Mathematics - 2023-06-07 (page 1 of 2)

noballpointpen

00:07

No, I don't like that product cell...

Better would be to justify that $\min$ must be "under" $y=x$ and $\max$ "above".

Yet, it is a shaky claim. The whole graph could be "under" $y=x$, and the fixed point just $f(0)=0$.

No, clearly not better.

Please, do not make a hint! The supplements say this problem becomes very easy if you see something.

shintuku

00:40

is there such a thing as the minimal ring containing $(1/2)\mathbb Z$?

hm, the intersection of all rings containing that set

shintuku

01:15

"there is no way to sum and multiply elements of $(1/2)\mathbb Z$ to obtain a rational number with a denominator that is not a power of $2$", do all algebraic arguments for this consist of showing $\{ \frac p q : 2 \nmid p, q = 2^n; n \in \mathbb Z\}$ is a ring?

Thorgott

@Balarka there's always an open $U\hookrightarrow X\times_YX$ s.t. the diagonal factors through a closed immersion $X\rightarrow U$ (fix an open affine cover $(U_i)_{i\in I}$ of $X$, then $U=\bigcup_{i\in I}U_i\times_YU_i$ does the job) and closed immersions are local on the target, so if the image of the diagonal is closed, you can use the cover $X\times_YX=(X\times_YX\setminus\Delta_X)\cup U$ to deduce the diagonal is a closed immersion then

sorry if this is a redundant remark, just skimmed through the convo

leslie townes

shin, if S is a subset of a ring that contains its 1 and its -1, then testing whether S is itself a ring (with the operations taken, where possible, from the "known" ring) amounts to checking whether S is closed under the addition and multiplication of the 'known' ring

shintuku

hm right, the issue is that $(1/2)\mathbb Z$ itself isn't a ring so i have to look for the next smallest ring

leslie townes

shin: so if you can do one of those things you're asking about, you can do the other. i don't know if it's helpful or necessary to envision the space of "all algebraic arguments" for a proposition. it might be possible to write some really goofy argument that, while equivalent to something normal, looks completely different

shin: all of this is sitting inside e.g. Q which is known to be a ring

shintuku

hm right

leslie townes

01:26

if S is a subset of Q and 1 and -1 are in S, then "there is no way to sum or multiply elements of S to obtain an element that is not in S" is equivalent to "S is a subring of Q" is equivalent to "S is a ring, when given the addition and multiplication and 1 from Q"

Thorgott

@ShaVuklia this is wrong. the identity morphism of any scheme is separated, so if it were true, it would imply all schemes are separated, which is not the case.

leslie townes

i'm slightly wondering about the context surrounding this question, e.g. why it would matter to focus on one phrasing and not another

shintuku

but i don't know which ring to test for being a subring :(

leslie townes

it certainly can get more confusing if there's no ring playing the role of Q here, i.e., there's no known ring that all of our constructions are playing out inside of

shintuku

well, $(1/2)\mathbb Z$ is a subset of $\mathbb Q$, but the issue is that $\mathbb Q$ doesn't have the property I want: having only elements who's denominators are powers of 2

there's some intermediate set between $(1/2)\mathbb Z$ and $\mathbb Q$ that has the property i want, and that's the one i need to test for ringness

(i think, if that's equivalent to the proof i want)

leslie townes

01:35

what do you mean by the notation "(1/2)"? where did this begin?

Sourav Ghosh

@noballpointpen Consider $g(x) =f(x) -x$ and then apply the Darboux property (/Intermediate value property) to $g$ to conclude $g$ has a zero in $[0, 1]$

shintuku

at leslie: $(1/2)\mathbb Z = \{\frac{z}{2}: z \in \mathbb Z\}$

leslie townes

oh, this is where i should have rendered chatjax :)

shintuku

but i'm testing out your approach, and verifying that $\{\frac{z}{2^n}:z \in \mathbb Z, n \in \mathbb N\}$ is a subring of $\mathbb Q$

noballpointpen

You should read messages before helping someone, Sourav. I asked to not give me any hints below. :(

shintuku

01:43

at leslie: yeah, it works!

