Mathematics - 2015-03-20 (page 1 of 3)

Karim Mansour

00:04

Hi guys

ᴇʏᴇs

00:14

Hi @KarimMansour

Karim Mansour

I am taking general topology next year I want to ace it :D I will read munkrees during the summer

ᴇʏᴇs

00:35

@KarimMansour Me too, we can read together

Karim Mansour

cool we can discuss some stuff during that time

I will usually be here in chat so we can discuss

ᴇʏᴇs

@KarimMansour I'm kind of slow, though

Karim Mansour

its np it is good to discuss ideas with people

Committing to a challenge

00:58

Anyone else noticed in the profile that where it once(up until two days ago I believe) said activities, it now says 'all actions'

Karim Mansour

01:08

brb I will go take a nap

Committing to a challenge

Enjoy your nap

ᴇʏᴇs

01:27

Hi @Committingtoachallenge

Committing to a challenge

Hi @ᴇʏᴇs @

(Does that ping you btw?)

ᴇʏᴇs

@Committingtoachallenge No

@Committingtoachallenge Yes

Committing to a challenge

Oh, you are the only one who appears when I type just the @

So to ping you I just type @ and click you

ᴇʏᴇs

@Committingtoachallenge That's just because I'm reading the chat most of the day lol

El'endia Starman

Hey guys. I just learned about rings in Abstract Algebra II this morning and am working on a homework problem. My goal is to find a subring $S$ of $\mathbb{C}$ that meets a specific criteria. Now, the question: I know $S$ has to be closed under addition, but does it have to be closed under multiplication?

For example, would $\left( \mathbb{Z}+\frac{1}{2}, +, \cdot \right)$ be a ring?

Semiclassical

01:32

if it's not closed, it's not a ring, and a subring has to be able to stand on its own as a ring

Committing to a challenge

What do you mean by $\Bbb Z + \frac12$?

El'endia Starman

The integers plus 0.5. Actually, that wouldn't be a group under addition.

The integers plus 0.5 and zero?

Wait, there's a better one.

Semiclassical

that would generate a group under addition, but it isn't one with just those elements

El'endia Starman

$\frac{1}{2}\mathbb{Z}$

Committing to a challenge

@Semiclassical A group under addition with $((\Bbb Z + \frac12)\cup\Bbb Z)$

Semiclassical

01:35

right. @Committingtoachallenge

El'endia Starman

@Committingtoachallenge I think that's the same as $\frac{1}{2}\mathbb{Z}$?

Committing to a challenge

Yep I just typed it differently

Semiclassical

that'll work for the addition operation, but not for multiplication

Committing to a challenge

Multiplication just needs to be a monoid

So associative

But yes, closure breaks down

Semiclassical

right

El'endia Starman

01:38

Yeah, the definition of a ring my professor gave didn't require that the set/group be closed under the second operation, multiplication in this case.

So would $\left( \frac{1}{2}\mathbb{Z}, +, \cdot \right)$ be a valid ring?

Committing to a challenge

Well a monoid maps $S\times S \to S$ so it is closed by definition

ᴇʏᴇs

What's a monoid

Committing to a challenge

A monoid is an associative binary operation with identity

ᴇʏᴇs

No inverse?

Committing to a challenge

Nope

Semiclassical

01:44

if everything had an inverse, it'd be a group

Committing to a challenge

Inverse makes it a group

Binary operation + Associativity + Identity -> Monoid and Monoid + inverse -> Group and Group + commutativity -> Abelian group

El'endia Starman

@Committingtoachallenge So, because I can do $\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \not\in \frac{1}{2}\mathbb{Z}$, my example above would not be a ring. Correct?

Committing to a challenge

@SemiC Can you confirm my answer to the above question being that yes, above isn't a ring

ᴇʏᴇs

Hi @JasperLoy

Semiclassical

right. $\frac12\mathbb{Z}\times \frac12\mathbb{Z}\to \frac14\mathbb{Z}$

Jasper Loy

01:46

@ᴇʏᴇs Hi, I feel much less anxious after taking the meds for two days.

