Mathematics - 2015-11-09 (page 2 of 5)

Karl Kronenfeld

07:01

@gxyd (mn)a is the sum of mn copies of a, while ma*na is the sum of mn copies of a^2 -- well, it doesn't seem to really undermine your proof, just a gap imo

gxyd

You are right. I have updated it.

Karl Kronenfeld

now, the problem appears to be that you do not actually derive (ma=0 for all a) or (na=0 for all a)

gxyd

I think i myself have found a flaw in my proof that is: If ma = 0, na \neq 0 for some of the elements and at the same same time may be na = 0, ma \neq 0 that does not imply that .

Karl Kronenfeld

instead you have (ma=0 or na=0) for all a

gxyd

Yes, indeed that's wrong what i did.

Karl Kronenfeld

07:09

@gxyd Also, as Sameer makes clear in their answer, you cannot conclude that P is prime; it is merely a power of a prime.

gxyd

@KarlKronenfeld In Sameer's answer after the line
> But then mxy=nxy=0mxy=nxy=0, contradiction

Can not i conclude it just here that P must be a prime. Since i reached a contradiction.

Karl Kronenfeld

It was hypothesized that There Exist m,n such that 1. P=mn, 2.gcd(m,n)=1 and 3. m>1 and n>1.
The negation of that is For All m,n either 1. P=/=mn, or 2. gcd(m,n)=/=1, or 3. m<=1 or n<=1.

user174558

@anon What is a good book on algebraic number theory?

Karl Kronenfeld

The OR connective can be expressed as an implication:
For All m,n If 1. P=mn and 3. m>1 and n>1, Then 2. gcd(m,n)=/=1.

That is all you can say. @gxyd

user174558

The nonlatex hurts my eyes.

Karl Kronenfeld

07:25

Ah, it looks so natural to me, since I do not use the chatjax thingy

user174558

Hey Karl, long time no see.

Karl Kronenfeld

Sup

user174558

The first time someone said that, I thought he wanted supper.

Karl Kronenfeld

Yeah, my hunger distracted me too much to finish typing that word.

user174558

@KarlKronenfeld What's a good book on algebraic number theory?

Karl Kronenfeld

07:27

I am not an expert on this matter, so I will refrain from answering.

user174558

One obvious choice is Lang.

Karl Kronenfeld

Very clever @skill

skill patrol

:D

user174558

@KarlKronenfeld What is your favourite algebra book?

Karl Kronenfeld

I don't play favorites, unfortunately.

user174558

07:30

I don't play computer games.

Karl Kronenfeld

Ah, so you think I'm playing games here?

This is all a serious matter to me

user174558

Computer games are boring.

skill patrol

true dat^

Karl Kronenfeld

You haven't found the right ones, I guess.

user174558

Play too much and you become stupid.

Karl Kronenfeld

07:31

Once was a computer game addict

skill patrol

:O

anon

@Jasper I don't know what level you want it to be at.

Karl Kronenfeld

@Jasper Not really offended

user174558

@anon Just name a bunch for me to look.

anon

anyway you can go shopping by looking through ToCs (or even pirated copies) quite easily

Karl Kronenfeld

07:32

@anon Thought that was a computer game reference

Balarka Sen

lol

Karl Kronenfeld

Just the print media game

anon

yeah, because of that one computer game with the levels...

Karl Kronenfeld

with the levels

user174558

A new edition of More Math Into LaTeX is coming out this year.

anon

07:33

anywho, I don't know books on AlgNT by heart or anything

user174558

I think I will get it to master LaTeX.

user174558

I am now running Linux Mint Debian Edition.

user174558

Using the Cinnamon desktop.

Karl Kronenfeld

Why master $\LaTeX$?

user174558

Why not?

user174558

07:36

I love typography.

Karl Kronenfeld

Ah, so it is just a step along the way toward higher goals in this typography hierarchy?

user147690

@Jasper What about topography?

user174558

@AlexClark Nope.

user174558

@KarlKronenfeld Yes. Also, I hope to be a mathematician eventually, so I will need it.

Karl Kronenfeld

You could pick up whatever latex you need along the way if you decided to focus on mathematics.

3

Damn, it was starred fixes grammar

user174558

07:41

I also need a latex pillow.

user174558

I am very particular about the pillow I put my head on, because my head is too heavy.

gxyd

@KarlKronenfeld Do you find any problem with Joy Gardenia's answer here math.stackexchange.com/a/732053/197214

?

