Mathematics - 2014-12-21 (page 4 of 6)

Ted Shifrin

17:00

no email yet ... :(

Hippalectryon

@Chris'ssis What of your other upcoming publications ? When are they due ?

Pedro Tamaroff

@TedShifrin What happened?

Ted Shifrin

Nothing @Pedro

But I got a Naruto hat to make @Mike jealous :P

Did you ever figure out my last exam question, @Pedro? :D

Pedro Tamaroff

@TedShifrin Ha. Cool.

@TedShifrin I didn't think about it. =D

Ted Shifrin

I can send you homework assignments, too, to keep you out of trouble :D

Pedro Tamaroff

17:02

@TedShifrin I can have some fun, sure. Summer break is upon us.

Ted Shifrin

You celebrating Chanukah/Christmas?

Seems so odd for that to be summertime :(

Nick

@TedShifrin These are fun to do. Thanks for the link. (I don't think I'm eligible to write it though, even if I fly over there)

Ted Shifrin

There are some very hard questions, and plenty of doable ones, @Nick. That should keep you busy for a long time :)

Jasper Loy

Wow @PedroTamaroff another new pic.

Ted Shifrin

No, you're not eligible :P

Nick

17:04

@TedShifrin I can finish them all of them in 2 weeks.

Mike Miller

@TedShifrin It'll be tough for you to make me jealous.

Ted Shifrin

As I said, @Nick, there are a number of combinatorial ones that I can't do, even reading the solutions :P

LOL, @Mike, well, I figured an inadvertent hat might make you jealous :D

Pedro Tamaroff

@TedShifrin Just send me a mail.

Nick

@MikeMiller It's easy for me to make you jelly. Especially on toast.

Pedro Tamaroff

@TedShifrin Christmas.

Mike Miller

17:05

@Nick But from scratch?

Pedro Tamaroff

We already have our tree up.

Ted Shifrin

@Pedro: I'll have to get copies off my school computer. They're in the old TeX format and I don't want to do the work to retypeset everything here.

Hippalectryon

@robjohn Please delete math.stackexchange.com/questions/tagged/prob (the tag)

Pedro Tamaroff

Of course, it is not a live tree.

Ted Shifrin

I prefer live trees :P

Mike Miller

17:06

Lawson is really making me work all the linear algebra I know, @Ted

Ted Shifrin

Oh, which do you have, @Mike?

Mike Miller

@Hippalectryon If a tag is not used by any questions for a week, it is automatically deleted. Just untag it.

@TedShifrin Spin geometry

Ted Shifrin

ah

Nick

@TedShifrin I'm somewhat weak in combinatorics too but it's probably worth doing.

Hippalectryon

@MikeMiller Ah ok

Ted Shifrin

17:06

Lots of analysis and everything else in there, @Mike.

Pedro Tamaroff

Hehe, some stupid flags appeared.

@MikeMiller Facebook, pls.

Mike Miller

I know, @Ted. But first I need to make it through the linear algebra :P

Ted Shifrin

@PedroTamaroff LOL at the order.

Pedro Tamaroff

@TedShifrin At what order?

Ted Shifrin

I linked it, @Pedro :D

Mike Miller

17:08

Sometimes they make assumptions without saying so out loud which is a bit annoying... I'm pretty sure at one point they made the assumption that our quadratic form is nonsingular without telling me so

Pedro Tamaroff

@TedShifrin I don't see any link!

Ted Shifrin

Oh, for Clifford algebras, @Mike? They might make a permanent ansatz on that.

It shows up here, @Pedro.

Mike Miller

They don't say it out loud, @Ted, because they occasionally talk about picking $q=0$ (so that they can analogize to the exterior algebra case). Now that I know we're making that assumption I'm fine, though.

Ted Shifrin

Hmm, Blaine is totally careful, ordinarily, @Mike.

Mike Miller

Maybe I wasn't careful and missed it somewhere... :)

Pedro Tamaroff

17:09

@TedShifrin I'm not following.

Ted Shifrin

We're talking about his reading Lawson's book Spin Geometry.

I relinked for you above, however, @Pedro.

Pedro Tamaroff

@TedShifrin Was it fun that I linked to spin geometry?

