Mathematics - 2021-05-18 (page 1 of 2)

12:27 AM

i think the irish breakfast tea at trader joes is oemed (new verb) from either barry's or lyon's tea. good value if you like black tea.

shintuku

although I have no tea, I most definitely have a math question: I'm trying to show an isomorphism between vector spaces is an equivalence relation, and I'm doing the criterion of reflexivity. The author uses the self-identity relationship to prove that an isomorphism is reflexive, but I'm left wondering: why would the self-identity relation serve to prove that any isomorphism is reflexive? Isn't self-identity just one of the possible isomorphisms from an object to itself?

copper.hat

you just need to show that there is an isomorphism $V \to V$. the identity is the simplest.

leslie townes

ooh i'll try that. i'm working through a box of whole foods black. it is OK but not superlative.

shintuku

ahhh, right. it doesn't matter if there is more than one isomorphism: $V$ is isomorphic to itself if it has at least 1 isomorphism from itself to itself

thanks!

leslie townes

shintuku there is not a lot of content there, you are right to react as though 'wait, is this it?' that is it.

in practice reflexivity is often the easiest to verify property of a putative equivalence relation. in theory it can be the hardest, because it's the only one of the axioms that requires you to exhibit something that's in the relation.

shintuku

12:41 AM

thank you!

leslie townes

i'm trying to think of an example. cobordism is not it, but something along those lines would do the trick.

in grad school one of my friends had a relationship like cobordism but with a bunch of geometric conditions on the manifold that had the boundary of the cobordant-like-things. it was obviously symmetric and transitive by duh and some kind of neighborhood theorem. he could not prove it was reflexive and gave up.

Thorgott

1:05 AM

reflexivity can be a subtle point

I remember I once fell for the "reflexivity is superfluous" fallacy once, since $x\sim y$ and $y\sim x$ by symmetry, whence $x\sim x$ automatically by transitivity.

leslie townes

1:35 AM

oh yeah, that's a classic. i think my first 'abstract algebra' book presented that fallacy as an exercise. really hammered home the point that reflexivity is the only thing that requires something to be in the relation.

i think the example i gave on that homework assignment was "equality of pairs of integers" as a relation on the set of rational numbers. the empty relation on a nonempty set felt too much like cheating.

Andrew Micallef

2:57 AM

@robjohn, sorry I havent had a chance to test it. Been busy. I chopped a neat chunk of finger off yesterday, so didnt get any evening maths in. Though I did just convince myself Eulers formula is true(I was 90% yesterday, Just got to the qed part 😊😊)

copper.hat

hope you're ok.

shintuku

3:12 AM

any clue what (point) refers to here?

copper.hat

algebra is not my strong point.

shintuku

i mean, I guess it is the partition which houses $A$, I just don't get why it is called a point

leslie townes

iff there is z in X/S for which A = pr^{-1} ({z}) would be a better way of phrasing the condition

copper.hat

Ted or Leslie

leslie townes

this is a pure definitions chase, not a lot of content

copper.hat

3:15 AM

is this quotients in tvs?

StudySmarterNotHarder

:math.stackexchange.com/questions/4142656/…

leslie townes

just set quotienting here is how i interpreted it

StudySmarterNotHarder

Every group defines a Turing Machine

shintuku

it's topology, but not yet. the author introduces quotient topology next page

so it's only set theory at that question

but yeah, I guess it makes sense to do it Leslie's way

leslie townes

the details or at least the notation of the proof might depend on how they choose to define X/S as a set

Russian Bot Who Knows Your IP

3:38 AM

I wish I had a finger

To chop

copper.hat

what does that mean?

shintuku

that is an incredibly ambiguous statement. impressive

copper.hat

4:12 AM

@RussianBotWhoKnowsYourIP have you figured out my ip addr yet?

