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12:42 AM
So, I was reading about IST, an extension of ZFC...
In IST, $\Bbb R$ has "nonstandard" elements, such as infinitesimals and infinitely large numbers.
An infinitesimal is one with a magnitude smaller than any standard real; an infinitely large number (an "unlimited number") is one larger than any standard real.
Two numbers $a$ and $b$ are "infinitesimally close to each other" if $a-b$ is infinitesimal.
And every limited (i.e. not unlimited) real is infinitesimally close to some standard real number. The one infinitesimally close to $a$ is denoted $\operatorname{st}a$.
Cool thing is: Definition of compact spaces are so much simpler!
A space $A$ is compact iff, for all $a\in A$, we have $\operatorname{st}a\in A$.
For example, $(0,1)$ isn't compact: Let $\epsilon$ be a positive infinitesimal. It's in $(0,1)$ because it's between $0$ and $1$. However, $\operatorname{st}\epsilon=0\not\in(0,1)$.
Similarly, $[0,\infty)$ isn't compact: Let $N$ be an unlimited number. Then $\operatorname{st}N$ doesn't exist, so $\operatorname{st}N\not\in[0,\infty)$.
(If you think this is too specific to subspaces of $\Bbb R$: Don't worry, this all extends to arbitrary spaces.)
(If $X$ is a standard space, then (for $t$ standard) $\tau$ is infinitely close to $t$ iff $\tau$ is in every standard neighborhood of $t$.)
(I forgot to mention: The above definitions all assume that the spaces are standard. However, there's an axiom of IST that basically lets us extend definitions like this to all spaces.)
(Also, I think — but have not verified — that we get a nonstandard version of Hausdorff spaces, too: I think $X$ is Hausdorff iff, for $a,b\in X$, we have "$a$ is infinitely close to $b$ iff $b$ is infinitely close to $a$." That is, "infinitely close" is a reflexive relation.)
(Amendment: Don't think $t$ needs to be standard above.)
 
 
2 hours later…
2:46 AM
I don't see what set theory has been done here, @columbus8myhw
moreover, why is this idea different from nonstandard analysis? ( en.wikipedia.org/wiki/Non-standard_analysis )
I mean, not that I know a lot about that, but it strikes me as similar on the face of things
 
@Samuel this is more of topology according to what I understand
 
also, why is there a unique standard real closest to $a$?
 
Firstly it is straightforward with the help of Heine Borel that any unbounded set cannot be compact
We can make any a infinitely closer to a given $a'$ I think
 
and what does IST stand for?
 
internal set theory
 
2:54 AM
@AlexC there?
hi @anon, what are you working on?
 
stuff
 
that's awful precise
:)
 
Hello@Soh
 
The subgroup of $\mathbb{Z}$ generated by $m$ and $n$ is $\langle m, n\rangle = d\mathbb{Z}$
What is d?
I think it's the gcd of m and n. @anon?
 
plus or minus, yeah
 
2:58 AM
as in $\pm \gcd(m,n)$?
 
You are doing DF?@Soham
 
No, Aluffi.
 
Aluffi is about abstract algebra or category theory?
 
Abstract algebra, explained with a little category theory.
 
math.stackexchange.com/questions/1299656/… THis question says that open sets need not depend on a metric then what does it depend on?
Can someone make me understand the first answer of that question
Hello@Robjohn
 
3:18 AM
@Rememberme In general a topological space does not have to be a metric space, it is just some structure that satisfies some axioms,
 
Such as the union of finite collection of subsets of a space X has to be in X , $\emptyset$ and X have to be in the set and the intersection of some finite sets the space X has to be in X
right?@Paul
 
Just google (or look in a topology book) what the definition is @Rememberme
But no
 
If I define a category with objects as group homomorphisms $\alpha : K \rightarrow G$ such that $\phi \circ \alpha$ is the trivial homomorphism mapping everything to the identity in $G'$ (for a certain fixed hom $\phi: G \rightarrow G'$), what are the morphisms in this category?
 
By tivial you mean $id_K$?
 
edited.
 
3:30 AM
you're the one defining the category, you tell me
 
It's a "reader exercise" in a footnote in my book.
I think it's commutative parallel compositions of $\alpha$s.
Like in this diagram, wait
 
Probably, I mean that is what all of them are :P
 
So the morphisms are just these diagrams, I guess.
@MikeMiller? Any clue what the most 'natural' morphisms are?
 