Semiclassical

so yeah, it's not a monoid since it's not a mapping from $S\times S$ to $S$

ᴇʏᴇs

Good @JasperLoy

Committing to a challenge

@JasperLoy That's great to hear. Sounds like things are looking good

@JasperLoy I reached 100 helpful flags today and 1000 profile views :) and 1100 rep

Helpful flags + profile views = rep :)

El'endia Starman

@Semiclassical Okay, thanks. That's all I needed to know. :)

Semiclassical

glad to help

Jasper Loy

01:50

@Committingtoachallenge Now you are becoming nuts.

Committing to a challenge

@JasperLoy ::::::::)

I must go now to do more work, bye!

Jasper Loy

@ᴇʏᴇs Did I tell you, that day when I sent out the email about my new email I sent it to the hospital as well by mistake. Luckily I did not say anything sensitive, lol. I was shocked when the hospital replied, and I told them it was a mistake, lol.

ᴇʏᴇs

@JasperLoy What did they say

Jasper Loy

@ᴇʏᴇs They wanted to update my email anyway, lol. I told them it was a mistake but had them update my email anyway, lol.

@ᴇʏᴇs I have now removed the hospital from my contact list to avoid this in future.

Clarinetist

02:17

Hello Math SErs

Jasper Loy

@Clarinetist So did you get the new job?

Clarinetist

Not yet. Waiting on an interview. :)

Next Wednesday

Jasper Loy

02:38

Hi @robjohn I feel better after taking the meds.

Jasper Loy

02:56

@ᴇʏᴇs Have you decided what courses to take?

Jasper Loy

03:10

Wow, this chat is so quiet.

Jasper Loy

04:01

Hello @KajHansen how is progress with the gym girl?

Mary Star

Hello @DavidWheeler!! Are you familiar with arc length??

Kaj Hansen

Haven't seen her in a while @JasperLoy, but that's alright

Jasper Loy

@KajHansen OK, there are many trees in the forest, lol.

Kaj Hansen

I'm feeling better recently due to a good topology test score, spring break, and getting accepted to an REU.

Jasper Loy

Good for you.

Kaj Hansen

04:06

@Committingtoachallenge, @Chris'ssis, I agree with you two. I know tons of people who take that shit, and I consider it cheating. I don't understand why people find it so difficult to sit down for 4 hours at a time or however long it takes to get their work done. Really makes me angry because, short term, I suffer for doing things the right way.

Paul Plummer

04:32

@Kaj I guess you think Erdos is a cheater and his existence makes you angry... I think that attitude towards drugs is strange, math isn't a sport or competition. Plus you imply that you have no difficulty sitting down and focusing for hours to get work done (not everyone has that ability even if that is what they want to do), so maybe you don't even have a "need" for such drugs (or your body produces such drugs at a higher rate or quantity than other people).

David Wheeler

05:01

Erdos, drugs, huh?

Jasper Loy

I think I can solve all my mental problems. I am 99 per cent confident now.

Kaj Hansen

@PaulPlummer, I actually do not think Erdos is a cheater. I agree with you that math is not a competition -- the opposite, really. There is competition, however, amongst high schoolers and undergraduates. Take the latter for example. I depend on funding to attend UGA. The availability of funding is determined by my GPA. The GPA cutoff is determined by average and median GPAs of people attending colleges and universities around Georgia over, say, the past decade.

Paul Plummer

@DavidWheeler here

Kaj Hansen

I sincerely believe that drug use is so rampant on campus that it makes a significant difference -- perhaps up to 0.5 GPA points -- in these statistics. So people who don't want to poison themselves with amphetamines are put at an unfair disadvantage.

Furthermore, possession of amphetamines is illegal. If you are, e.g., a member of a fraternity or sorority, where people have a ridiculous degree of immunity when it comes to being punished for breaking the law, then you are obviously at an advantage if you choose to take amphetamines (among the purposes these organizations serve, IMO, is as a front for distributing drugs to their members, distributing tests to their members before they take them, and so forth).

On the other hand, Erdos was neither an undergraduate when he was taking amphetamines, nor were amphetamines illegal for much of…

Jasper Loy

Is this an amphetamine discussion?

@kaj Are you for and against them?