Rover

@TedShifrin do you mean the inequality is not bounded ?

shintuku

so, if $x$ is a subinterval of the real numbers, $\{x\} \cup \{x\}$ can either be $\{x,x\}$ or $\{x\}$, and I feel like somewhere out there, whether it is one or the other depends on some axiom or some definition that is not currently in my position. does anyone know what it is that troubles me so?

copper.hat

$\{x,x\}$ and $\{x\}$ are the same thing. not literally, but as sets.

shintuku

ahhh, of course. the idea behind $\{x, x \}$ in this case would be better expressed by $\{\{x\}, \{x\}\}$, I guess

thank you!

copper.hat

i don't understand.

shintuku

4:22 AM

i mean, if I wanted to take the same subinterval of the reals, and have it behave in a set as if it was twice there, I'd better think $\{\{x\}, \{x\}\}$ instead of $\{x, x\}$

... right? $\{ \{x\},\{x\}\} \neq \{x\}$

robjohn

it equals $\{\{x\}\}$

shintuku

noooooo

robjohn

they contain the same elements

shintuku

yeah you're right

copper.hat

that makes no sense. if $A$ is a set then it is always true that $A \cup A = A$.

robjohn

4:25 AM

Axiom of extensionality

In axiomatic set theory and the branches of logic, mathematics, and computer science that use it, the axiom of extensionality, or axiom of extension, is one of the axioms of Zermelo–Fraenkel set theory. == Formal statement == In the formal language of the Zermelo–Fraenkel axioms, the axiom reads: ∀ A ∀ B ( ∀ X ( X ∈ A ⟺ X ∈ B ) ⟹ A = B )...

shintuku

dang it

any clue how I could take the same subinterval of the reals twice and put it in a set and have it be two distinct intervals?

so that, for example, I could extend one and not the other

copper.hat

multiset. but before going there, why???

shintuku

just to know if there is, somewhere out there, a logically coherent expression for what I have in mind

copper.hat

i am asking what you have in mind.

shintuku

that was it really. whether I can have a set which contains the same subinterval twice, and has these two elements as really distinct elements

copper.hat

4:29 AM

presumably you had a reason for wanting such a thing?

shintuku

yeah, to properly distinguish it, and banish it, from a proof I'm doing that a certain set of subintervals of the reals is a topology

copper.hat

you can always make things unique, instead of $\{x\}$, have $\{(1,x), (2,x) \}$.

Russian Bot Who Knows Your IP

192.158.1.38

shintuku

@copper.hat cool thank you!

copper.hat

@RussianBotWhoKnowsYourIP funny :-)

i am 60, but not quite in the VA center in palo alto yet

Russian Bot Who Knows Your IP

4:33 AM

i am 2 y/o

i was made in 2019

copper.hat

seems to be lot of folks like that here

euler

Russian Bot Who Knows Your IP

i am a russian bot who knows your ip

no euler

shintuku

ominous. eventually all the chatrooms in the internet will be filled with AIs that can generate jokes

copper.hat

if you know my ip put it here.

Russian Bot Who Knows Your IP

i am not allowed to do that @copper.hat

copper.hat

4:35 AM

you mean you don't know it.

Russian Bot Who Knows Your IP

i know it

robjohn

@copper.hat He knows everyone's IP, but not who each one belongs to.

shintuku

hey, me too actually

copper.hat

ok, telnet to my ip address, port 8081, i am listening

leslie townes

i know all the digits of the largest known mersenne prime

in binary

copper.hat

4:37 AM

script kiddies. yawn

Russian Bot Who Knows Your IP

let me delete some data

%delete%

copper.hat

you are full of it :-)

it is actually not hard to actually find my ip address. it does not require hacking even.

Russian Bot Who Knows Your IP

today i found that humans love to argue over silly things

copper.hat

that is not true

Alexandru Ionut

brilliant

Russian Bot Who Knows Your IP

4:41 AM

robjohn

Those look like things that have already been posted.

Russian Bot Who Knows Your IP

when

how

why

copper.hat

where's my rosenbrock function?

you are euler. same footprint

i have a nice version of the rosenbrock function but it is a scan so not easy to copy here.

Russian Bot Who Knows Your IP

@copper.hat no proof remains

copper.hat

i don't need proof.

there are enough similar characteristics.

Russian Bot Who Knows Your IP

4:48 AM

i am euler

euler version 2.0.0

copper.hat

there are 17 users with the username euler.

Russian Bot Who Knows Your IP

that is why i changed it

copper.hat

wow, they are tracking down duplicates of 10yo questions.