Ted Shifrin

STAHP @Pedro.

Pedro Tamaroff

shrugs

Ted Shifrin

@Chris'ssis @Jasper: Still no email.

Chris's sis

17:12

@TedShifrin take it

Jasper Loy

@PedroTamaroff No offence, but your face seems extremely long to me.

Pedro Tamaroff

@Chris'ssis That's one line?

@JasperLoy Maybe you're short faced, Jasper.

Jasper Loy

@TedShifrin That is very weird.

Chris's sis

@PedroTamaroff Yeah.

Ted Shifrin

Got it, @Chris'ssis. I consider that neither one line nor elementary (using Fubini plus other stuff) for high school students.

Chris's sis

17:14

@TedShifrin The essence can be written in one line.

robjohn

@Hippalectryon Since there are no longer any questions using it, it should disappear in 24 hours

Hippalectryon

@robjohn ok

Ted Shifrin

You're silly, @Chris'ssis. Even your very first "can be gotten elementarily" takes work.

Chris's sis

@TedShifrin It depends on how good you are.

Ted Shifrin

But it's consistent with your usual approach to problems. I doubt it's totally new.

Pedro Tamaroff

17:16

@Chris'ssis Why are you obsessed with one liners? What is there to gain from this?

Ted Shifrin

This is far from a one-liner. To fill everything in requires sophistication and a lot of advanced analysis knowledge :P

But it's cute.

Hippalectryon

^ never expected Ted to say "cute"

Pedro Tamaroff

@Chris'ssis There's no shame in admiting something requieres work.

Mike Miller

@Hippa Everyone says cute.

Pedro Tamaroff

I find it less respectful when people degrade mathematics with "obvious", "trivial", "basic", or "direct."

Ted Shifrin

17:17

It reminds me of the double-integral solution that UserX was obsessed with, which appears various places.

Pedro Tamaroff

One can admit the material is not complicated, but it has it's own value.

Chris's sis

@TedShifrin I've never ever seen before something similar to it. I'd be very curious to see if something similar exists.

Ted Shifrin

It's of a common thematic flavor with other proofs I've seen.

Jasper Loy

@PedroTamaroff Well, but sometimes it is.

Pedro Tamaroff

@Chris'ssis Euler already did that, Chris.

Chris's sis

17:19

@PedroTamaroff Euler did it like I did?

Pedro Tamaroff

21

A: Different methods to compute $\sum\limits_{n=1}^\infty \frac{1}{n^2}$

I'll post the one I know since it is Euler's, and is quite easy and stays in $\mathbb{R}$. (I'm guessing Euler didn't have tools like residues back then). Let $$s = {\sin ^{ - 1}}x$$ Then $$\int\limits_0^{\frac{\pi }{2}} {sds} = \frac{{{\pi ^2}}}{8}$$ But then $$\int\limits_0^1 {\frac{{{{...

The idea is the same.

Integrate twice.

And it is even simpler. =)

robjohn

@Chris'ssis The two methods I was considering did not work.

Ted Shifrin

Euler also did it by considering $$\int_0^1\int_0^1 \frac{dx dy}{1-xy},$$ which I believe UserX was working on.

Chris's sis

@PedroTamaroff My way can be understood easily by any student, good or not that good.

Jasper Loy

@PedroTamaroff That page has so much code it won't load.

Ted Shifrin

17:22

@Chris'ssis, "understood easily"? No. Stop kidding yourself. You continue to think that third-year university mathematics is high-school level.

Chris's sis

@robjohn OK

Pedro Tamaroff

@Chris'ssis Well, no.

Ted Shifrin

OK, lunchtime for me.

Chris's sis

@PedroTamaroff That one is a proof that contains things put in a hurry. I can make it bright like a star.

Pedro Tamaroff

@Chris'ssis And that will take some pages.

Chris's sis

17:23

@PedroTamaroff No.

Pedro Tamaroff

@Chris'ssis It will.

Chris's sis

@PedroTamaroff No, it shouldn't take so much space.

Pedro Tamaroff

Your obsession with these kind of things eludes me, Chris.

Who cares if it is 5 or 8 pages?

Chris's sis

@PedroTamaroff The idea is to make everything very easy and very fast.