it gives us a very fast argument for the proposition that there is no way to sum and multiply elements of $(1/2)\mathbb Z$ to obtain a rational number whose denominator is not a power of 2: first note $(1/2)\mathbb Z \subseteq \{\frac{z}{2^n}: z \in \mathbb Z, n \in \mathbb N\} \subseteq Q$, and show that the middle set is obviously closed under substraction and multiplication

thanks, neat stuff

leslie townes

02:08

i dunno why this focus on (1/2)Z, but you could certainly say more than "is not a power of 2" for that specific set. the denominators of sums or products of things in there can be taken in {1,2,4}

it's a goofy set, though, e.g. because it's not closed under multiplication

is this coming up in some exercise or just random self-study

shintuku

i'm studying $\mathbb Z/4\mathbb Z[\frac{1}{2}+4\mathbb Z]$, and i'm trying to figure out what exactly the denominators of its members are

turns out, thanks to that above exercise, that the denominators are exactly those powers of 2

shintuku

02:55

any quick way to see $\mathbb Q/n\mathbb Z$ is a ring?

eh, i'll just start accepting things are rings when they have multiplicative and additive closure

hm just to be sure i should still check inverses

heck let's just test the axioms

robjohn

General Rings do not need inverses, as far as I know.

shintuku

yeah you're right, unity i meant. inverses are there for fields

one potato two potato

no for division rings

shintuku

sounds right, i haven't had the need to distinguish fields from division rings so far

copper.hat

03:26

Hello!

Rngs don't necessarily have identities...

one potato two potato

I don't remember Stone-Cech compactification explicitly but I remember it uses some dual concept during the construction. I remember it's interesting, although I've never seen some interesting application of that.

shintuku

best way to pronounce rng is to sound like you're choking a bit

one potato two potato

prof. once said that a nonunital ring is very hard and almost impossible to study. I remember he was talking about dvr or something

copper.hat

chokng

shintuku

03:43

is there an algebraic structure closed under taking fractional powers?

nvm that question

shintuku

04:03

nvm

Ted Shifrin

@shintuku Do you still not recognize this as $\Bbb Z[1/2]$?

shintuku

huh, i had not seen that

Ted Shifrin

Read Artin (or my book) about adjoining elements to rings :)

shintuku

will do, thanks for the tip

nice: if we have $a \in R$, the isomorphism theorems give us that the image of $R[x] \to R: p(x) \mapsto p(a)$ is a subring of $R$

hm but that's not what we need here to have $\mathbb Z[1/2]$ ring

oh, take a ring extension $R'$ of $R$ instead, and $a \in R'$, and only now consider $R[x] \to R':p(x) \mapsto p(a)$

we're free to take $R' = \mathbb Q$ here, guaranteeing $\mathbb Z[1/2]$ is a ring

cool cool

.... no the above is wrong

robjohn

04:23

Are you just trying to ascertain that $\mathbb{Z}[1/2]$ is a ring?

shintuku

yeah

or, an arbitrary extension by a rational

Ted Shifrin

It’s the image of a homomorphism to $\Bbb Q$.

Doesn’t need to be rational. Works for any algebraic number.

Or any complex number (then map to $\Bbb C$).

Balarka Sen

I was going to put an example in my thesis because I think the example is very insightful but I am realizing the example can only be drawn and not written

shintuku

oh, i see

Balarka Sen

Well, I can always write "schloop spinny zigzags" but something tells me this would not be acceptable exposition.

copper.hat

04:38

i am fascinated by the idea that your example can be drawn but not written...

Ted Shifrin

Schmoopy schloopy?

leslie townes

05:27

that would make for an interesting thesis defense. "unfortunately, nobody can to be told what this next example is. you have to see it for yourself." passes out pills

robjohn

06:01

@leslietownes Doood! You’re so righteous!