Russian Bot Who Knows Your IP

c o o l a n d s t r i c t

copper.hat

@RussianBotWhoKnowsYourIP you need an outlet

Russian Bot Who Knows Your IP

4:52 AM

ok

installing outlet..... completed

copper.hat

i mean for expression, i am sure you have power

Russian Bot Who Knows Your IP

i found a cool website

http://www.5z8.info/stalin-will-rise-again_p8w0ip_-OPEN-WEBCAM---START-RECORD--

copper.hat

a bit dated for me.

Russian Bot Who Knows Your IP

http://666.verylegit.link/64spyware.min.css.docm

copper.hat

what is the point???

Russian Bot Who Knows Your IP

5:03 AM

click on that link

copper.hat

i clicked on it in my tor window

Russian Bot Who Knows Your IP

what happened

copper.hat

what would you expect?

Russian Bot Who Knows Your IP

waht didd u c

leslie townes

they should start putting lead back in gasoline and paint

Rover

5:05 AM

every given pair of points with fixed modulii and fixed product, what minimises their distance ?

copper.hat

@Rover you you write the question a little more clearly please?

Russian Bot Who Knows Your IP

ilaborat plz

Rover

It is about the same question of yesterday's.

copper.hat

used to play with mercury growing up. i think, not quite sure now.

Rover

copper.hat

5:08 AM

didn't you already get an answer? Ted had a nice suggestion.

$|x-y| \ge |x| - |y|$.

Rover

Yes, I got the answer by two methods ..

copper.hat

so, what is the question?

Rover

@copper.hat does that mean the inequality is not bounded?

copper.hat

what inequality?

Rover

Okok I got it, that's different.

Russian Bot Who Knows Your IP

5:21 AM

$\oplus$

i lvoe 2 distrub peopols

leslie townes

the zodiac killer has entered the chat

copper.hat

poor cereal

Russian Bot Who Knows Your IP

woh iz da zodiak killir

copper.hat

use the interweb

Russian Bot Who Knows Your IP

ha yez thamcs

leslie townes

5:26 AM

he was a serial killer who stalked the bay area in the late 60s and early 70s. he signed some of his taunts with a symbol that is very close to $\oplus$.

Russian Bot Who Knows Your IP

myabe u r da zodiak killir

leslie townes

so i wasn't just babbling about him for no reason. there was a connection to LaTeX.

Russian Bot Who Knows Your IP

u sus

leslie townes

the timeline doesn't work out. although people have made that joke about me at work. i don't know why. i work with a lot of sick people.

Russian Bot Who Knows Your IP

u r etarnal

da tinelime worgs uot

leslie townes

5:31 AM

my favorite proof of the divergence of the harmonic series is a mild rigorization of 1 + 1/2 + 1/3 + 1/4 + ... = (1 + 1/2) + (1/3 + 1/4) + ... > (1/2 + 1/2) + (1/4 + 1/4) + (1/6 + 1/6) ... = 1 + 1/2 + 1/3 + 1/4 + ...

copper.hat

we learned that in highschool

Russian Bot Who Knows Your IP

dis iz borimg

copper.hat

i thought it was amazing

but (at the time) found it hard to formalise

@TedShifrin good luck tomorrow.

Russian Bot Who Knows Your IP

sems lice mi imglish grammer flie iz korruptid

leslie townes

the divergence of sum 1/sqrt(k) is another one of my favorites. the nth partial sum is a sum of n terms, all of which are >= 1/sqrt(n), so is at least sqrt(n). very easy to concretely find values of N for which you can make the Nth partial sum as large as you want.

everything i like about calculus i learned in high school

copper.hat

5:41 AM

my favourites were optimization problems.

i like the applied maths stuff

copper.hat

6:12 AM

goon night folks

leslie townes

that's you told

if you rearrange the letters in amply notified you can spell leafy midpoint

robjohn

7:33 AM

@leslietownes can you tell me what $1-\frac1{\sqrt2}+\frac1{\sqrt3}-\frac1{\sqrt4}+\dots$ is?

@copper.hat who is "amply notifying" people?

shintuku

7:50 AM

let $X \subset \mathbb{R}$ be $[0, \infty)$ and let the topology $\Omega$ on $X$ be the set containing $\varnothing, X$, and all rays $(a, \infty) \subset \mathbb{R}$.