Pedro Tamaroff

@Chris'ssis Usually fast and easy don't go together.

Chris's sis

17:25

@PedroTamaroff With some brilliance one can do it.

Pedro Tamaroff

They usually are inverse proportional quantities for students.

Jasper Loy

@Chris'ssis I think most of the things you think are simple are actually hard. It's because you are a genius, like I said many times before.

Pedro Tamaroff

@Chris'ssis With some X one can always do X.

skullpatrol

We are different.

Chris's sis

@PedroTamaroff I did it for the Au-Yeung series. Any kid in high school can easily understand my proof there.

Pedro Tamaroff

17:27

@Chris'ssis No. Don't be delusional.

Most high school kids don't know what a derivative is.

skullpatrol

don't be judgemental

Balarka Sen

hem-hem

Jasper Loy

@Chris'ssis Maybe the standard of high school is very high in Romania.

Pedro Tamaroff

Are you planning to publish a book for high school students?

Else this obsession with high school level seems uncalled for.

Chris's sis

@PedroTamaroff It's a fact.

@PedroTamaroff That book I wanna publish can also be used by high school students (partly). Did you see Ovidiu's book? I wanna publish something like that.

Hippalectryon

17:28

@Chris'ssis As much as I love your work, I'd have to agree with @PedroTamaroff on that point : most High School kids barely know what a derivative is

Jasper Loy

@Chris'ssis I think the level of your book is around upper undergraduate level.

Pedro Tamaroff

@Chris'ssis I take Ovidiu is someone from Romania.

Chris's sis

@PedroTamaroff springer.com/mathematics/analysis/book/978-1-4614-6761-8

@PedroTamaroff He has a very beautiful book on integrals, series and limits.

Jasper Loy

@Chris'ssis Which is also the title of your book!

Chris's sis

@JasperLoy Well, it will contain these words ... :-)

Balarka Sen

17:32

what was the proof, @Chris'ssis?

can you link me to it? i wasn't here when you posted it.

Pedro Tamaroff

@Chris'ssis I see many of the problems you share come from this book.

Balarka Sen

ok.

Chris's sis

@PedroTamaroff I wouldn't say that. There are some, but not many.

Venus

Speaking of high school kids

Brilliant answer as usual & more important it is easy to understand for a high school student like me, +1 ≧◠‿◠≦✌ — Anastasiya-Romanova 秀 Oct 27 at 15:23

Pedro Tamaroff

@Chris'ssis It's a nice book.

I prefer Polya and Szego's book, though.

It has more material, and solutions.

Venus

17:34

Maybe this kind of kid that can understand your problem easily @Chris'ssis :D

Pedro Tamaroff

But this one is very nice.

Chris's sis

@PedroTamaroff Yeah, it's very nice.

Balarka Sen

interesting

that's actually a good proof @Chris'ssis

Chris's sis

@BalarkaSen thank you :-)

Balarka Sen

although swapping the integrals takes a bit of justification

Daniel Fischer

17:35

@robjohn I just came across it again today, and I wondered. If you find something, don't forget to notify me.

Balarka Sen

i wouldn't agree it's elementary though

Chris's sis

@BalarkaSen Tonelli's theorem

Balarka Sen

You mean Fubini?

skullpatrol

@DanielFischer did you see my post about scientific notation?

Chris's sis

@BalarkaSen No. I mean Tonelli's theorem.

Balarka Sen

17:37

no idea what that is. just doing fubini suffices.

Hippalectryon

@BalarkaSen Tonelli = Fubini for "infinite" sums/integrals

Balarka Sen

oh ok.

Daniel Fischer

@Chris'ssis If you find it interesting enough to continue thinking about it, that's nice. If not, don't feel pushed, I'm just curious, nothing breaks if no closed form is found.

Balarka Sen

i call the later Fubini too then. =P

Hippalectryon

Fubini's theorem

In mathematical analysis Fubini's theorem, introduced by Guido Fubini (1907), is a result which gives conditions under which it is possible to compute a double integral using iterated integrals. One may switch the order of integration if either order yields a finite answer when the integrand is replaced by its absolute value. As a consequence it allows the order of integration to be changed in iterated integrals. Fubini's theorem implies that the two repeated integrals of a function of two variables are equal if the function is integrable. Tonelli's theorem introduced by Leonida Tonelli (1909...