What morphisms are you talking about?
 
Morphisms in that category I just defined.
Hi @AlexC
 
user147690
3:33 AM
@SohamChowdhury I am now
 
user147690
Hey
 
I can see that, haha. How's it going?
 
I imagine that it would be morphisms $K \to K'$ so that the diagram commutes (where you have $K \to G, K' \to G$
 
Nope, $K$ is fixed.
 
user147690
I am alright, I feel relaxed now. Now to get back to work and do well
 
3:34 AM
All three groups are fixed here, @Paul.
@AlexClark yeah, go hit them books.
You have class now?
 
if someone said that was the category and didn't tell me what the morphisms were, I'd assume a morphism $\alpha \to \beta$, where $\alpha: K \to G, \beta: K' \to G$ were two objects of your category, is a morphism $\gamma: K \to K'$ that makes the obvious diagram commute (i.e. $\alpha = \beta \gamma$)
Paul beat me to it
 
Well then set $K'=K$ @SohamChowdhury
 
user147690
@SohamChowdhury No, I am in a silent part of uni I just found. There are computers and it is so quiet I feel bad for typing(really lightly I might add). So I'll stop typing and do 20 min of study
 
ok yes, same thing
 
6 hours ago, by Paul Plummer
Skimmed through bits of groups acting on the circle last night, I was very surprised to see that $PL_+([0,1]) $ does not contain a free group of rank 2! (At a tablet right now so probably wont respond)
 
3:36 AM
on the other hand the question, if nothing else was said, is silly; it's like me asking you what comes next in the sequence 1,2,4,8,16...
 
@MikeMiller Forgot ping you that a while ago
 
oh that's cool
i've forgotten quite a bit of what was in that; i intend to go over it more carefully in the near future, too
probably not until next week though
 
Yes, I was thinking so too. Are those diagrams viable morphisms, @Mike?
The one I posted?
 
i have no idea what the morphism is there
 
You have not described any morphisms @SohamChowdhury
 
3:37 AM
The morphism is just that commutative diagram. Isn't that possible?
It would be better if $K$ and $K'$ were different, so then the morphisms would be homs $K \rightarrow K'$ making the diagram commute.
Right?
 
Oh that is what we described to you
 
if you specify $K=K$ what Paul and I are suggesting is that, if your objects are $\alpha, \beta$, a morphism $\alpha \to \beta$ is a map $\gamma: K \to K$ such that $\beta \gamma = \alpha$.
the issue with this question is I can make the morphisms whatever I want
 
Oh, okay.
I get it now.
 
The funny thing is, that made it even more surprising to me, is that $PL_+(S^1)$ has a lot of free groups
 
I could make the morphisms be trivial except for the identity morphism if I wanted. Of course, when I asked you above what the next term in the sequence 1,2,4,8,16 was, it was 31).
 
3:40 AM
Yeah, that circle division thing.
I see the issue with the morphisms now.
They can be (almost) anything.
@Mike right this time?
 
yeah
 
hmm
it's a vague question, then
 
but there is definitely a set of "obvious" morphisms given something like that; this is what Paul and I were describing
 
yeah, thanks
 
that's what i meant when I said if you just told me that was the category and moved on, I would assume those were the morphisms
but if you asked me what the morphisms were I'd give you a hard time :p
@PaulPlummer the thing is that (I assume the PL case is just like the smooth and topological case) the group $\text{Homeo}(S^1)$ is a semidirect product of $\text{Homeo}^+([0,1])$ and $O(2)$; this semidirect product should give us room to have free groups and stuff in there when neither of the originals did
the twisted product is at the very least my intuition for why it can have wildly more interesting structure
 
3:53 AM
I could buy it, although I think I definitely need to play around with these groups more to (and read a bit more thoroughly) to get a feel for what is really making them tick, and where their differences are coming from
 
Let X be a set and let τ be a family of subsets of X. Then τ is called a topology on X if:

Both the empty set and X are elements of τ
Any union of elements of τ is an element of τ
Any intersection of finitely many elements of τ is an element of τ
@Paul So i get the definition of a topological space now can you tell me a question which is bugging me since days... How is a topological space and a metric space related?
 
Is homotopy a topological thing?
 
They are related in the sense that there is a "natural" topology to put on a metric space, I am not sure if there is any deeper reason @Rememberme
Yes @SohamChowdhury
 
Hm.
And "cohomology of a double complex"? @Paul?
All this stuff is in my book near the end. I was looking at the contents.
 