Kaj Hansen

05:07

Against, obviously. But only insofar as people making their way through undergrad. @JasperLoy

Jasper Loy

OK. I don't know anything about them. I am not sure whether they are legal in my country, but I think not.

Kaj Hansen

As far as what people do in their own time with their own money, I don't care. People should be free to act as they wish so long as they aren't hurting anyone.

David Wheeler

amphetamines always made me think in circles

not very effective for math n stuff-wound up solving the same problem repeatedly

robjohn

@JasperLoy I'm glad you're seeing a doctor and getting meds. I hope you feel much better.

Jasper Loy

I just answered a low hanging fruit, please upvote.

Jasper Loy

05:35

@robjohn I will keep my current SE account for life, I promise.

Paul Plummer

@Jasper How did you let your other account die, forget to feed it?

LeGrandDODOM

@DanielFischer Thanks, I'll take a look

Jasper Loy

@PaulPlummer I just choose to delete accounts now and then. I think I have deleted about 30k of rep by now.

Paul Plummer

@Jasper Why?

Jasper Loy

@PaulPlummer I am weird.

robjohn

05:43

@JasperLoy That will be nice :-)

Paul Plummer

@Jasper Do you like to make furniture/attire out of you previous account?

Jasper Loy

@PaulPlummer What does furniture mean?

Paul Plummer

Like chairs, couches, tables, etc

Jasper Loy

Haha, I don't get your joke.

Paul Plummer

@Jasper Are you like a serial killer that likes to wear your accounts you have killed skin, and sit on them when bending the bodies into chairs?

Jasper Loy

05:49

@PaulPlummer No. Your joke is too sophisticated for a banana like me.

Paul Plummer

@Jasper I am not sure if I would go so far as sophisticated, its probably closer to to not funny

Jasper Loy

@PaulPlummer OK, I like stupid jokes, like saying that I am a banana.

Paul Plummer

@Jasper I is just you mentioned you didn't keep your account alive, personifying your account, and I am vaguley referencing movies like Silence of the Lambs and Texas Chainsaw Massacre where they killed people for the reason, in part, to do things like wear their victims skin. Just seeing if you have a similar reason for letting your accounts die.It

Jasper Loy

@PaulPlummer Ah, I think I watched those movies.

Sayan Chattopadhyay

06:54

hi guys back and running

user61230

07:05

Anyone have a moment for a Fourier transform problem? My math prof, Wolfram, and by-hand calculation are giving me three different answers.

user61230

It's the FS of $\delta(t+7)$.

Jasper Loy

Fourier transform has different definitions, lol.

Sayan Chattopadhyay

hi @JasperLoy

Jasper Loy

@SayanChattopadhyay Hi.

user61230

My teacher's notes say it's $e^{7iw}$, Wolfram says $\frac{e^{-7iw}}{\sqrt{2\pi}}$, and $\frac{1}{2\pi}\int_{-\infty}^{\infty}\delta(t+7)e^{-iwt}dt$ spits out $\frac{1}{2\pi}e^{-7iw}$.

Sayan Chattopadhyay

07:08

how are you hows your health

Jasper Loy

@Emrakul Like I said, Fourier transform has different definitions. Look up Wiki first.

@SayanChattopadhyay Just started taking meds again. I hope I can sort out my thoughts soon.

Sayan Chattopadhyay

@JasperLoy what happened to a beautiful mind

Jasper Loy

@SayanChattopadhyay Nothing.

Sayan Chattopadhyay

and jasper the question which balarka gave me i got the answer for that@JasperLoy

Jasper Loy

@SayanChattopadhyay OK, you check with him then.

user61230

07:14

I... am confused. I'm not entirely sure where I'm going with this, but the prof. has written to use the table of transform properties, but the table disagrees with their result.

user61230

I'm honestly just not sure how to make sense of it.

user61230

...oh wait, I'm an idiot

Nick

07:48

@Emrakul No, you're not an idiot. Everyone errs but the only thing that matters is learning from it :D Keep on learning. Have a great life!

user61230

@Nick ...thanks! I actually needed that. :]

Nick

@Emrakul :D Everyone needs someone to say eveythings okay, buddy. Just pass on the kindness over to those who need it.

user61230

I'll, yeah. I'll remember it.