Balarka Sen

17:40

[link](link), @Hippa. ugh.

Chris's sis

@DanielFischer Is something that makes you believe there is such a closed-form? Maybe you have some reasons to think of such a possibility. The similarity between the series you posted and the very known alternating one makes me strongly believe that if there were a closed form then it would be known.

Hippalectryon

@BalarkaSen It was intended

Daniel Fischer

@TedShifrin Expanding $\sin$ and switching the order of summation, I get this, but it's not very closed.

Balarka Sen

i'd like to study open forms of integrals @DanielFischer :P

Daniel Fischer

@Chris'ssis I have no reason to believe there is one (but no reason to believe there is none, beyond the statistical argument that most series don't have a nice closed form), but I thought maybe you know something about such series.

Vrouvrou

17:43

Hello, I have $d(x,y)=|x-y|^2$ and i want to prove that $d(x,y)\leq d(x,z)+d(z,y)$

How to do please

Hippalectryon

@Vrouvrou Triangular inequality

Vrouvrou

that what i want to prove

Hippalectryon

@Vrouvrou How do you define |...| ?

Vrouvrou

to see that d is a distance on R

Hippalectryon

What space are you working on

Oh R

Khallil Benyattou

17:45

$d(x,y)$ is the distance between $x$ and $y$ on the real line, right @Vrouvrou?

Vrouvrou

yes

robjohn

@DanielFischer I tried a couple of things that did not produce any results.

Behaviour

@Hippalectryon The number in the title is the total number of reviews listed in the sidebar. This number refreshes every 100 seconds, as on the page. The button "clear" works for me (although I don't see much need for it myself, clicked only to test it now).

Studentmath

@Vrouvrou do you know Cauchy-Schwartz inequality?

Hippalectryon

@Behaviour "Hide" doesn't work. "Clear" works

robjohn

17:47

@Chris'ssis Did you post your solution to the Basel problem? I missed it, if so.

Behaviour

@Hippalectryon I don't see any button "hide" there.

Chris's sis

@DanielFischer I see. I'll let you know if I find something useful in the future.

Vrouvrou

for (.,.) yes! @Studentmath

Hippalectryon

@Behaviour At the right panel's items' right

Chris's sis

@robjohn See my deleted messages above. Can you find it? If not I posted it again.

Studentmath

17:48

@Vrou use it, it should produce what you need

robjohn

@Chris'ssis I will look...

Balarka Sen

killing a fly with a gun, @Studentmath?

Chris's sis

@robjohn take it

Studentmath

@Balarka You'll be surprised that's the standard proof in many books

Oh wait I am stupid, he is working in $\Bbb R$ not $\Bbb R^2$

Hippalectryon

@BalarkaSen No need for a gun. A rocket launcher is enough. :3

robjohn

17:50

@Chris'ssis I got if from the previous post.

Studentmath

@vrou I take it back.

Chris's sis

@robjohn OK

Vrouvrou

I don't understand how to do , i don't obtain a result

Behaviour

@Hippalectryon Screenshot? Because I don't get what we are talking about. The Review+ extension does not have a "hide" button.

Vrouvrou

is there something easier ?

Khallil Benyattou

17:51

Are we assuming that $0 \leqslant x \leqslant z \leqslant y$, @Vrouvrou?

Hippalectryon

@Behaviour gyazo.com/c7a461b55a1a7977e2d0f79aca360fd8

Vrouvrou

non, just let $x,y,z\in \mathbb{R}$

Khallil Benyattou

D'accord.

Hippalectryon

@Behaviour Also as you can see it's displaying (3) whereas only one item is here gyazo.com/ff1aff28b63edee08483ea897bf7ea37

Khallil Benyattou

Is there any way you can manipulate the RHS of the inequality?

That seems like it'd be the most straightforward way.

Vrouvrou

17:53

but if we are in $\mathbb{R}$ we have that $x\leq y\leq z$ for example @KhallilBenyattou

@KhallilBenyattou

user153330

0

Q: Retarded function?