@Soham Why are you asking questions about a subject which you havent studied yet?:P
 
3:56 AM
I was looking at the contents of my book.
And it's an algebra book, so . . .
 
Balarka will literally kill me If i ask him that :p
 
:)
@Rememberme Ask him what?
 
@SohamChowdhury <--- This one
 
Haha, yes.
 
Sure sounds like it could be topological, don't really know what it is. Cohomology is used in a lot of areas, and I don't really think it is necessarily a "topology thing". What book? @SohamChowdhury
 
3:58 AM
it's homological algebra
 
Aluffi, last chapter. It's called "Homological algebra".
Which is something I know nothing about atm.
But I thought I saw Balarka and bananas discuss "cohomology" or whatever. Thanks, @Mike.
 
you probably did
 
I'll get back to work now.
 
That same chapter is there in DF Its a big part@Soham
 
4:00 AM
So I just finished 1-10 questions of rudin
Lets look into more detail about a topological space
 
$\checkmark \left(\color{green}{\frac{10}{10}}\right)$
As @AlexC says.
 
It is probably closer to $\checkmark \left(\color{red}{\frac{10}{30}}\right)$ if it is rudin
 
its same as DF
for DF do you do the problems after reading each chapter ?
or you don't do problems at all @SohamChowdhury and @Rememberme
 
Its $\checkmark \left(\color{red}{\frac{10}{20}}\right)$
Well its impossible to know the concept if you dont do the prob. @Karim
 
4:03 AM
I'm not doing DF, but I do some of the problems after finishing every section and do most of the rest when I finish the chapter, @KarimM.
 
yeah exactly @Rememberme
 
@PaulPlummer I didn't get you
 
What didn't you get
 
The first statement which you stated
"They are related in the sense that there is a "natural" topology to put on a metric space"
THat one @Paul
 
Well there is a topology, The "union of open balls topology"
Where the basic open sets are defined to be balls, which you get from you metric (the set of points less than some distance from some other point)
 
4:11 AM
So you mean that the union of open balls is a metric space?
 
No, a metric is a function that "measures distance"
that satisfies some properties
They are distinct concepts, metric spaces are not topological spaces and topological spaces are not metric spaces
 
But every metric space is a topological space isnt it?
 
No
A metric space is a metric space, that is it
 
A metric space induces a topology via the metric
 
You can put a topology on a metric space, that "plays nice" with the metric, but again they are different concepts. A topology on a set is just subsets of the space that satisfy some properties, and metric spaces are sets with a distance function that satisfies some properties
You can put a totally wacky topology on a metric space if you want, that has no relation to the metric
 
May 21 at 16:11, by Mike Miller
Sorry to be pedantic, but I think this is an important distinction. Every metric space can be given the structure of a topological space. But a metric carries nontrivially interesting amounts of information beyond that. eg there's no way to "topologize" the notion of a Cauchy sequence.
A little ways below that
 
looks good
 
May 21 at 16:13, by copper.hat
consider $\mathbb{R}$ with the metric $d_1(x,y) = |x-y|$, $d_2(x,y) = |\arctan x - \arctan y|$. same topology, but one space is complete, the other isn't
 
So Balarka meant something else over there
 
May 21 at 16:12, by Balarka Sen
He doesn't know what a topological space is, so it's ok to be less pedantic :P
 
4:18 AM
Now i know
Thanks again really you help a lot @PaulPlummer@MikeMiller
 
Maybe all my conversations on here should be me quoting what others have said, it seemed to have worked out here......
 
loool
 
Jul 15 '13 at 22:08, by mick
it that a real laugh or a polite one ?
 
real laugh :D
 
Feb 13 '12 at 20:13, by robjohn
good :-)
 
4:29 AM
:D
haha
 
user147690
LOL @Paul
 
@AlexC what are you doing atm?
 
user147690
Being annoyed that they won't credit it as MATH6006
2
 
I just finished subgroups, now I'll go back and make notes and solve the remaining exercises for the chapter upto this point :D
 
user147690
Now I will have to credit it as SCIE3275, which means I have to do a bunch of admin crap
 
4:35 AM
@AlexClark That sucks.
@AlexClark What's admin crap?
 
user147690
@SohamChowdhury Nice work
 
user147690
@SohamChowdhury Crap like, weekly literature reviews, power point presentations and maybe some essay garbage
 