Sayan Chattopadhyay

08:04

hi @robjohn

Jasper Loy

08:16

@robjohn By the way, which of my messages were flagged recently? I would like to know if possible.

Fermé somme

@JasperLoy What do you recommend as a supplement for Rudin's Real and Complex Analysis?

Jasper Loy

@Fermésomme Why do you ask me?

Fermé somme

@JasperLoy You know a lot about books, right?

Jasper Loy

@Fermésomme What topic do you want to learn?

Fermé somme

@JasperLoy I was planning to learn Adult Rudin.

I wish I could find a book with more exercises.

Jasper Loy

08:26

@Fermésomme I mean what topic you want to learn so that I can recommend you alternatives, since Rudin deals with many topics.

Fermé somme

@JasperLoy I'm starting with abstract measure theory (first few chapters of rudin).

Jasper Loy

@Fermésomme I recommend Folland's Real Analysis.

Fermé somme

Does it contain a good number of exercises?

Jasper Loy

I think it does, but it has many typos.

Fermé somme

@JasperLoy Okay, I will take a look at it. Thanks!

Jasper Loy

08:29

@Fermésomme Did you have another username?

Fermé somme

@JasperLoy. Yes, many usernames.

Jasper Loy

@Fermésomme Can you tell me?

Fermé somme

Last one was Oracle.

Jasper Loy

Oh I see.

Fermé somme

@JasperLoy Are you feeling better?

Jasper Loy

08:32

@Fermésomme Only time will tell.

NikolajK

08:46

What's a (simple) statement in the language of Peano arithmetic, which can only be proven once we add the schema of induction?

Balarka Sen

09:06

@SayanChattopadhyay So, what's the answer?

Sayan Chattopadhyay

$$(n-1)n$$ i think so

Balarka Sen

no

that's wrong.

Sayan Chattopadhyay

why....

Balarka Sen

well, tell me how you arrived at that result, i can point you out the flaw. it's definitely wrong though - the answer is a wee bit more complicated.

Sayan Chattopadhyay

see if there are n letters that there always will be one way we can not put it into the envelope that is the ith envelope. now this will be the case for all n envelopes

Jasper Loy

09:10

Hi @Chris'ssis.

Chris's sis

Greetings

@JasperLoy Hi

Sayan Chattopadhyay

$$(n-1).........(n-1)$$ n times @BalarkaSen

Jasper Loy

@Chris'ssis I feel much better after taking the meds, at least for now.

Chris's sis

@JasperLoy Great. :-)

Balarka Sen

@SayanChattopadhyay i don't understand your logic. you have to put n letters into n envelopes such that no i-th letter goes into the i-the envelope. how're you planning to do it?

and how do you even get (n-1)?

[i think you are on the right track, but i am not getting what you are trying to do. it'd be nice if you can explain your ideas more clearly :)]

Sayan Chattopadhyay

09:13

see lets take there there is no idea of ith letter going into the ith envelope then there are n ways of putting it into envelopes.

robjohn

@JasperLoy this one. It was because they assumed you were talking about the fake user's comment

Sayan Chattopadhyay

now since we have the idea of the ith letter not going into the ith envelope there will one way in which we cannot put the letter into the envelope that is the ith letter not going into the ith envelope. this will similarly go for all n envelopes

ADG

hello

Balarka Sen

@SayanChattopadhyay not true. if there is no conditions imposed, then there are $n!$ ways to do it.

ADG

hello

Jasper Loy

09:16

@robjohn LOL

@robjohn My toes are laughing now.

ADG

anyone watching AUS vs PAK?

Sayan Chattopadhyay

me

Balarka Sen

you see the flaw in your logic, then, @Sayan?

ADG

PAK may win

and then MAUKA MAUKA !! :D

Sayan Chattopadhyay

therefore then there will be $$(n-1)!n!$$

Balarka Sen

09:17

not true either

Jasper Loy

@robjohn This is really too funny for me to even laugh.

ADG

let experts solve your problem @SayanChattopadhyay

(^_^)

Balarka Sen

You should read up chapter 3 of part I in Hammack carefully, then, @Sayan. You haven't done so, apparently.