Koro

06:13

@Jakobian yes, I understood it now. The idea is that if X is a space with cofinite topology and $Y\subset X$, then the subspace topology on Y coincides with the cofinite topology on Y.

Dannyu NDos

Does the convolution theorem work for discrete-time Fourier transform?

Koro

So if X is infinite, then it is connected: if not, then X= U\cup V. U and V are non empty, disjoint and open in X. \emptyset= U\cap V. So $U\subset V^c$ and $V\subset U^c$, i.e., U and V are both finite. $X= U\cup V$ is finite. contradiction. So $X$ with cofinite topology is connected.

So the only connected subsets of R are singletons and infinite sets.

empty set too as per one's taste.

If instead we consider Sorgenfrey line R, then I claim that it is totally disconnected: Let $F\subset R$ be non empty, finite with atleast two elements. Take any a in F. Set d= min (|x-a|) as x varies over F. {a}= F\cap [a,a+d/2). So {a} is open in subspace topology. F is a discrete space in subspace topology. $F$ can't be connected.

If S\subset R is infinite, then take any s in S such that s> inf S. $S= ((-\infty, s)\cap S)\cup ([s,\infty)\cap S)$ is a separation so not connected.

So the only connected subsets are the singletons.

@Jakobian: Could you please verify the following?: I think that I didn't have to consider F with subspace topology above. Why? Because connectedness is a topological property so it doesn't depend upon subspaces. F is finite so discrete hence disconnected? Is it correct to say so?

but this looks wrong as finite is not always discrete: X= {a,b,c} and the topology on X is T= {\emptyset, X, {a,b}}.

therefore, I had to use subspace topology on F.

I think I answered my own question.

冥王 Hades

07:39

Million dollar advice: Never put your hand in burning hot cooking oil out of curiosity.

I may or may not have done that today...

robjohn

08:25

hot cooking oil is worse than hot water in that it sticks and does not boil away.

@冥王Hades If you did, how was the ER?

Curio

What is the laplace transform of x(t)*step(t)? Is it X(s)/s ?

one potato two potato

why man usually do math or science better than women?

PlaceReporter99

08:43

@onepotatotwopotato WRONG!

one potato two potato

why

PlaceReporter99

@onepotatotwopotato gender has negligible, if not no effect on cognitive ability.

@冥王Hades BILLION dollar advice: never put water in burning cooking oil.

one potato two potato

why? Look around any math class in college. majority of them are men. most of the faculty are men. Many famous mathematicians are mostly men.

PlaceReporter99

Burning oil + water = ^^^

Koro

@onepotatotwopotato I don't know almost every faculty at my college here is man and they are all garbage as teachers. I didn't find even one of them to be good or average. All of them are lethargic and just wait for end of the month for salary credit to their account.

there was a woman teacher too who 'taught' us last semester. She was also no different.

PlaceReporter99

08:55

@Koro what do you expect from teachers if they aren’t paid enough?

Koro

I think they are overpaid.

hence the tantrum

I have no respect for them.

we were like 40 when we joined the college. Now only 30 +/- 2 of us are left.

PlaceReporter99

Still a lot

one potato two potato

@Koro They are basically researchers not professional lecturers

Peter

@Koro In Germany , surely !

Koro

I refuse to believe that they were not 'good' students. They left in frustration. The teachers could not hold their interest.

one potato two potato

09:00

good researcher and good advisor are also different matters basically.

Koro

09:14

the bare minimum expected from a teacher is that they should not lie-atleast not in front of students. But the ones here have no shame or honour. So one of the teachers here convinced the students that most of the questions in some exam will be from pre-midsem but guess what - nothing was asked from pre-midsem. I felt very bad for those who wrote the exam. I saw the question paper: if I wrote it, I don't think I would pass it. Covering spaces of Lens space something...

but this is somewhat reasonable because the teacher himself doesn't decide the question paper. It goes for some review by some committee and they may make some changes in that nullifying what the teacher wanted in the exams.

one potato two potato

anyway that's nothing to do with my question and I now regret asking here.