How was the sign of the retarded green function chosen? A friend told me it had something to do with the Fourier Transform, but I didn't really understand him. The same explanation might apply to the non-retarded function, I'm not sure. Thanks for the help! Edit: Possible explanations could in...

Khallil Benyattou

Ah, okay!

user153330

^^^ I'm surprised this question survived for 9 months

Behaviour

@Hippalectryon Maybe you are also running my older extension "task-specific review indicators"? Because it had those "hide" links. I imagine that using both extensions at the same time would conflict.

Balarka Sen

LOL

Vrouvrou

17:55

@KhallilBenyattou $|x-z|^2+|z-y|^2$ what i can do for this

Behaviour

There is no code for "hide" in the current Review+.

Khallil Benyattou

What comes to mind is that the absolute value function is defined as $|a| = \sqrt{a^2}$ where $a \in \mathbb{R}$, @Vrouvrou.
Accordingly, you'd have that the RHS is equal to $(x-z)^2 + (z-y)^2$.

Hippalectryon

@Behaviour Oh, you're right.

@Behaviour Can't I run both ?

Studentmath

@Balarka how's the work on fundemantal groups progressing?

Behaviour

@Hippalectryon Do you need both? I can look into making them compatible.

Hippalectryon

17:57

@Behaviour Well I do use both, it would be great if you could make them compatible :)

Anyhow I gtg, bbl i guess

Balarka Sen

@Studentmath you mean studying alg topo?

Studentmath

@Balarka yep

Balarka Sen

fine I guess. just going through covering spaces. then am gonna look at Hatcher.

I've got a homeomorphism problem for you.

Studentmath

shoot

Balarka Sen

OK, sorry got disconnected.

@Studentmath Prove that $\Bbb R$ and $\Bbb R^2$ are not homeomorphic.

of course patently obvious, but not quite trivial to prove :P

Don Larynx

18:04

obvious but not trivial = contradiction

@Balarka

Balarka Sen

not really @DonLarynx

Don Larynx

obvious ~ clearly // not trivial ~~ not clear...

Balarka Sen

consider the Jordan Curve Theorem.

Don Larynx

clear·ly

/ˈklirlē/

adverb

adverb: clearly

in such a way as to allow easy and accurate perception or interpretation.
@Balarka

Balarka Sen

Want a hint, @Studentmath?

Studentmath

18:07

@Balarka not yet, perhaps in a minute

Balarka Sen

:P

Don Larynx

The Jordan curve theorem isn't very long @Balarka but I believe it takes some work to prove.

Balarka Sen

it takes a lot of work @DonLarynx

a LOT

Studentmath

Is there a reason to consider the homeomorphism between a sphere minus a point in $R^2$ and $R$?

Balarka Sen

you can't really fully prove it without homology, AFAIK.

@Studentmath Maybe, maybe not.

Khallil Benyattou

18:12

Have you heard of left and right inverses of mappings, @Balarka and @Studentmath?

Pedro Tamaroff

@BalarkaSen Wrong.

@KhallilBenyattou Cool people call them sections and retractions.

Balarka Sen

I dunno @Pedro. I have only seen a "nice" case proved using some covering spaces and fundamental groups.

Never read the proof, maybe I will sometime this week.

Don Larynx

@Pedro, but those who call them sections and retractions are not necessarily cool ;)

Khallil Benyattou

Ah! I guess I'm not cool, @Pedro. U_U

Pedro Tamaroff

@KhallilBenyattou Your logic is flawed there.

Balarka Sen

18:14

@PedroTamaroff C'mon. Section? Retractions? :P

Those are highly topological terms.

Pedro Tamaroff

@BalarkaSen Yes and no.

Studentmath

@Balarka ah take a point of $R$ and $R^2$, $R$ is no longer connected

Balarka Sen

Right @Studentmath

Studentmath

$R^2$ is

Pedro Tamaroff

They have been absorbed by algebra, too.

Khallil Benyattou

18:15

My logic is flawed?

Mike Miller

What did you get treasure hunter for, @Behaviour?

Pedro Tamaroff

@MikeMiller HA HA HA.

Behaviour

in Tavern on the Meta on Meta Stack Exchange Chat, 11 hours ago, by Behaviour

I roomba-fed about a hundred old questions on meta.SE... and I only did that to get Electorate... and I only did that to get the pirate hat... and still no hat. >:[

Balarka Sen

@Pedro I rather visualize sections (in short exact sequences of split sequences) as sections on Cayley graphs.