For the same question.

Karl Kronenfeld

@gxyd Yeah, a great answer.

As in super-clever

user174558

Finally, I got over 100 rep. Time to retire.

user147690

@Jasper Nah you have to do it fairly by unofficially deducting the upvotes Twink gave you

user174558

07:46

@AlexClark I see you have been spying on everyone.

user147690

Wut, no he said it while I was here

Karl Kronenfeld

@Jasper I was only criticizing that particular rationale, not your choice. In fact, my opinion is that having a broader knowledge (fluffier pillows, if you like) can only help you as you strive toward whatever particular long-term goals you have.

user174558

Yes.

user174558

I noticed that Springer sells paperback and hardback at same prices on their website.

Karl Kronenfeld

That's interesting.

user174558

07:49

Well, production costs are not that different.

user174558

But some silly publishers jack up prices to ridiculously high.

user174558

Springer books are reasonably priced.

gxyd

08:02

Thanks @KarlKronenfeld for your help.

Karl Kronenfeld

@gxyd You're welcome. I probably would have lingered in chat doing nothing otherwise. :D

skill patrol

Not to mention lingering around the sites :-)

Vrouvrou

08:25

hello, please why if $\phi(t)t$ is increasing in $(0,+\infty)$ then we have that $\phi(|x|)x-\phi(|y|))(x-y)>0, \forall x,y\in\mathbb{R}^N,x\neq y $

please

@skillpatrol are you there ?

skill patrol

@Vrouvrou I am, but I don't know sorry.

Try the main site :-)

Vrouvrou

08:45

ok thank you

maybe @robjohn knows

felipa

09:09

hi... math.stackexchange.com/questions/1517103/… Can anyone who knows about linear algebra tell me why my question is getting no love?

@MartinSleziak hi... comment added

skill patrol

5

Vrouvrou

@skillpatrol (y)

skill patrol

thanks pal :-)

Martin Sleziak

@felipa Thanks for that. I think this is always good to do this if you are cross-posting.

It would be better if I were able to give you also some advice related to the actual content of your post. Let's hope you will have more luck with some other user, who knows about the area you are asking about.

felipa

09:38

@MartinSleziak thanks.. I am surprised by the lack of upvotes/comments etc.

as it doesn't seem such a strange question

Paradox 101

09:58

Can anyone give a hint on approching this problem? I don't have any idea.

Antonio Vargas

11:25

@Paradox101 Hint: Riemann integrable functions are bounded.

Paradox 101

@AntonioVargas yes but how do I use this?

I mean if this is bounded then f< M

Antonio Vargas

In fact $|f| < M$

Now try bounding your integral.

Let me know if you get stuck @Paradox

Paradox 101

@AntonioVargas so then we can say that the integral of f from 0 to 1 is bounded and is less than the integral of M from 0 to 1?

Antonio Vargas

@Paradox101 are you familiar with the triangle inequality?

Paradox 101

@AntonioVargas yes

Antonio Vargas

11:35

That's all you need, and if $f \leq g$ then $\int f \leq \int g$ follows from the definition of the Riemann integral

Paradox 101

@AntonioVargas but isn't the triangle inequality this : $|x+y| \leq |x|+|y| $?

Antonio Vargas

@Paradox101 I meant the one for integrals: math.stackexchange.com/q/447460/5531

Alec Teal

@robjohn did you get math.stackexchange.com/questions/1516347/… in the end?

Paradox 101

11:55

Antonio Vargas

@Paradox101 No, it seems to be a bit convoluted and I don't agree with the logic in the conclusion.

Here's how I would write it:

Since $f$ is Riemann integrable we know that $|f| \leq M$ for some $M \in \mathbb R$. Then $$\begin{align}\left| \int_0^1 x^n f(x)\,dx \right| &\leq \int_0^1 |x^n f(x)|\,dx \\&= \int_0^1 x^n |f(x)|\,dx \\&\leq \int_0^1 x^n M\,dx \\&= M \int_0^1 x^n\,dx \\&= \frac{M}{n+1} \\&\to 0\end{align}$$ as $n \to \infty$.

where in the third line we used that fact that $\int p \leq \int q$ if $p \leq q$.

here $p = x^n |f(x)|$ and $q = x^n M$.

and in the second line we used the fact that $x \geq 0$ to conclude that $|x^n| = x^n$.