Skimming through everything is very easy. Grasping the concepts is hard. That's why I said learning set theory may take a few years.

[Hint : To solve this problem, you need to apply inclusion-exclusion principle]

Sayan Chattopadhyay

wait a second then.... $$n(n-1)(n-2)...............3*2$$ for the first envelope@BalarkaSen

Balarka Sen

what d'you mean "for the first envelope"?

Sayan Chattopadhyay

09:27

the letter number 1 for the envelope number 1@BalarkaSen

Balarka Sen

and what are you doing with the letter number 1? you are counting the number of ways you can put letter #1 into an envelope, is that what you're doing?

Sayan Chattopadhyay

yup

Balarka Sen

then false. it's $n$.

Sayan Chattopadhyay

with not counting the 1st envelope for letter 1

Balarka Sen

then it's $n-1$.

Sayan Chattopadhyay

09:30

then isnt what i said right $$(n-1)n$$

Balarka Sen

no

figure out why ;)

zed111

hey what is the meaning of $\subset_{+}$ symbol?

Sayan Chattopadhyay

there are $$(n-1)$$ ways for a letter to be put into envelope

Balarka Sen

right

Jasper Loy

@zed111 Context.

Sayan Chattopadhyay

09:32

there are total n letters and n envelopes

zed111

A $\subset_{+}$ B where A and B are sets

Balarka Sen

correct

zed111

@JasperLoy $\subset_{+}$ B where A and B are sets

@JasperLoy: A $\subset_{+}$ B where A and B are sets

Sayan Chattopadhyay

so if $$n-1$$ for one letter it will be for all the letters

gtg cming back in a minute

Balarka Sen

no : this is because $|A \cup B| \neq |A| + |B|$ for arbitrary sets $A, B$. ;)

and the corrected formula is precisely what we know to be the inclusion-exclusion principle.

@Sayan Here's a hint : why don't you start with counting the number of ways to put n letter into n envelopes such that the i-the letter does go inside the i-th envelope for at least one i?

then you can subtract this from n! to get the desired result (complementary sets). You can steadily use inclusion-exclusion if you do this.

robjohn

10:05

@JasperLoy what? that people are confusing new and old?

Jasper Loy

@robjohn If they want to flag they can flag what ABM said. Why flag that line they flagged? LOL

David Wheeler

Jasper you are too obsessed with flags.

Jasper Loy

I wonder who the flaggers are. Can you confess?

robjohn

@JasperLoy why should they?

David Wheeler

I don't flag, unless someone really gets offensive. And it's hard to offend me, since IDGAF.

Jasper Loy

10:12

@DavidWheeler Flagged for saying F.

David Wheeler

Whatever.

Daniel Fischer

@JasperLoy You can't say "Flagged" without saying 'F', so ...

David Wheeler

I'm not sure where this is going....should I call roto-rooter?

Nick

I pledge my allegiance to the flag. (I'm not the flagger though)

Jasper Loy

I got 2 downvotes for a correct answer.

Nick

10:18

Ugh, I say the darnest things

@JasperLoy Sometimes they can't handle the truth.

Jasper Loy

-2

A: explain |a+b|≤|a|+|b| in simple terms please

$a\le b$ means $a$ is less than or equal to $b$, so we have $3\le 3$ and $3\le 4$ for example, since $3$ is equal to $3$ and $3$ is less than $4$.

Sayan Chattopadhyay

so @BalarkaSen think of something like this:

Jasper Loy

I think the downvoters think they are very smart when I actually have interpreted his question correctly. He doesn't understand the meaning of the symbol.

I think there are people who downvote the answers without reading the question carefully, lol.

Sayan Chattopadhyay

if there are 3 letters and 3 envelopes then the total number of ways of putting the letters into the envelope such that the ith letter does not go into the ith envelope. therefore the total number of ways will be 6 ways acoording to the question. this does go with what i found that is $$(n-1)n$$ ways@BalarkaSen

If i am wrong can tell me where i am wrong

Balarka Sen

You miscounted. It's not 6. It's 2.

{1, 2, 3} --> {2, 3, 1} and {1, 2, 3} --> {3, 1, 2}.