Don Larynx

@Balarka is this because, say remove point $a$ from both sets, but yet $(a, b)$ still exists in R^2?

Balarka Sen

18:16

@DonLarynx Well a homeomorphism is a continuous bijective map from R \to R^2

and a cont. image of a point is a point.

Pedro Tamaroff

@BalarkaSen ORLY.

Behaviour

@MikeMiller tl;dr Had no chance to get it on this site, so got it for voting elsewhere.

Pedro Tamaroff

@BalarkaSen The image of a point is always a point.

Mike Miller

@Behaviour Ah, I see. I'm slated to get one on academia on the 29th UTC, and on MSE on the 3rd UTC.

Pedro Tamaroff

Regardless of anything.

Khallil Benyattou

18:17

$(p \implies q) \iff (\neg q \implies \neg p)$ is logically correct, @Pedro!

Don Larynx

Thanks for the remark @Pedro

Balarka Sen

fair nuff @Pedro

Pedro Tamaroff

@KhallilBenyattou Hehe, you're right.

Khallil Benyattou

^_^

Pedro Tamaroff

I just didn't mean to imply anything.

Balarka Sen

18:19

so now when you remove a point from R, it gets disconnected but removing a point from R^2 leaves it connected @DonLarynx

Don Larynx

@Balarka: But if you remove a point $a$ from $R$, then you remove a whole line $(a, b) \forall b \in R^2$ correct?

Balarka Sen

@DonLarynx no

Pedro Tamaroff

@DonLarynx WAT.

Balarka Sen

an image of a point is a point, as @Pedro said

Pedro Tamaroff

@BalarkaSen What are you doing here? Proving $\Bbb R$ and $\Bbb R^2$ are not homeomorphic?

Don Larynx

18:20

consider $f: R \Rightarrow R^2$. , if you remove $3$ from R,, then you remove $(3,1), (3,2), (3,1.5)$ from R^2

Balarka Sen

@PedroTamaroff i know how to do it. i was asking @Studentmath, who recently learnt what a homeomorphism is, to prove it.

Pedro Tamaroff

@DonLarynx Eh, no.

Don Larynx

true or false, and if false, why? @Balarka @Pedro

Balarka Sen

@DonLarynx you know what a homeomorphism is?

Pedro Tamaroff

If you remove $3$ from $\bf R$, you remove $f(3)$ from ${\bf R}^2$.

Balarka Sen

18:21

it's a map.

ok, what @Pedro said

user153330

can someone help me with this question??

1

Q: A treatise on Probabilistic arguments and Laplace/Fourier transforms to solve limits/integrals from basic calculus.

I've seen in some answers in Brilliant.org to some very complicated limits and integrals that uses probabilistic arguments (Let $X$ be a random variable from $[0,1]$... some examples are in those answers, see also this for an example that has to do with evaluation of limit of a series) or some us...

calculus probability integration limits reference-request

Don Larynx

Right, so isn't the map $f: \mathbb{R} \to \mathbb{R^2}$ defined by $f: a \to (a, b)$?

Balarka Sen

@Pedro you covering space bro?

if so, i have a problem for you

Pedro Tamaroff

@BalarkaSen No.

Balarka Sen

ok.

Studentmath

18:23

@Don Not really, that's not a map from $\Bbb R$ to $\Bbb R^2$, I think. You need to map a point to a point, here you are mapping a point to an entire line.

Balarka Sen

R^2 consists of points, not lines. you can map a to (a, b) for some particular b, say

but not all b

otherwise it'd not be a function

Studentmath

What would it be though?

Balarka Sen

i dunno and i duncare.

@Studentmath similar ideas can be used to show R^2 and R^3 are not homeomorphic, but some nontrivial notions must be introduced to do that.

Don Larynx

@Balarka but aren't all elements in $\mathbb{R^2}$ of the form (a, b)?

Balarka Sen

sure @DonLarynx

Studentmath

18:29

Well, I barely know connectedness yet - just know the very general definition and that it is a topological propert/invariant, so I doubt I have the tools to do that just yet.