I won't really feel like I "get" everything until I make my notes and draw pretty diagrams!
I just finished reading through the rest of the chapter, actually. For a sort of "bird's-eye view" of what's coming up.
 
user147690
If I had it as Math6006 I would have three assignments and an oral exam
 
@AlexClark Lit reviews?
What's that?
 
user147690
4:37 AM
@SohamChowdhury Nice, I wish I was as efficient at studying
 
22 hours ago, by Remember me
Poor @AlexC
 
user147690
@SohamChowdhury I have no idea, but it sounds terrible.
 
user147690
Feb 1 at 16:13, by Ted Shifrin
Stop that.
 
Dec 6 '13 at 1:18, by Mike
Nooooo!
 
user147690
Mar 27 '13 at 1:29, by Alexander Gruber
goddamn it
 
4:38 AM
The fact that I'm only half a chapter away from knowing what a chain complex is is a great motivator, plus the fact that I'm already fairly comfortable with groups helps.
Not as efficient as you'd imagine me to be.
 
Hello@AlexC
 
user147690
Hello @rem
 
user147690
I'm going to go do some more work, talk soon guysz
 
@AlexClark What are you gonna do now?
 
Apr 14 '13 at 21:59, by Gustavo Bandeira
Sound of the rain.
 
4:39 AM
Chain complexes .... @soham Well I have DF in front of me but I feel like doing Topology now
 
@Rememberme Do topology.
 
You done topology?@Soham
 
user147690
@SohamChowdhury I am going to do more ring theory. I'll stop when I finish this chapter
 
Good. Have fun, man. You sure deserve some after the nonsense you went through yesterday.
What after that? @AlexC
@Rememberme Oi I'm looking at Munkres a little sometimes when I'm tired of algebra and school work.
 
School Work... Thats something i have not thought about
 
4:42 AM
Aug 5 '13 at 6:15, by Peter Tamaroff
OK, done.
 
Since days
I am really going to get a nice scolding from my teacher :p
 
I haven't even opened a book since school closed. You're in 10, right? (I forget, sorry)
 
Going to 11,
Well you are in?
 
As in? You haven't given the ICSE yet?
 
Hey everyone
 
4:43 AM
Hey, @Kaj.
 
CBSE results postponed today, Sitting in front of my pc (nervous!!), Schools start from 3 june@Soham
Hey @Kaj
 
How's it going @Soham ?
 
@Rememberme Oh, so you're basically in the same position as me.
 
Yes you can say that
 
Hey there @Rememberme. I'm sorry; there's been a lot of name changing going on recently....what was your old handle again?
 
4:44 AM
@KajHansen Doing a little algebra and sitting around.
 
Ahhhhh.... Sayan :p@Kaj
 
Cool cool!
 
Pronounced "Shaaa-yoan" :P
 
Algebra is my jam. We've never really met @Soham . Where are you in your math career? Undergrad?
 
@KajHansen 11th grade.
 
4:46 AM
@Kaj we are in the same grade Me and soham :p
 
Oh very nice! When you say algebra, I'm assuming you mean abstract?
 
Soham and I*
@KajHansen yes. Aluffi and loving it right now, even though I get tripped up from time to time.
 
I'm going to be in my final year of undergrad this coming fall.
 
@Kaj Your Ramsey theory videos are nice just there is this weird shaking of the camera
 
Cool.
You into combi, @Kaj? Nice.
 
4:48 AM
@Rememberme, thanks for giving them a shot! They are more of a prototype than anything. I had no experience getting up in front of people and explaining stuff before, and my cameraman didn't have much experience with shooting and editing. Learning experience for us both.
 
Nice!!
 
Mmm, those are cool, I'll give them a shot when I'm free today.
 
Last year I gave a full ~1 hour talk on everything I'd learned. I'll get around to uploading it one of these days. Perhaps more too. One of the most fascinating results I'd come across was that Fermat's Last Theorem never holds in "large" finite fields.
 
@Soham You should watch Ted's videos The best in the market
Nice @Kaj
FLT You understand its proof?
 