These are the only ways to put 3 letters into 3 envelopes without letting any of the i-th letter to be inside the i-th envelope. If you disagree, show me another one.

Sayan Chattopadhyay

10:28

why first letter can go to (3,2) envelope second letter to (1,3) envelope and third can go to (1,2)

Balarka Sen

what?

what do you mean by "(3, 2) envelope"? that doesn't make any sense.

Jasper Loy

@Nick Will you upvote my answer?

Sayan Chattopadhyay

yes letter 1 can go to 3 envelope or 2 envelope

Jasper Loy

In fact, it is the only answer that answers the question, lol.

Balarka Sen

you just show me another way to do this other than the two ways i mentioned above without speaking gibberish :P

Jasper Loy

10:31

I think these days people answer just reading the title without reading the body, which is what I do to understand the real problem.

Sayan Chattopadhyay

listen to me letter 1 can go to envelope 2 and 3 right @BalarkaSen

Balarka Sen

sure.

4 mins ago, by Balarka Sen

{1, 2, 3} --> {2, 3, 1} and {1, 2, 3} --> {3, 1, 2}.

i have already enlisted the two cases.

Sayan Chattopadhyay

similarly letter 2 will go to envelope 1 and 3 ,letter 3 wil go to envelope 1 and 2 therefore the total way will be six

Balarka Sen

recommend you stop this stupid argument and show me the 6 cases.

Sayan Chattopadhyay

(1,3) (2,3) (1,2)

thats the six cases

Nick

10:34

@JasperLoy Ok, I upvoted it, but the thing is your answer only offers an explanation to how it is true. It is good intuition which is as you say, correct. But if I were the OP, I'd be after the motivation and would be truly satisfied upon a decent proof common to textbooks.

Balarka Sen

@SayanChattopadhyay letter 1 can't go to two envelopes at once.

it'll go to either the 2nd envelope or to the 3rd envelope.

Jasper Loy

@Nick His confusion is about the symbol, not the proof.

Balarka Sen

you show me your six cases using my {...} --> {...} notation.

maybe i am just not getting your notation.

Sayan Chattopadhyay

fine now i get the loophole in my proof

Balarka Sen

yes :P

Jasper Loy

10:37

@Nick The downvoters clearly think they know the proof and I don't.

Nick

@JasperLoy exactly but seeing how both the initial answers were downvoted at the same time, it is likely that the OP himself was a downvoter.

Sayan Chattopadhyay

then if i take n envelopes

Balarka Sen

@SayanChattopadhyay now use the hint i gave above and try to work out this problem. it needs some set theory.

Sayan Chattopadhyay

there are n ways of putting one with the ith letter going in the ith envelope

The Substitute

Hi, off topic. If $X$ is a set, is the Borel sigma algebra generated by the singletons in $X$ simply $\lbrace \emptyset, X\rbrace$?

Balarka Sen

10:40

no, there's just 1 way to put the i-th letter going in the i-th envelope.

Jasper Loy

@Nick Anyway I am not too concerned about rep, lol.

Sayan Chattopadhyay

i mean for all the letters and the envelopews

Nick

@JasperLoy Anyhow, yeah you're right. What you said is no different than what @crash who has 2 upvotes said. So, the moral of this story is that presentation is often crucial in communicating to the layman.

Jasper Loy

@Nick The moral of the story is that voters cannot think, lol.

Balarka Sen

@SayanChattopadhyay still not correct. to put every i-th letter to the i-th envelope, there is only one such way. {1, 2, ..., n} --> {1, 2, ..., n}.

Nick

10:42

@JasperLoy Such is democracy and why I prefer to enter politics only as a dictator.

Jasper Loy

@Nick It's good not to think too much, or one will go mad.

Sayan Chattopadhyay

then it will be $$n!-1$$

Nick

@JasperLoy A certain ten starred message convinces me that people don't think as much as they ought to.

Balarka Sen

nope

Sayan Chattopadhyay

is this right @BalarkaSen

Jasper Loy

10:44

@Nick I wonder who ABM is.

Balarka Sen

@SayanChattopadhyay would you like to try it, or should i reveal it for you?