Koro

yeah sorry for diverting the topic. I'll stop talking about teachers at my college now.

I've already talked too much about the teachers at my college. Even I don't like to talk about them anymore.

Prithu Biswas

10:31

@onepotatotwopotato I don't think "The majority of people studying math and science are men" implies that "Men will inherently do better than women in math and science".

Jakobian

@onepotatotwopotato part of this is social pressure coming from statements like "men are better at math"

Which I don't agree with btw*

Koro

10:47

Show that if $0< m(A)\le \infty$,then for each positive $q< m(A)$, there is a perfect subset of A with measure $q$. Here, $A\subset R$ is given to be Lebesgue measurable.

I can find a measurable subset of measure q.

But not sure how to get perfect subset.

冥王 Hades

@robjohn Well for one it was extremely painful and it still hurts despite having it treated and bandaged properly. I was told that I'm stupid by one of the nurses when she was told why it happened.

I didn't take that too kindly. If I were stupid I'd fail my math classes horribly.

Jakobian

webmd.com/parenting/features/math-myths-boys_girls-

冥王 Hades

@PlaceReporter99 I wanna try this now

Jakobian

@Koro a closed set is union of perfect set and countable set

Koro

yes, I know.

Jakobian

10:51

Now find a subset of measure q+1 say, then of measure > q by regularity

Koro

So suppose that I have a subset $B\subset A$ of measure q.

Jakobian

By intersecting with closed intervals we should get a closed subset of measure q I believe

Koro

By regularity, I can find closed subset $F_n\subset B$ such that $|B|-|F_n|<1/n$. Now what next?

Jakobian

Using the previous result we can make it perfect

Koro

yes. But from |F_n|, how do you get closed set of measure q?

ohh, you're saying something different.

you want to approximate q from above?

by finding subset of measure $q+1$ then of $q+1/2$ etc.

?

Jakobian

10:56

1. Find a compact B c A with q < m(B) <= m(A)

Koro

I can find a closed such B.

but assuming A is bounded, ok.

Jakobian

2. Find t for which f(t) = m((-inf, t] \cap B) = q

@Koro lebesgue measure is semifinite or however it was called

Koro

@Jakobian ok. So I'll deduce the unbounded case later.

using the bounded case.

Jakobian

I mean is there a need to split it

Koro

@Jakobian it should turn out to be continuous and increasing. So for some t, we'll get q by IVT.

Jakobian

10:59

Lebesgue measure m(A) = sup_{K c A, K compact} m(K)

Koro

hmm, I think it should be 'closed' instead of 'compact' in this formula?

Jakobian

3. Write obtained closed subset, say C, of A of measure q as C = D u S where D is perfect and S is countable, m(C) = m(D)

@Koro definitely compact

Koro

because we have: given $\epsilon>0$, there exists a closed subset $B\subset A$ such that $m(A\setminus B)<\epsilon$.

(where A\subset R is measurable.)

Jakobian

That's only for bounded A

For compact B

But I wrote a sup

Regular measure

In mathematics, a regular measure on a topological space is a measure for which every measurable set can be approximated from above by open measurable sets and from below by compact measurable sets. == Definition == Let (X, T) be a topological space and let Σ be a σ-algebra on X. Let μ be a measure on (X, Σ). A measurable subset A of X is said to be inner regular if μ ( A ) = sup { μ ( F ) ∣ F ⊆ A , F compact and measurable...

Koro

@Jakobian no

it is for any subset A.

Jakobian

11:07

There's no compact K with m(R\K) < 1

Koro

I call $A\subset R$ Lebesgue measurable if there is a Borel set $B\subset A$ such that m(B\A)=0.

this is equivalent to

4 mins ago, by Koro

because we have: given $\epsilon>0$, there exists a closed subset $B\subset A$ such that $m(A\setminus B)<\epsilon$.

@Jakobian oh you mean the one with compact.

Jakobian

Anyway the sup is the same over compacts as it is over closed sets

Koro

As I expressed earlier, I don't know much about that.