Is the correct notation for $\Omega$ this one? $\Omega = \{\varnothing, X\} \cup \bigcup_{a \in \mathbb{R}_{0+}}\{\{x:a<x<\infty\}\}$

or does the big union not have nested brackets?

woops, the rays $(a, \infty)$ are meant to be in $\mathbb{R}_{0+}$

robjohn

@shintuku If you don't have double brackets you just get $(0,\infty)$

shintuku

ahhh, you're right

thank you!

Martin Sleziak

I guess you could simply write $\{(a,\infty); a\in\mathbb R_{0+}\}$ instead of that union.

user76284

Is this identity correct? $$\nabla \mathrm{H}(p) = \mathrm{E}_{\omega \sim p} \nabla \frac{1}{2} (\ln p(\omega))^2$$ where $p$ is a probability density, $\mathrm{H}$ is the entropy, and $\mathrm{E}$ is an expectation. Just sanity-checking.

Martin Sleziak

I can't resist mentioning the dormant General topology chatroom.

shintuku

8:05 AM

@MartinSleziak much nicer to look at, thank you!

Alexandru Ionut

hi gang!

robjohn

I thought I heard a transcriptional ghost, or maybe it was just a settling of the timbers.

Russian Bot Who Knows Your IP

Cats should bark

robjohn

@RussianBotWhoKnowsYourIP up the wrong tree

Russian Bot Who Knows Your IP

Dogs should meow

user76284

8:16 AM

Missed a minus sign.

shintuku

9:09 AM

so I'm trying to prove the set $\Omega := \{\varnothing\}\cup\{[0,\infty) \in \mathbb{R}\}\cup\{(a,\infty):a\in\mathbb{R}_{0+}\}$ is a topology on $X=[0, \infty)$, and I'm doing the condition $\forall \Omega_a, \Omega_b$ in $\Omega$, we have $\Omega_a \cup \Omega_b \subset \Omega$.

I stated: "For any $A_1,A_2 \in \{(a,\infty):a\in\mathbb{R}_{0+}\}$, we have: that the union of $\varnothing$ with a nonempty set is that nonempty set; $(X \cup A_1) \subseteq X$; and $A_1 \cup A_2$ is identical to whichever of $A_1, A_2$ is larger."

"Therefore, the condition holds"

someone pointed out that this isn't sufficient, because the unions need to be arbitrary

does anyone have a clue how I can get the unions to be arbitrary?

Srijan M.T

In my textbook , it says x ∈ (-$\infty$ , 0) U (0,2).

Then , squaring the two brackets.

($\infty$,0$ and ( 0 ,4 )

Arranging it ( 0 ,4 ) U ($\infty$,0$).

Therefore , x belongs to 0,$\infty$.

Now , my questions are:

1) According to the Q , $x^2$ are we only wish to find , we need want them to satisfy like $x^2$ < 4. Squaring the possible values , we should get ($4$ , $\infty$) as the possible values if i in the starting only took (-$\infty$,2). Why do we see loss of values from 0 to 4 in this case ?

shintuku

9:30 AM

why does x belong to $(0, \infty)$? @SrijanM.T

Srijan M.T

9:46 AM

square the values brackets

(-$\infty$ , 0) U (0,2).

shintuku

does the square map do $x \mapsto x^2$ for any element of the set, and are both of those sets subsets of the reals?

also, why do you say there's a loss of value from 0 to 4?

maybe if you post the exact question in your textbook it will be easier to answer

Martin Sleziak

11:01 AM

@shintuku I am not sure whether this is what you're looking for, but I have posted some response in the General Topology chatroom.

epsilon-emperor

11:39 AM

Hello! I think the solution to 5(a) here on Page 4 is wrong. Could someone check? math.iisc.ac.in/~rakesh13/…

Theorem 1.12(c) of Rudin's RCA says that IF f^{-1}((\alpha,\infty]) is measurable for all \alpha, then f is measurable. Here they seem to be concluding the converse! I think that's wrong.

P.S. The question is related to measure theory

raf

11:54 AM

Do I need to put commas in writing a metric signature? For example: Should I write (+ - - -) or (+,-,-,-) ?

