00:00 - 16:0016:00 - 00:00

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.

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?

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

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

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!

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.

12:41 AM
thank you!

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.

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.

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.

1 hour later…
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 😊😊)

hope you're ok.

3:12 AM
any clue what (point) refers to here?

algebra is not my strong point.

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

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

Ted or Leslie

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

3:15 AM
is this quotients in tvs?

just set quotienting here is how i interpreted it

Every group defines a Turing Machine

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

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

3:38 AM
I wish I had a finger
To chop

what does that mean?

that is an incredibly ambiguous statement. impressive

4:12 AM
@RussianBotWhoKnowsYourIP have you figured out my ip addr yet?

@TedShifrin do you mean the inequality is not bounded ?

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?

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

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

i don't understand.

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

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

noooooo

they contain the same elements

yeah you're right

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

4:25 AM
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 )...

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

multiset. but before going there, why???

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

i am asking what you have in mind.

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

4:29 AM
presumably you had a reason for wanting such a thing?

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

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

192.158.1.38

@copper.hat cool thank you!

@RussianBotWhoKnowsYourIP funny :-)
i am 60, but not quite in the VA center in palo alto yet

4:33 AM
i am 2 y/o

seems to be lot of folks like that here
euler

i am a russian bot who knows your ip
no euler

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

if you know my ip put it here.

i am not allowed to do that @copper.hat

4:35 AM
you mean you don't know it.

i know it

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

hey, me too actually

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

i know all the digits of the largest known mersenne prime
in binary

4:37 AM
script kiddies. yawn

let me delete some data
%delete%

you are full of it :-)
it is actually not hard to actually find my ip address. it does not require hacking even.

today i found that humans love to argue over silly things

that is not true

brilliant

Those look like things that have already been posted.

when
how
why

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.

@copper.hat no proof remains

i don't need proof.
there are enough similar characteristics.

4:48 AM
i am euler
euler version 2.0.0

there are 17 users with the username euler.

that is why i changed it

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

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

@RussianBotWhoKnowsYourIP you need an outlet

4:52 AM
ok
installing outlet..... completed

i mean for expression, i am sure you have power

i found a cool website

a bit dated for me.

what is the point???

5:03 AM

i clicked on it in my tor window

what happened

what would you expect?

waht didd u c

they should start putting lead back in gasoline and paint

5:05 AM
every given pair of points with fixed modulii and fixed product, what minimises their distance ?

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

ilaborat plz

It is about the same question of yesterday's.

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

5:08 AM
$|x-y| \ge |x| - |y|$.

Yes, I got the answer by two methods ..

so, what is the question?

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

what inequality?

Okok I got it, that's different.

5:21 AM
$\oplus$
i lvoe 2 distrub peopols

the zodiac killer has entered the chat

poor cereal

woh iz da zodiak killir

use the interweb

ha yez thamcs

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 $\oplus$.

myabe u r da zodiak killir

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

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.

u r etarnal
da tinelime worgs uot

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 + ...

we learned that in highschool

dis iz borimg

i thought it was amazing
but (at the time) found it hard to formalise
@TedShifrin good luck tomorrow.

sems lice mi imglish grammer flie iz korruptid

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

5:41 AM
my favourites were optimization problems.
i like the applied maths stuff

6:12 AM
goon night folks

that's you told
if you rearrange the letters in amply notified you can spell leafy midpoint

1 hour later…
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?

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 $\mathbb{R}_{0+}$

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

ahhh, you're right
thank you!

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

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.

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

8:05 AM
@MartinSleziak much nicer to look at, thank you!

hi gang!

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

Cats should bark

@RussianBotWhoKnowsYourIP up the wrong tree

Dogs should meow

8:16 AM
Missed a minus sign.

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?

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 ?

9:30 AM
why does x belong to $(0, \infty)$? @SrijanM.T

9:46 AM
square the values brackets
(-$\infty$ , 0) U (0,2).

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

1 hour later…
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.

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

11:54 AM
Do I need to put commas in writing a metric signature? For example: Should I write (+ - - -) or (+,-,-,-) ?

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

12:13 PM
Okay, thank you.

12:32 PM

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

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

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)

@shintuku Yes
but where is 0,4 to infinity

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

@shintuku Yes

step 2: therefore, x^2 is in (4, infinity)
where is the loss?

1:06 PM
@shintuku When you took - infinity , 0
U 0,2

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

This also means - infinity,2 right ?
- infinity , 0 U 0,2 = - infinity ,2
Agree ?
U is union

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

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.

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

1:11 PM
@shintuku Yes. Right
I rearranged it afterwards
Check
Infinity , 0 U 0 , 4. Right
Then it’s final is 0 , infinity

there is no rearranging moment
try x in (-100, 0)
what is x^2?

@shintuku 0 , 100^2

right, make sure to add brackets
0, 100^2 is ambiguous
(0, 100^2) makes it clear it is an interval

Right.
@shintuku K.

{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

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

so, we have x in (-infinity, 0]U[0, 2)
this means x is also in (-infinity, 2)

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

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

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

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

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

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

Ok.
@shintuku For this , you meant 0,infinity?
@shintuku That’s the answer in textbook

ah, actually you're right
ok, yeah I made a huge mistake

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.

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)

1:30 PM
@shintuku Nooo bro.
(-infinity, 2)
What is square of 2

And -infinity

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)

Right. So , you cant solve it like square of just 2 and - infinity
That is because of 0

also because of -2 and 2
and all similar numbers

1:32 PM
Ok.

because they both square to 4
if you remember the shape of a graph f(x) = x^2

Right. In between them , 0. Which is still 0.
@shintuku Parabola

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

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

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

1:33 PM
good luck

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

@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.

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!

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

what's +ve
also, it is difficult to read you without the brackets

@shintuku 5 th line ?

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

@shintuku positive.
Positive values
But no , it can -ve also. Total expression can be -ve
And so can x be

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

2:47 PM
what if the inside of the square root is complex?

@Onir Those are imaginary numbers
I don’t think they want that
I’m not sure just think

oh

@shintuku See
What if x is - 2
You still +ve numbers right

\sqrt{-6} is complex

@shintuku This expression , doesn’t have -6.
What do you guys think of the answer as

2:52 PM
the possible values of $x$ such that $\sqrt{x^2 - 4}$ has a real solution?

@shintuku I think that’s what it wants. But I’m guessing
This is the answer and I don’t approve of it.

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

why doesn't it get your approval?

@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

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

2:56 PM
But the square root function is usually considered to be the one that gives you the positive one

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

oh

I just solved one equation and it still gave right answer.
@shintuku No.sqrt of x^2 - 4

horribly put exercise

@Thorgott fully agree

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.

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

@shintuku Ok.

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

Why is textbook so wrong

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

23

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...

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

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

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

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

are L1 functions simple ?

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

@leslietownes thanks for the find

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

3:35 PM
@leslietownes I am not sure I understand the idea about simple functions in terms of L1 spaces. Would you please explain.

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

@leslietownes thanks a lot for the answer!

@leslietownes Thank You!

hello what can I do for you

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)

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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