Paradox 101

Ohhh ok. I get it now. Mine was fairly wrong and I failed to integrate it properly. Thanks a lot :)

Antonio Vargas

Sure thing, glad to help :)

jukka.aalto

12:31

How do I find the least $N$ that satisfies $(3+4i)\frac{n}{n+3}\le\frac{1}{1000}$? I have problems isolating $n$ because of the fraction \frac{n}{n+3}

Antonio Vargas

@jukka.aalto Is $i$ the complex number?

jukka.aalto

@AntonioVargas Yes

Antonio Vargas

@jukka.aalto Then the inequality doesn't make sense. The complex numbers aren't ordered, you can't make one less than another.

jukka.aalto

what if the complex number is replaced with the variable $a$?

Antonio Vargas

Try multiplying both sides by $n+3$ then isolating $n$

Remember me

13:36

@Chris'ssistheartist Here is an integral for you I have been trying :
$$\int_{0}^{\infty}\frac{\ln(x)}{x^2+2x+2}$$

jukka.aalto

Why does the signs change in fleablood's answer here? : math.stackexchange.com/questions/1513651/…

@AntonioVargas, if you have the time :) ^

I meant equality sign instead of sign!

I've got to go, but feel free to help out in the question :)

Remember me

13:54

@jukka.aalto Do you mean this part :
$|\frac {n-7}{n+7} - 1| = 1 - \frac {n-7}{n+7}$

If you mean this then see that $\frac {n-7}{n+7} - 1$ is always negative. So by definition of modulus $|\frac{n-7}{n+7}-1|=-(\frac{n-7}{n+7}-1)=1-\frac{n-7}{n+7}$

Paradox 101

Can anyone explain how to prove a sequence of functions uniformly converges to $f$?

Chris's sis the artist

14:17

@Rememberme What kind of proof do you want me to show you? A simple (usual) one or a mind-blowing one?

Remember me

anything is fine until I understand it :)

0celo7

@anon Right before the referenced post, he said that his username "user" got banned for 12 hours. He said this less than 12 hours after that happened. Is evading a ban by switching accounts allowed?

Chris's sis the artist

@Rememberme Now, I tell you what to do and then you write it down. Note that $x^2+2x+2=(x+1)^2+1$. Let the variable change $x+1=y$. What you get?

Remember me

okay

$y^2+1$

Jake1234

Guys, in the definition of a T3 topological space, why is it defined (in most books) that it is regular and T1 ? It seems like regular and T0 would be enough.

Chris's sis the artist

14:20

@Rememberme Write the new integral.

Remember me

Do this. Tell me the proof I will read it and then tell if I have any doubts

Chris's sis the artist

@Rememberme $$\int_1^{\infty } \frac{\log (y-1)}{y^2+1} \, dy$$

Right?

Remember me

Good

Chris's sis the artist

Now we let $y=1/z$ and then we get $$\int_0^1 \frac{\log(1-z)-\log(z)}{z^2+1} \ dz$$

Right?

Remember me

Shouldn't we be having a $z^2$ in the numerator ?@Chris'ssistheartist

Chris's sis the artist

14:25

@Rememberme you have $y dy = -1/z^2 dz$, abd then that $z^2$ from numerator cancels.

Remember me

oh okay. Yes . Sorry my bad

Chris's sis the artist

Now, the brilliant step comes in place, and we write all as follows

Remember me

I can feel the excitement :P

Chris's sis the artist

$$\int_0^1 \frac{\log(1-z)-\log(1+z)+\log(1+z)-\log(z)}{z^2+1} \ dz$$

Remember me

fine...

Chris's sis the artist

14:28

that is $$\int_0^1 \frac{\log((1-z)/(1+z))}{z^2+1} \ dz -\int_0^1 \frac{\log(z)}{z^2+1}\ dz+\int_0^1 \frac{\log(1+z)}{z^2+1} \ dz$$

Remember me

fine

Chris's sis the artist

@Rememberme note the first and the second integrals cancel out. To see that, let the variable change $(1-z)/(1+z)=w$

All reduces to the famous well-known integral $$\int_0^1 \frac{\log(1+z)}{z^2+1} \ dz$$

For this one, you have below some proofs

47

Q: Evaluate the integral: $\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} \mathrm dx$

Compute $$\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} \mathrm dx$$

Q.E.D.

