted: we're given that the function is defined and continuous over $[a,b]$, isn't it impossible for it to have an image $[0,1)$?

Xander Henderson

@shintuku Yes...

Do you know how to prove that?

Also, it isn't necessary to say "defined". If you say that $f$ is continuous on (not over) $[a,b]$, then you are implicitly saying that it is defined on that interval.

Ted Shifrin

To prove impossibility requires proof. And I meant that as the domain, not the image. The image could be $[q,\infty)$ and still be closed.

shintuku

right that proof would be the EVT, but doesn't it amount to saying the image must be closed? (compact?)

i.e., we can't have $[0,1)$, we must have $[0,1]$

Ted Shifrin

You keep ignoring my comments about unbounded intervals.

Xander Henderson

@shintuku I mean, you could say that the proof is the EVT, but that would be a pretty trivial proof.

shintuku

21:37

@TedShifrin we're given $f$ is bounded

Xander Henderson

If the image is $[0,1)$, then the function has not attained its supremum, which contradicts the EVT.

Ted Shifrin

OK. So what exactly has been proved and what do you need to prove? This is all muddled.

shintuku

We're given $f$ is bounded and continous on $[a,b]$, we need to prove that it has an absolute maximum, so we begin by considering $M = \sup\{f(x):x \in [a,b]\}$ and showing $f$ hits $M$, but I can't distinguish this from saying $\{f(x):x \in [a,b]\}$ is closed (when we're already given it is bounded)

Xander Henderson

@shintuku So you are trying to prove the Extreme Value Theorem?

shintuku

@XanderHenderson yes

Ted Shifrin

21:41

But you don’t know the image is closed.

shintuku

right, the proof would amount to that

Ted Shifrin

We’ve been going in circles. So you’re going to have to use the defn of sup.

shintuku

yeah after all, i'll end up using sup, but I was just trying to understand whether closedness of the image was what could fail (and therefore be the reason for the need to prove something)

copper.hat

$f([a,b])$ is compact. That follows almost immediately from the definition. Hence it is closed.

Ted Shifrin

Correct. Without continuity everywhere or without the domain being a closed interval, it can fail.

He doesn’t know compactness, copper.

shintuku

21:45

i'm a philistine

Ted Shifrin

It’s very hard to know what you know and what you don’t.

You do not provide context.

shintuku

there's the known knowns, the known unknowns, the unknown knowns and then the unknown unknowns

-Defense Secretary Donald Rumsfeld, fighting for the Free World

Xander Henderson

@shintuku Not helpful.

shintuku

sorry

Xander Henderson

What tools do you have?

In most analysis texts, the assumption is that you have a definition of continuity, and that you know the properties of the real numbers (specifically, the least upper bound property).

Are these the tools you have?

shintuku

21:52

we're done, i just needed that doubt cleared. suppose $M$ is sup and is not attained. then $sup\{1/(M-f)\}=\infty$ will make you have a sad day, breaking the bounds of friendship it had agreed to because 1/(M-f) was continuous. so to bringten things $f$ will meet with $M$

Xander Henderson

...

Hopeful Whitepiller

whats up gamers

Xander Henderson

The sky.

alternative punchline: An outward facing normal vector to the surface of the Earth.

geocalc33

22:30

What's up? The Miami Heat are up 3-0 against the Boston Celtics

shintuku

is dealing with ideals more easy when they're considered to be subschemes?

Xander Henderson

@geocalc33 Ice hockey?

No... wait... the Celtics are a cricket team, right?

geocalc33

@XanderHenderson basketball

No team has ever come back from 0-3 in NBA (I think it's happened in the MLB once)

mick

23:47

0

Q: solutions to $x = \sqrt{a+\sqrt{a+\sqrt{a+\cdots}}}\;\;(\text{mod}\;p)$

Fix an integer $0<a<p-1$, and an odd prime $p$. Define $$S(a,p)=\sqrt{a+\sqrt{a+\sqrt{a+\cdots}}}\;\;(\text{mod}\;p)$$ to be the set of all integers $x\in\{0,...,p-1\}$ such that, for some positive integer $n$, $$x\equiv \sqrt{a+\sqrt{a+\sqrt{a+\cdots + \sqrt{a+x}}}}\;\;(\text{mod}\;p)$$ with ex...

number-theory prime-numbers modular-arithmetic asymptotics nested-radicals

any ideas ?

need sleep

bye

Transcript for

$Mathematics$ Mathematics