Jakobian

The usual definition of regularity uses compact sets

Koro

@Jakobian ok. I understand it now.

If A is bounded, then its closed subsets are compact.

Jakobian

11:09

m(F) = sup_n m([-n, n] cap F) is a sup of compact sets

So sup over compact and closed subsets agree

Using compact set here has an advantage that we don't need to pick a closed subset with finite measure > q

That it's finite is implicit

PM 2Ring

@onepotatotwopotato That's primarily due to social / cultural factors that discourage many girls from developing their mathematical skills & interests in STEM fields.

Koro

@Jakobian I understood it now. Thank you so much. But I propose few modifications:

PM 2Ring

Here in Australia, there was a time when married women were prohibited from working as scientists for the CSIRO, a government science organisation. One brilliant scientist who worked with her husband kept their marriage secret for 3 years. After she got fired, she became a highly respected maths teacher at a girls high school. The ban was lifted a few years later, but she decided to remain a teacher. Many of her students went on to have careers in STEM.

Koro

11:26

Case 1: A is bounded. In this case, $A\subset [a,b]$ for some a,b. There exists a closed set $B\subset A$ of measure m(B) such that $m(B)\in (q, m(A)]$.
Step 2: Here $f(t)= m([a,t]\cap B)$. In your case, $a=-\infty$ but in my case a is a real no. $f$ is continuous and increasing so must attain q at some c. So $C=[a,c]\cap B$ is a closed set of measure q.

@Jakobian then $C= D\cup S$ gives $m(C)= m(D)=q$.

monoidaltransform

0

Q: Equivalent metrics on compact metric space

Let $X$ be a compact metric space with two metrics $d,d'$ inducing the same topology on $X$. Show or find a counterexample to: For every $\epsilon>0$ there exists a sequence $\delta_n>0$ such that $\delta_n\rightarrow 0$ and if $d(x,y)<\epsilon$ then $d'(x,y)<\delta_n$. Thoughts: (1). Since $d,d'...

Koro

Case 2: A is not bounded. In this case $\cup_{n\in N} A_n =A$, where $A_n= [-n,n]\cap A$ so by continuous of Lebesgue measure, there is some n such that m(A_n)>q.

By case 1 on $A_n$ we get a perfect subset D' of measure q.

Jakobian

works equally well

PM 2Ring

@onepotatotwopotato You should read about some of the great women in maths & science from earlier times, when it was really hard for women to get a math / science education, and to be treated seriously. Eg, en.wikipedia.org/wiki/Sophie_Germain en.wikipedia.org/wiki/%C3%89milie_du_Ch%C3%A2telet en.wikipedia.org/wiki/Emmy_Noether en.wikipedia.org/wiki/Henrietta_Swan_Leavitt en.wikipedia.org/wiki/Margaret_Burbidge en.wikipedia.org/wiki/Jocelyn_Bell_Burnell

Danica McKellar is an actress and a mathematician. She has written numerous books to encourage girls to develop their mathematical skills & interests. danicamckellar.com

Sha Vuklia

11:42

@BalarkaSen Also, funny that you mentioned Hartshorne, because I've avoided Hartshorne like the plague. I used to read in Mumford, simply because any other book than Hartshorne was already an upgrdade to me (and for the scheme part of our course, we follow Mumford anyways). However, last year I found Liu's Algebraic Geometry, and that book is just incredible. For me, it's the scheme theory/alggeo equivalent of Lee's stellar Intro to Smooth Manifolds

shintuku

13:09

@onepotatotwopotato couple of studies show women underperform on math tests when an obviously social or cultural intervening factor distabilizes them, e.g., they perform better when the test is not called mathematical, or when they are not told men perform better

you can easily find those on google scholar

ZaWarudo

13:23

A function $f:E \to \mathbb{R}$ is measurable if for each $t\in\mathbb{R}$ the set $\{x \in E \ | \ f(x)>t\}$ is a measurable set. To prove that if the characteristic function $\chi_E$ of a set $E$ is measurable, then $E$ is measurable I tried this: since $\chi_E$ is measurable, for each $t \in\mathbb{R}$ the set $\{x \in E \ | \ \chi_E (x)>t \}$ is measurable.