Remember me

Cool! @Chris'ssistheartist You should write an answer then. There is this question on MSE

Chris's sis the artist

@Rememberme :D

Less time for writing answers. if you want you can write it and get some points. ;)

user174558

@Chris'ssistheartist Points are useless. I cannot impress Laura with them.

Remember me

14:33

I have just one more doubt@Chris'ssistheartist

Chris's sis the artist

@Rememberme Which one?

0celo7

@Jasper who dat

Remember me

Since I am doing the variable change shouldn't the last integral also be effected@Chris'ssistheartist

user174558

@0celo7 Just a joke. I like the actress Laura Ramsey.

Chris's sis the artist

@Rememberme you have these ones, right? $$\int_0^1 \frac{\log((1-z)/(1+z))}{z^2+1} \ dz -\int_0^1 \frac{\log(z)}{z^2+1}\ dz+\int_0^1 \frac{\log(1+z)}{z^2+1} \ dz$$

0celo7

14:36

@Jasper yes, she is quite attractive

Chris's sis the artist

@Rememberme you do that variable change for the first integral only ((1-z)/(1+z)=w). Let the other 2 untouched.

0celo7

why are people doing horrible integrals

user174558

@0celo7 Twink got you suspended by flagging your harmless message.

0celo7

@Jasper Nice!

Thanks for that news :)

For how long? I don't think he's suspended right now.

Remember me

Oh you can do that? I thought if we do a variable change it implies to all other integrals which have that variable

0celo7

14:38

@Rememberme No, they're independent!

user174558

@0celo7 I mean you were suspended, not him.

0celo7

@Jasper Oh

I'm bad at reading.

Remember me

Okay.

Chris's sis the artist

@Rememberme No. You are specific and apply the variable change to only one integral (in that case).

0celo7

Well I knew that, obviously :P

Remember me

14:38

@Chris'ssistheartist Yes I get it . Thanks a bunch!!

user174558

@0celo7 That was very bad of him, so I will ignore him again, I think.

Chris's sis the artist

@Rememberme :D

Paradox 101

@Chris'ssistheartist can I ask you a question concerning convergence of sequence of functions?

Remember me

I will also be bringing a few multiple integrals to you. I have been doing them since last week@Chris'ssistheartist

Chris's sis the artist

@Rememberme OK

0celo7

14:40

@Jasper ok.

Balarka Sen

I just watched The Sixth Sense. Brilliant.

Chris's sis the artist

Back to my work for my first book.

Remember me

Sixth sense ? @Balarka

user174558

14:57

Why is Dummit so popular?

Remember me

I have it myself @Jasper I guess it is because of the amount of questions

They have some good questions

user174558

I don't like rings to be defined without 1, lol.

Huy

@BalarkaSen: You're young for that movie.

Balarka Sen

@Huy That's BS.

user174558

@BalarkaSen You are BS.

Balarka Sen

15:00

:|

Huy

*too

pretty sure it requires you to be 16

Balarka Sen

I'm 16 years - 2 month, dude.

Huy

?????????????????????????????

I thought you were like 14

user174558

@huy My first book to study will be O'Leary: Mathematical Logic and Set Theory.

Mike Miller

@Huy He is, and always will be.

A rather unfortunate scenario.

Balarka Sen

15:03

morning @MikeMiller.

Remember me

@AGoogler You can see my and Chris sis discussion on your integral

A Googler

@Rememberme Yep , I am reading that.

Balarka Sen

@MikeMiller If $f : S^n \to S^n$ is a homeomorphism, can we always extend it to a homeomorphism $\tilde{f} : D^{n+1} \to D^{n+1}$? I think not. What if there is a crazy self-homeomorphism of the Alexander horned sphere?

Hmm. Can't we just take cone over $f$?

Mike Miller

Yes.

Balarka Sen

Right. I was being silly.

Mike Miller

15:14

Can you prove the inverse function theorem yet?

Balarka Sen

No, but I'll let you know when I do. Sorry for being slow though :(

I'm onto chapter 6 (where the pf of IVT is), but I need to review a few exercises from 5 first and ask Ted for an exam.

Mary Star

Hello!! Could someone of you take a look at my question:

0

Q: Normal unit vector

I am looking at the following exercise: I have done the following about the second part, about the signed curvature of $\iota$ : The signed curvature of $\gamma$ is different from the signed curvature of $\iota$, right? So, let $\kappa_s$ be the signed curvature of $\gamma$ and $\kappa_...

differential-equations vectors curves involutions

A Googler

What book are you studying?