Balarka Sen

the essential idea is this : in R^2 - a point, take a point x_0. then there is a loop based at x_0 that can't be continuously "tightened" to a point.

however, in R^3 - a point, you can always do that

Studentmath

@Don consider the (not injective nor surjective) function $f:\Bbb R \to \Bbb R^2$, $f(x)=(0,0)$. Anyhow, any function from $R$ to $R^2$ will have to map points to points, as Balarka said.

Don Larynx

Remove $3$ from $\mathbb{R}$...then remove $f(3)$ from $\mathbb{R^2}$. Here $f(3)$ is defined by some point $(3, k)$ for some $k \in \mathbb{R}$. But then $(3, k-1)$ is still an element of $\mathbb{R^2}$. This does not have an inverse, and so it's not homeomorphic.

@Balarka? @Studentmath?

Balarka Sen

@DonLarynx depends on what is $f$.

Studentmath

Not always, $f^{-1}(3,k-1)$ might exist, might even be 0, 2, 1 or whatever.

user153330

18:33

Just stopping by to say this user is awesome. Great to have helpful folks like this around.

10

Studentmath

@Balarka I think I get the idea - kind of at least.

Balarka Sen

at least that one is not fake @user153330

Don Larynx

@Studentmath: Do this for all $o \in \mathbb{R} : o \neq k$. Then you find that $(3, o)$ maps to every point except $3$, as you say. What about $(2, o)$? This has to map to somewhere on the real line. But it can't (if we want to preserve homeomorphism) because the line $o \in \mathbb{R} : o \neq k$ did this for us already.

@Balarka: $f : \Bbb{R} \to \Bbb{R^2}$

Khallil Benyattou

As a very quick question, a non-injective map is one that is obviously not injective and can either be surjective or neither injective, nor surjective like the map from $A$ to $B$ below, right?

Studentmath

I am not sure what's the fallacy here (if there is), but trying to show it isn't surjective/injective probably shouldn't work since $|\Bbb R|=|\Bbb R^2|$

Khallil Benyattou

18:40

Chris's sis

I'm going to show you something REALLY AMAZING (a problem I just created)

Khallil Benyattou

Blow. My. Mind. @Chris'ssis.

user153330

can you help me with my question @Chris'ssis ????

Chris's sis

@user153330 springer.com/mathematics/analysis/book/978-1-4614-6761-8

@user153330 It's good to support the writer and buy the book (if possible). That might encourage him to publish further.

Don Larynx

@Studentmath: Great remark, but I think the aleph number of aleph-1 squared is aleph-2

user153330

18:44

@Chris'ssis i have that book but it doesn't have probabilistic methods and laplace/fourier transforms to find integrals/limits

Don Larynx

i lied

its not

i lied.

Chris's sis

@user153330 No, but we live in the same country. :-)

Studentmath

Plus we aren't sure if $|\Bbb R|$ is aleph-1 or not

Mike Miller

@DonLarynx also note that $|\Bbb R| = \aleph_1$ is the continuum hypothesis

(but nonetheless for infinite sets, $|X| = |X|^2$

Don Larynx

@Studentmath why not?

Studentmath

18:46

It's the continuum hypothesis, based merely on ZFC it's undecideable (or at least hasn't been proved/disproved yet - not sure which is true)

Chris's sis

@robjohn you don't have to miss the integral above :D

Studentmath

I am glad I actually remember set-theory. I have to thank Asaf for that as well.

Mike Miller

Undecidable, @Studentmath

Studentmath

Thanks

Mike Miller

Godel shows you can't disproved it in ZFC and then way later (someone who I don't remember) showed you couldn't prove it using forcing

and thus forcing was born

not that I understand what forcing is

Studentmath

18:50

Paul Cohen

Mike Miller

What a guy

Studentmath

Say, if $f$ is a bijection, and $g=fu$, then $f^{-1}g=u$

Right?

Mike Miller

what're g and u

functions (from the appropriate domains to the appropriate codomains)?

if so, then sure, just write down what they do to elements...

Studentmath

Yes, for some reason I felt unsure

evinda

Could someone maybe explain to me this proof:

http://math.stackexchange.com/questions/1076864/the-fuction-is-an-embedding

?

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