Since this holds for each $t\in\mathbb{R}$, it holds for $0 \le t_0<1$ and for this values $\{x \in E \ | \ \chi_E(x)>t_0\}=E$. Hence, $E$ is measurable.

Could this work?

Jakobian

Yeah

I mean, no need to wonder what t_0 works, I'd just take 1/2

ZaWarudo

Thank you!

Jakobian

13:41

I feel like I'm constantly forgetting things

in the sense that I have some kind of thing that I do frequently, as if forming a habit
and then after it seems like I did form a habit, it disappears

Mad

14:02

I have a little inquiry about summation in a set of numbers that is periodic.
Suppose {1,...n} is periodic, such that n+x = x for x from 1 to n.
how can summation be presented?
\sum_1^n x_i , sum_2^n+1 and so on

if the index is peridoic

Koro

If $A$ is measurable subset of R of measure >0, then there is a d>0 such that $A\cap (A+x)$ is non empty whenever |x|<d.

How to prove this?

Jakobian

you're doing Rudin right

Koro

Suppose $f(x)= m(A\cap (A+x))$. Then I want to show that f is continuous at 0.

@Jakobian no

Jakobian

oh, I thought this is a standard exercise from Rudin

Koro

ohh

$f(x)- f(0)= |A\cap (A+x)|-|A|\le |A|-|A|=0$

but how to show that $f(0)- f(x)$ is also small when x is small?

Jakobian

14:16

$|A\cap (A+x)| = \int 1_A(y) \cdot 1_{A+x}(y) = \int 1_A(y)\cdot 1_A(y-x)$

@koro

I think you should try to think of it like this

shintuku

is there such a thing as modular arithmetic over rationals?

we don't have $a \equiv b \mod n \implies ca \equiv cb \mod n$ unlike the integer case

Jakobian

modular arithmetic is the study of quotient rings of $\mathbb{Z}$, quotient rings of $\mathbb{Q}$ are $\mathbb{Q}$ and the trivial ring

shintuku

hm, that's important to know, thanks

Jakobian

it's just my view of it, maybe someone acquainted with number theory can tell you better, I just don't know the ideas they use there

Koro

14:50

@Jakobian that looks cool but not sure how to go from here.

I was thinking of using translational invariance of Lebesgue measure.

PM 2Ring

15:00

@shintuku Not really. But take a look at pi.math.cornell.edu/~hatcher/TN/TNpage.html particularly the introductory stuff about The Farey Diagram and Linear Fractional Transformations.

shintuku

@PM2Ring cool stuff, thank you!

PM 2Ring

15:20

@shintuku No worries. Farey sequences are pretty cool. I worked through the first part of that book a couple of years ago, but I got lost towards the end. :)

Shaun

15:40

Hi :) I have an idea I'm not sure how to articulate into mathematics. It requires analysis I haven't studied in a while.

Suppose you have an $n$-dimensional surface coloured in different shades of the same colour, that vary continuously.

Suppose further that the surface is of a shape that is "closed", the same way a finite three dimensional shape is; I think I just mean "finite" here . . .

Can you always find a nontrivial, continuous path on that surface with all the same shade?

I couldn't think of a counterexample.

It seems like the sort of thing someone would have studied before.

noballpointpen

@Ted, can you elaborate on your Mobius strip comment? I am curious. Here are my questions you didn't see: [click](https://chat.stackexchange.com/transcript/message/63752265#63752265)

Shaun, hello! How it's been going?

Shaun

However, with this not being my type of mathematics, I don't know what to search for.

noballpointpen

Now even the link formatting doesn't work!

Shaun

@noballpointpen Hi! I'm slowly recovering from COVID. I got to sleep last night, after a few nights up.

noballpointpen

Yet had been doing math even before you could sleep?