Mary Star

?

A Googler

@MaryStar I was asking balarka sen.

Balarka Sen

15:17

@AGoogler Multivariable Mathematics, Theodore Shifrin.

A Googler

thanks

Mike Miller

@BalarkaSen: OK. Persuant to your previous question, can you do this for PL automorphisms? Diffeomorphisms?

Américo Tavares

24

Q: What is the purpose of this site?

What is (and what should be) the purpose of math.SE? As far as I can say, various users have different views on this question. Some users view it as a repository of knowledge. Some users approach it as teaching opportunity. Similar to previous, but slightly different: It could be understoo...

discussion

Balarka Sen

@MikeMiller Hmm. For the smooth category, taking cone would not help.

Agawa001

15:41

@AGoogler have you already posted ur yersterday problem

A Googler

@Agawa001 yep

Agawa001

@AGoogler too good, 4 answers, too bad my answer in similar to one of them

user174558

@BalarkaSen Where did you get it?

Agawa001

16:00

i think my about-to-be-posted answer looks same like rememberme's

Remember me

@Jasper I guess amazon

It is costly there

Mary Star

16:15

Could you take a look at my question: math.stackexchange.com/questions/1513665/normal-unit-vector ? Do you have an idea? @robjohn @Huy

user174558

@MaryStar You are a star.

Mary Star

:D @Jasper

anon

@0celo7 no

Remember me

@anon math.stackexchange.com/questions/1520926/… Can we do this by using the universal property of quotient spaces?

0celo7

@anon Ok, well you can read for yourself what he said. I'm just letting you know.

user174558

16:34

By the way, is Serge Lang gay?

0celo7

Why would you think that?

I have no clue either way.

TIL he was a socialist.

user174558

On some days, I think all his books are good. On others, I think they are all bad.

Mary Star

Are you familiar with differential geometry? @anon

anon

not very

user174558

Differential geometry can mean too many things.

Mary Star

16:43

Is the unit vector $n$ of a curve $\gamma$ the same as the unit vector $\tilde{n}$ of a reparametrization $\tilde{\gamma}$ of $\gamma$ ?

anon

I assume by n you mean unit normal vector, in the 90 degree counterclockwise sense in the plane. yes, you can define this unit normal purely in terms of the set of points the curve inhabits, independent of its parametrization.

Mary Star

16:59

We have the curve $\gamma(t)$ and we define $\gamma_a (t)=a\gamma (t)$, where $a$ is a non-zero constant. Let $n$ be the unit normal vector of $\gamma$ and $n_a$ the unit normal vector of $\gamma_a$. We have that $n_a(t)=sgn(a) n(t)$, right?

Then we get the arc length reparametrization of both curves, $\tilde{\gamma}(s), \tilde{\gamma}_a(s_a)$, where $s$ is the arc length of $\gamma$ and $s_a$ the arc length of $\gamma_a$.

Which is the relation of $n(t)$ and the unit normal vector of $\tilde{\gamma}(s)$ (where now the variable is $s$ and not $t$ ) ?

Paradox 101

Can anyone please explain how to go about this?

Huy

he just told you before @MaryStar

Mary Star

But now we have an other variable... So is it $\tilde{n}(s)=n(t)$ ? @Huy

anon

You mean $\gamma(at)$ presumably. Yes $n_a(t)={\rm sgn}(a)n(t)$ (didn't think about changing the direction of the curve). There are only two unit normal vectors to a curve at any given point, so there are only two possible unit normal vector functions. This should be geometrically obvious.

@Paradox101 well, f(x) looks like it's discontinuous, so maybe consider nbhds of a point of discontinuity

namely +/-1

Mary Star

But when we reparametrize the curve as for the arc length we have $\tilde{\gamma}(s)=\gamma(t(s))$, so now the variable is $s$. So is it $\tilde{n}(s)=n(t)$ ? @anon I got stuck right now...

Paradox 101

17:05

@anon but we've already defined the function on a set that doesn't contain +/-1

anon

you can still consider nbhds of +/-1

@MaryStar Do you think so? If you do, then prove it. Compute both and find that they're equal?