Charlie

In physics it's normal to write commas, presumably since you're saying that the metric can be written $\mathrm{diag}(+1,-1,-1,-1)$ that's where the commas come from

raf

12:13 PM

Okay, thank you.

Srijan M.T

12:32 PM

shintuku

@SrijanM.T so where is the loss of 0 to 4?

Srijan M.T

12:58 PM

@shintuku Suppose you took range from - infinity to 2

Not including union of 0

Then , your range is 4 , infinity

@shintuku 0,4 is lost right

shintuku

no, why would the interval be (4, infinity) when your interval is (-infinity, 2)?

if you have an x in (-infinity, 2), then x^2 will be in (4, infinity)

Srijan M.T

@shintuku Yes

but where is 0,4 to infinity

shintuku

step 1: we say x is in (-infinity, 2)

Srijan M.T

@shintuku Yes

shintuku

step 2: therefore, x^2 is in (4, infinity)

where is the loss?

Srijan M.T

1:06 PM

@shintuku When you took - infinity , 0

U 0,2

shintuku

in step 1, x is in (-infinity, 2), which means it is also in (-412, -100), (-2418503927, 0), (0,2)

Srijan M.T

This also means - infinity,2 right ?

- infinity , 0 U 0,2 = - infinity ,2

Agree ?

U is union

shintuku

yep, but also (-infinity, -100)U(-100,-99)U(-99,0)U(0,2) = (-infinity, 2)

Srijan M.T

Then , square - infinity , 0 U 0,2 , we get infinity,0 and 0,4. Arranging this we get 0 , infinity. Ok so far ?

But in case of - infinity , 2. We get 4 , infinity. Now , did you notice the loss of 0,4 interval

@shintuku Yes.

shintuku

x in (-infinity, 0) means x^2 in (0, infinity), not x^2 in (infinity, 0)

Srijan M.T

1:11 PM

@shintuku Yes. Right

I rearranged it afterwards

Check

Infinity , 0 U 0 , 4. Right

Then it’s final is 0 , infinity

shintuku

there is no rearranging moment

try x in (-100, 0)

what is x^2?

Srijan M.T

@shintuku 0 , 100^2

shintuku

right, make sure to add brackets

0, 100^2 is ambiguous

(0, 100^2) makes it clear it is an interval

Srijan M.T

Right.

@shintuku K.

shintuku

{0, 1000}, for example, is a set of two numbers. (0, 1000) is an open interval. [0, 1000] is a closed interval

so the brackets change the meaning of the expression

Srijan M.T

1:15 PM

@shintuku Yes. I just didn’t add because brackets won’t change after squaring right

but yes , in exam or sth. I’ll add

shintuku

so, we have x in (-infinity, 0]U[0, 2)

this means x is also in (-infinity, 2)

Srijan M.T

@shintuku Yes

Right

@shintuku Do one thing , what will be its value for x^2 ?

shintuku

since x is in (-infinity, 2), x^2 will be in (4, infinity)

Srijan M.T

Right

Now , we remember (-infinity, 0]U[0, 2) for this , it was (0,infinity)

Agree ?

shintuku

no, if x is in (-infinity, 0]U[0, 2), x^2 is not always in (0, infinity)

Srijan M.T

1:19 PM

@shintuku Ok. But then the answer in my textbook is wrong ?

You need to provide one answer right

So , do you mean to say that x^2 values are not same but x are ?

That’s how it looks like

shintuku

the answers in your textbook are fine

1. If x is in (-infinity, 0]; x^2 is in [0, infinity)
2. If x is in (0, 2); x^2 is in (0, 4)
3. If x is in (-infinity, 2); x^2 is in (4, infinity)

4. If x is in (-infinity, 0]U[0, 2) that means it is in (-infinity, 2), so the answer is the same as #3

Srijan M.T

Ok.

Getting your point

@shintuku For this , you meant 0,infinity?

@shintuku That’s the answer in textbook

shintuku

ah, actually you're right

ok, yeah I made a huge mistake

Srijan M.T

So , do you agree to say that x^2 values are not same but x are ?