Shaun

15:49

I haven't done much serious maths but I have had fun learning about stuff. I quite like combinatorial game theory and all its weird numbers!

noballpointpen

Also, I wanted to ask you about your saying "always think about math". How do you define it? :)
Now I am shut, so someone could see your question.

Shaun

@noballpointpen Within reason, I guess. There are times that call for concentration outside of the subject; they're important. Rests are part of that too.

I watched the film, "The Blind Side" yesterday, for instance. It's over two hours long. It has nothing to do with maths, but I had fun.

shintuku

16:26

is there a way to characterize all rings $R$ such that, given an arbitrary $R'$, there is at most one ring homomophism $R \to R'$?

Soumik Mukherjee

@ShaVuklia Is Liu's Algebraic Geometry book available online?

D.C. the III

I know I'm going to be excoriated for this, but why is the range not a subset of the domain in this image?

shintuku

what's your function? does it have $\mathbb R \to \mathbb R$?

or are you doing $\mathbb R^2 \to \mathbb R^2$?

D.C. the III

the function is the red line

R to R

shintuku

is it $\mathbb R \to \mathbb R$?

draw lines from each end of your function to the corresponding axis, check whether they correspond to one another

D.C. the III

16:34

that's what I was thinking initially

shintuku

also, very important: do you have range = codomain or range = image?

D.C. the III

It was a vague question I had seen, but the way it was used I'm assuming image

shintuku

here's a statement that applies to your graph: the image of your function is not a subset of the domain

D.C. the III

so it's more so what is happening is that the domain of my function is not in the domain of the line $y=x$ completely

@shintuku that's what I was trying to figure out

shintuku

look at the left end of your function, and draw one perpendicular line to the x axis and one to the y axis

you will see that the image begins on the y-axis at a point where the domain has not yet begun on the x-axis

we have to suppose the axes are the same number line and have the same scale

you can see this by drawing a circle centered at the origin whose radius touches the perpendicular line on the y axis

D.C. the III

16:41

let me draw this quick.

shintuku

imgur.com/a/I7wziIs

so the intersection point of the circle with the x-axis defines an interval on the x-axis where the image is defined but the domain is not

D.C. the III

Ahhh.. I got it now....I was thinking along these lines, but the idea was buried deep and forgot it

Thanks. And Ted being here means I'm going to be excoriated now for forgetting such a basic concept.

shintuku

17:30

anyone know anything about the set $S$ of rings $R$ such that, given an arbitrary ring $R'$, there is at most one homomorphism $R \to R'$?

Ted Shifrin

@noballpointpen It was meant mostly for humor, but I said Möbius strip only because I was pretty sure you would know what that is. Most objects with more complicated topology — non-orientability was not really the point — will make your intuition wrong. There are usually other routes from here to there other than crossing the diagonal line. Turn the square into a cylinder, for example. Can you see how to do it?

@shintuku Do your ring homomorphisms have to take $1$ to $1$?

If so, you should be able to answer your own question.

I'm not sure why it's an interesting question, but ...

aarbee

Hi. How to type a column matrix?

D.C. the III

446

A: MathJax basic tutorial and quick reference

Matrices Use $$\begin{matrix}…\end{matrix}$$ In between the \begin and \end, put the matrix elements. End each matrix row with \\, and separate matrix elements with &. For example, $$ \begin{matrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \\ \end{matrix} $$ produces: $$ \begin{matrix} 1 & x

noballpointpen

Should that cylinder-from-square have caps?

shintuku

@TedShifrin yeah, I'm thinking it is the set of rings $R$ s.t. any $r \in R$ is a finite sum of $1_R$

Unknown x

17:39

0

Q: What guarentees the existance of scalar function $λ?$

Suppose all the tangent lines of a space curve pass through a fixed point. What can we say about the curve? Proof starts like this in the textbook Without loss of generality, we take the fixed point to be the origin and the curve to be arc length-parametrized by $\vec α$. Then for every s we have...

differential-geometry vector-analysis

Ted Shifrin

D.C. Re your question, the image is always a subset of the range = codomain. Even if the domain and range are both $\Bbb R$, the codomain and domain are different.