Antonio Vargas

@Paradox101 First calculate $f$. Then if $U$ is a closed subset of $S$ then there is a $\delta > 0$ such that $|x-1| > \delta$ and $|x+1| > \delta$ for all $x \in U$ (prove this if you need to). Use this to get a bound for $|f_n - f|$ on $U$ that is independent of $x$.

Paradox 101

I don't get it. I mean for uniform convergence of sequence of functions we have to find an epsilon right? So then this function is pointwise convergent as $f(x)$ becomes zero as $n$ approaches infinity. But then how do I choose epsilon for the actual uniform convergence? @anon

Antonio Vargas

Hint 2: $f$ is not $0$ everywhere.

anon

I was talking about showing f(x) is not uniformly convergent on all of S.

If you want to show f(x) is uniformly convergent on closed subsets of S, it suffices to show it is so on closed subsets of (-\inf,-1), (-1,1) and (1,inf) individually. (because sets on which functions are uniformly convergent are closed under finite unions I think.)

Huy

17:15

@anon: I might have asked you this before, but do you know about the fundamental domain of $SL_2(Z) \backslash SL_2(R) / SO_2(R)$?

Mary Star

We have that
$$\tilde{n}(s) \bot \tilde{\gamma}(s) \Rightarrow \tilde{n}(s) \bot \gamma (t(s)) $$
and
$$n(t) \bot \gamma (t)$$

Do we conclude then that $\tilde{n}(s)=\pm n(t)$ ? @anon

jukka.aalto

@Rememberme ok, thanks. But why does he write "less than 1/1000" instead of "less or equal to 1/1000"?

anon

@Huy {z:|z|>=1 and |Re(z)|<=1/2} is one

should be able to find them googling

Paradox 101

@AntonioVargas $f$ will be zero as $n$ approaches infinity? Can you explain your second point again as I didn't get it? The part that states: Then if $U$ is a closed subset of $S$ then there is a $\delta > 0$ such that $|x-1| > \delta$ and $|x+1| > \delta$ for all $x \in U$ (prove this if you need to).

Antonio Vargas

@Paradox101 "$f$ will be zero as $n$ approaches infinity?" No. Re the second point: perhaps you should take a while to try to understand it.

Paradox 101

17:23

@AntonioVargas what will the value of $f$ be? Isn't $f$ equal to the limit of $f_n(x)$ as n approaches infinity?

anon

have you graphed f_n(x) for a few n?

do so

once you figure out what's going on, you should have no trouble reasoning why it does so

Antonio Vargas

@Paradox101 I am more than confident you can find this for yourself :)

Paradox 101

@anon the peak of the graph becomes flatter with increasing $n$ and the graph becomes flatter

@AntonioVargas thanks for showing more faith in my capabilities than I have myself :)

I've studied the theorems but I have no idea how to actually apply them

@anon as n approaches infinity, the graph will be zero between the intervals -infinity to -1 and 1 to infinity. But between -1 and 1 the graph always approaches 1

anon

right

how one could get this without graphing: consider lim (x^2n) depending on if |x|<1 or |x|>1

Paradox 101

17:38

@anon so does that mean that if we look at it overall it doesnt converge uniformly but if we take separate closed intervals for instance between -1 and 1 or -infinity to -1 or 1 to infinity it will be uniformly convergent on those subsets alone?

anon

yes

Paradox 101

Ohh ok. I get it visually. Now I just have to deal with the harder part

@anon if |x|<1 then the lim (x^2n) is 0 and if |x|>1 then it's infinity?

anon

yes

Agawa001

ofcourse :///

Paradox 101

so then overall the limit for |x|<1 will be 1 and for |x|>1 it will be 0? @anon

anon

17:44

yes

Paradox 101

While proving this is it necessary to utilize the actual definition for uniform convergence of sequences including choosing an epsilon? Or is it sufficient to simple consider the different cases for values of x? @anon

Antonio Vargas

Damn, not a yes/no question :)

anon

(:

tbh analysis bookkeeping is something I avoid thinking about 99% of the time

0celo7

the worst part is checking that what you've constructed after 20 minutes of work is actually a homeomorphism and not just an isomorphism

Huy

wtf

nub?

Balarka Sen

17:55

you mean homomorphism, not homeomorphim

Mike Miller

@anon: but it's so fun

anon

@0celo7 do you mean homeomorphism and not just a continuous bijection?

Balarka Sen

the typo with the e in homeomorphism is much more likely.

anon

only if there's also a typo in order of words

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$Mathematics$ Mathematics