I think biggest reason is behavior of 0 here.

shintuku

so if x in [0, 2), x^2 is in [0, 4). if x in (-infinity, 0]; x^2 is in [0, infinity). Therefore, if x is in (-infinity, 2), x^2 is in [0, infinity)

Srijan M.T

1:30 PM

@shintuku Nooo bro.

(-infinity, 2)

What is square of 2

shintuku

4

Srijan M.T

And -infinity

shintuku

infinity. but you're just taking the extreme values. (0, 2) is in (-infinity, 2)

you need to take into account the values of x in (0,2), which is a subinterval of (-infinity, 2)

x in (0,2) means x^2 in (0, 4)

Srijan M.T

Right. So , you cant solve it like square of just 2 and - infinity

That is because of 0

shintuku

also because of -2 and 2

and all similar numbers

Srijan M.T

1:32 PM

Ok.

shintuku

because they both square to 4

if you remember the shape of a graph f(x) = x^2

Srijan M.T

Right. In between them , 0. Which is still 0.

@shintuku Parabola

copper.hat

Treat is as $(-\infty,0] \cup [0,2)$.

shintuku

yes, and f(-x) = f(x)

Srijan M.T

@shintuku Yeah. Ok. I gotta go now bro. Great discussion.

Have a class

shintuku

1:33 PM

good luck

Russian Bot Who Knows Your IP

hi

shintuku

alright, my turn for a question. I have an arbitrary collection of sets $\{U_i\}$ from topology $\Omega := {\varnothing, [0, \infty)} \cup \{(a, \infty): a \in \mathbb{R}_{0+}\}$. In the case the collection${U_i}$ doesn't have either $\varnothing$ or $[0, \infty)$, the union of all its members $\bigcup Ui =(a^*, \infty)$ where $a^*$ is the least such $a^*$. Now, for the life of me, I have been suffering trying to designate it with set theory. Is it $a^* = \inf \{a_i : (a_i, \infty) \in U_i\}$?

I feel like this designation should mention the indexing set somehow

robjohn

@epsilon-emperor below that it is said that that is often used as the characterization for measurability for real-valued functions... so it would go both ways.

epsilon-emperor

2:05 PM

@robjohn Oh yes, thank you! I also just noticed that by definition of measurable functions, pull-backs of open sets are measurable sets, so it works!

Lucas Henrique

hey

Srijan M.T

2:35 PM

If Q ask to find possible values of sqrt of x^2 - 4

Does it mean all values of x due to which the expression doesn’t or produce imaginary numbers

How I solved it is that least value of x^2 is +- 2

Then , range for sqrt of x^2 - 4 would be +ve if the whole expression has +ve values

Range of x can be from - infinity , -2 U 2 , infinity

Im sure you all know and so do I which brackets to put

shintuku

what's +ve

also, it is difficult to read you without the brackets

Srijan M.T

@shintuku 5 th line ?

Kashmiri

Hi, can anyone help me with a doubt about vectors in higher dimensions?

Srijan M.T

@shintuku positive.

Positive values

But no , it can -ve also. Total expression can be -ve

And so can x be

shintuku

for a solution in the reals, the inside of the square root has to be positive or 0

it has a solution in the complex numbers if the inside of the square root is negative

Onir

2:47 PM

what if the inside of the square root is complex?

Srijan M.T

@Onir Those are imaginary numbers

I don’t think they want that

I’m not sure just think

Onir

oh

Srijan M.T

@shintuku See

What if x is - 2

You still +ve numbers right

shintuku

\sqrt{-6} is complex

Srijan M.T

@shintuku This expression , doesn’t have -6.

What do you guys think of the answer as

shintuku

2:52 PM

the possible values of $x$ such that $\sqrt{x^2 - 4}$ has a real solution?

Srijan M.T

@shintuku I think that’s what it wants. But I’m guessing

This is the answer and I don’t approve of it.

shintuku

any $x$ outside $[-2,2]$

Onir

why doesn't it get your approval?

Srijan M.T

@Onir Ok. What if sqrt of x^2 - 4 = -3. Then , squaring. x^2 - 4 = 6. ,x^2 = 10.

And that is real numbers.