@shintuku So what is such a ring isomorphic to?

shintuku

the infinite case is isomorphic to $\mathbb Z$, the finite case is isomorphic to $\mathbb Z/n \mathbb Z$, maybe... i think...

Ted Shifrin

@Unknownx If the origin is on every tangent line, that means that $\vec 0 = \vec{\alpha}(s) + \lambda(s)\vec T(s)$ for some scalar $\lambda(s)$. Right?

shintuku

i need to convince myself that these should be the only possible members of that set

aarbee

To type column matrix, I tried \begin{bmatrix}1\\1\\1\end{bmatrix} but it didn't work. I tried \hline1\\1\\1, it didn't work either

Ted Shifrin

17:42

It worked fine.

You need to enable the rendering of ChatJax on your screen.

D.C. the III

I agree with the image being a subset of the codomain. I was just trying to think of a simple counter example to the contraction mapping principle.

speaking of which. Is there something special I'm missing in order to show $f(x) = \sqrt{x^2 + 1}$ has no fixed point? I wanted to show it is not a conraction mapping.

Ted Shifrin

Oh, so the domain and codomain are both $X$, and so it makes sense to think of them as the same set. So you're trying to draw a graph of a function $\Bbb R\to\Bbb R$, say, that does not intersect the diagonal.

D.C. the III

yes

Ted Shifrin

Watch out for your logic. Does not contraction mapping imply no fixed point?

D.C. the III

Funny....I did ask myself that question

Ted Shifrin

17:45

You're getting too old to be screwing up that logic.

Didn't the exercise ask both questions? Why is it not a contraction? And show by algebra that it has no fixed point.

D.C. the III

IT asks to show no fixed point, that $|f'(x)|< 1$ and why this does not contradict the theorem

Damn so we do got to show by algebra.....I was hoping it was a logical thing

Ted Shifrin

It's totally trivial algebra. Be serious.

You need to get past applying wrong logic.

Plenty of non-contraction maps have fixed points. You'd better understand that.

D.C. the III

understood..

shintuku

at ted: the question is more or less interesting to me because if the unique homomorphism is injective, such rings R have that, given some other ring R', there is a unique way to define an extension from R to R'

Unknown x

@TedShifrin like this. right? particularly, when a=0. we get the above result

am I correct?

D.C. the III

17:55

is the algebra so trivial that all I needed to do was take $f(x) = x \Rightarrow \sqrt{x^2 + 1} = x$? From which I solve and get $1 = 0$.......which is a problem.....

Unknown x

sorry $\lambda(s)\vec{T}(s)$

Ted Shifrin

Indeed @D.C.

D.C. the III

Too trivial I suppose that I worked myself into a tizzy thinking I needed more.

I'm gonna need you to stick around for about 3 more decades to talk me down from buildings in my math thinking

Ted Shifrin

@Unknownx What is $a$ in your picture? If it's $\vec\alpha$, then you can't say it's $0$.

D.C. the III

or I could just get better at it. :)

Ted Shifrin

17:58

Indeed. :)

D.C. the III

lol

Unknown x

@TedShifrin a fixed point in which all tangent passes.

Ted Shifrin

So how do you write a point on the tangent line to $\alpha$ at $\alpha(s)$?

noballpointpen

Sourav's hint was acutally that "very brief" hint stated in supplements that makes the problem very easy... Was I in the right direction of arguing yesterday, at least? Could it be possible to turn my claims into rigorous? Without more sophisticated definitions and theorems that are not learnt in basic real analysis.
I know what is Mobius strip shape (in popular sense), I think I just didn't get your joke and that meaning with intuition. If we have a unit square shape where the graph lies, turning it into cylinder makes the diagonal like like this >. In Mobius case, uh... I think I need 3D …

Transcript for

$Mathematics$ Mathematics