Sayan Chattopadhyay

no keep the anxiety

Balarka Sen

maybe you should revise a bit of inclusion-exclusion first if you're planning on doing it.

Jasper Loy

@BalarkaSen Actually giving him this exercise now is a bad idea, since this is more for a combinatorics course. Basic set theory will do, and this is not exactly that.

Nick

@JasperLoy I don't wonder. Seriously, a truly beautiful mind is the hero of its user.

Balarka Sen

10:46

@JasperLoy Hammack introduces basic combinatorics.

Jasper Loy

@BalarkaSen In fact I think he might be better off starting with Apostol, lol.

Balarka Sen

And I personally think this exercise is gives a solid understanding of inclusion-exclusion.

If he can't do this, he didn't really understand it.

@JasperLoy Apostol has a set theory book?

Jasper Loy

@BalarkaSen No, but one can start learning about proofs from a calculus book can't they?

iwriteonbananas

@TheSubstitute no that set isnt even a sigma algebra

Nick

You guys always talk about good books but I have to ask you this,how do you manage to get through em?

iwriteonbananas

10:48

(unless $X=\mathbb{R}$)

Balarka Sen

@JasperLoy i'd think better learn counting first before doing something as abstract as calculus.

anyway, he claims to know calculus.

iwriteonbananas

the complement on the empty set isnt in there

Jasper Loy

@BalarkaSen Well counting can be very hard.

Sayan Chattopadhyay

yes i know calculus

Balarka Sen

yes, but this is kind of basic.

Jasper Loy

10:49

@Nick The two people who downvoted me probably think that I am an idiot, lol.

Balarka Sen

not so hard, i.e. needs some thinking though.

anyway, you can give up if you want @Sayan.

Sayan Chattopadhyay

i showed you the proof for the calculus question you asked

Nick

Never give up. Never give in.

@JasperLoy DGAF bout em. They're big nobodies who could't even think to read even for a split second.

The Substitute

@iwriteonbananas how isn't that a sigma algebra? It contains the empty set, it's closed under complements and arbitrary unions.

iwriteonbananas

@TheSubstitute i was confsued because you wrote BOREL sigma algebra generated by...

it has nothing to do with the borel sigma algebra

but it is indeed a sigma algebra in $X$

The Substitute

10:53

@iwriteonbananas is it the sigma algebra generated by the collection of singletons in $X?

Nick

@TheSubstitute Let $\Bbb C$ be the set of all singleton subsets of $\Bbb R$. Then σ($\Bbb C$) is the set of all countable sets of real numbers and their complements. Any uncountable set with uncountable complement, for example $(0,1)$, will not be in σ($\Bbb C$), even though it is generated by union of elements of $\Bbb C$, because the union is uncountable.

iwriteonbananas

no it is not in general

Nick

@TheSubstitute If the set can only be generated by uncountable operations, then it does have to be explicitly included, since the axioms of a σ-algebra won't get you to uncountable unions.

iwriteonbananas

if $X=\{ x_0 \}$, then $\sigma( \{ X \} ) = \{ \emptyset, \{x_0\}, \{x_0\}^c, X\}$

The Substitute

@Nick ah that makes sense.

@iwriteonbananas I agree but that is the sigma algebra generated by a single element.

David Wheeler

10:56

@Nick Never gonna give you up, never gonna let you down...

Jasper Loy

Going to take a nap.

I hope my miracle comes soon.

iwriteonbananas

@TheSubstitute yeah, $X$ could have only one singleton

Nick

@JasperLoy Night ol sport :)

Balarka Sen

@Sayan note that you need just to compute the number of ways to put the letter in the envelopes such that at least one $i$-th letter goes inside the $i$-th envelope, not for all $i$.

Nick

@JasperLoy Want a miracle? BE THE MIRACLE

iwriteonbananas

10:57

or do we know anything else about $X$? what is it?

Sayan Chattopadhyay

@BalarkaSen is it $$n!-n$$

Nick

Never gonna run around and desert you...
Never gonna make you cry

Balarka Sen

nope :)

@Nick your piece on $\sqrt{3}$ is quite nice.

Nick

@SayanChattopadhyay Warmer. Permutations are involved.

Transcript for

$Mathematics$ Mathematics