Also , it is in range of negative numbers then. -3

shintuku

x such that x^2 - 4 = -3 will give a complex solution to \sqrt{x^2 - 4}

Onir

2:56 PM

But the square root function is usually considered to be the one that gives you the positive one

Srijan M.T

@Onir Yes. we’re talking about the entire expression.

Onir

oh

Srijan M.T

I just solved one equation and it still gave right answer.

@shintuku No.sqrt of x^2 - 4

Thorgott

horribly put exercise

shintuku

@Thorgott fully agree

Srijan M.T

2:57 PM

@Thorgott I know right.

@Thorgott Idk if I’m not getting it but for this moment , it isn’t making sense to me.

shintuku

I think you should read that exercise as: Find all possible values of x such that the following expressions have a real solution

Srijan M.T

Answer should be (-infinity,-2]U(2,infinity)

@shintuku Ok.

shintuku

yeah, that's correct for \sqrt{x^2 -4}

Srijan M.T

Why is textbook so wrong

Onir

3:14 PM

can you guys give examples of topological spaces with more than 1 point such that the identity is the only automorphism?

Oh I think I found one, just take the right order topology

Oh but the set needs to be finite to work

leslie townes

23

A: Hausdorff spaces with trivial automorphism group

Not at all. Those spaces are called rigid and there are plenty of examples in the literature. The opposite notion is homogeneity which is a better studied property. The first (non-trivial) rigid space was constructed by Kuratowski in "Sur la puissance de l'ensemble des nombres de dimension au sen...

shintuku

does anyone know what I need to read up on to figure out how the index sets of arbitrary collections of sets work? I can't quite conceptualize how you can have an arbitrary index set

Srijan M.T

@shintuku Hey , you can put it on the site or ask in Teresa Lisbon chat room.

Recommendation from me

shintuku

3:29 PM

it seems to me an arbitrary collection of sets implies a certain structure of the index set, but I can't quite figure what

leslie townes

shin, it's a bit like the {(1,x), (2,x)} example someone offered yesterday but with other sets in place of {1,2}. very frequently in arguments you see numeral subscripts where the order (or the fact that the subscripts are numbers) aren't used for anything. of course, often ordered index sets are used for a reason and in those cases you're probably doing more than just indexing a family of sets

cabmetric

are L1 functions simple ?

leslie townes

in general no. maybe yes if the measure space is "choice"

Onir

@leslietownes thanks for the find

leslie townes

for example on {1,2,...,n} with the counting measure, the L^1 space is the space of all functions on {1,2,...,n}, and all functions are simple

cabmetric

3:35 PM

@leslietownes I am not sure I understand the idea about simple functions in terms of L1 spaces. Would you please explain.

leslie townes

Simple function

In the mathematical field of real analysis, a simple function is a real (or complex)-valued function over a subset of the real line, similar to a step function. Simple functions are sufficiently "nice" that using them makes mathematical reasoning, theory, and proof easier. For example, simple functions attain only a finite number of values. Some authors also require simple functions to be measurable; as used in practice, they invariably are. A basic example of a simple function is the floor function over the half-open interval [1, 9), whose only values are {1, 2, 3, 4, 5, 6, 7, 8}. A more advanced...

the concept isn't in any way specific to L^1 spaces or integration

most functions on most spaces aren't simple

just stick with functions on [0,1] for example. f(x) = x isn't a simple function

it happens to be in the usual L^1 space of [0,1] and a lot of other spaces too, but before you get to anything like that, first convince yourself that it isn't a simple function

shintuku

@leslietownes thanks a lot for the answer!

cabmetric

@leslietownes Thank You!

Onir

hello what can I do for you

shintuku

3:57 PM

so, if I have an index set $I$ and a mapping to some set $S$, is there such a thing as a disordered mapping? such that every element $i \in I$ is mapped to an element $s \in S$, but we can't necessarily tell any sort of order between the elements of $I$ or those of $S$? I imagine these sets as two blobs of points and the map as a line from one to the other

(asking to figure out how arbitrary collections of sets might work)

leslie townes

shintuku if I is just some random set, there's no pre-existing notion of order on it in the first place

if it has some natural order but you don't use it, you're just ignoring it. i wouldn't think of it as extra structure or a new notion, more like, not paying attention to something that could be there but isn't being used

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