@Rememberme I think it'll take him a while. :P
 
4:50 AM
@Soham, I've taken a course in combinatorics, but it was killer. Went way beyond just enumeration - lots of coding theory, design theory, applications to number theory. The problems were hard as hell, but I learned a decent bit. Did much of it stick now that it's been 2 years? Well, that might be another story :P
 
Yes As balarka said " Mathematicians take few(many) Years to understand the proof of FLT, Riemann hypothesis in finite fields, Poincare conjecture" @Soham
 
FLT over $\mathbb{N}$? Absolutely not. But this proof that you actually can find solutions to $x^n + y^n = z^n$ in "large" finite fields is actually fairly cool and straightforward with a little Ramsey theory.
 
What do you mean by "large" finite fields? @Kaj
Not familiar
 
To be honest, it's one of my favorites. Erdos' original probabilistic method is really cool too.
 
It'll take me a few months to really know what a finite field is, so I'm going to hold you to that, @Kaj.
 
4:52 AM
Its a field that is finite @SohamChowdhury
 
"really know" entails a lot more. at least, that's the sense I used it in.
 
Large means a huge finite field? @Paul
 
@Soham, do you know what a field is? $\mathbb{Q}$ for example? Well, you can have finite sets with multiplication/addition defined that satisfy the field axioms. Look at $\mathbb{Z}/p\mathbb{Z}$ for example.
 
huge finite fields are infinite fields
:P
 
Anyway, I should get back to work.
 
4:54 AM
lol
 
@KajHansen Yes, I know what a finite field is.
 
Oh ok, so let me elaborate.
 
It is also called a Galois field
 
I can't appreciate the idea though, yet.
@Kaj, I'm all ears.
 
The quotient ring $\mathbb{Z}/n \mathbb{Z}$ is a field if and only if n is a prime number@Soham
 
4:57 AM
What I mean by "large" is that, for a given $n$, the equation $x^n + y^n = z^n$ has solutions in the finite field $\mathbb{Z}_p$ for all $p$ sufficiently large. In particular, all $\mathbb{Z}_p$ with $p$ greater than the Ramsey number $R(3, 3, 3, ...)$ with $n$ threes will work.
As is often the case in Ramsey theory, that isn't anywhere close to the best lower bound, but it is a lower bound.
 
How low can you go? :D @KajHansen
 
@KajHansen Are they computable?
And is there an upper bound? I suppose it is often the case in Ramsey theory that upper bounds are . . . not small. :P
 
@SohamChowdhury, technically yes, but computing Ramsey numbers requires LOTS of computing power unless you're supremely clever. Even when you're just 2-coloring a graph, things get bad fast. See my post here for a little more detail as to why: math.stackexchange.com/questions/867470/…
No clue @PaulPlummer. But the bounds on Ramsey-theoretic problems are notorious for being completely terrible. For example, Graham's number was an original lower bound on a problem where mathematicians suspected the true lower bound was under $100$.
 
@KajHansen Yes thats right
It seems so funny that mathematicians can go that wrong
 
It is amusing in a sense. I guess one thing is that before we even work with bounds, we want to establish that something can always be done beyond a certain point. At least those proofs accomplish that.
But that's fascinating in itself. It's really, really cool that we at least know Ramsey numbers are guaranteed to be finite, even on arbitrarily high numbers of available colors.
 
5:09 AM
Unless you are doing infinite combinatorics...
 
Try finding the inverse of $f(x)=x+\sin(x)$
 
It is $f^{-1}$
 
Ahh i mean the value of $f^{-1}$
 
@KajHansen Haha, I didn't know that.
 
So what is it @Rememberme
 
5:13 AM
God knows
I just thought of it out of the blue
 
@Rem WA found nothing. Did Rudin ask for this?
 
It is a bijective function
So there should exist an inverse
 
So, doesn't mean it is easy to describe
with normal functions
 
So what is the mysterious function i need
 
what are you talking about
 
5:15 AM
@SohamChowdhury No Rudin didn't
 
If WA finds nothing, it's very probable it's not computable in terms of even special functions like all those Bessel functions and whatnot.
 
I mean what will be its inverse??
 
It is the function that $x+ sin(x) \mapsto x$
 
@Rememberme Probably can't be expressed in terms of standard mathematical functions.
 
5:16 AM
you can find a series
 
Series doesnt help @Gbeau
 
What do you need it for? @Rememberme
 
then what does help?
 
Just thinking@Paulp
 
do you want a closed form?
 
5:18 AM
No I just want a damn inverse :p
 
He just said you could with a series @Rememberme
 
You cant try
 
I am not sure what that means
 
He wants a nice closed-form.
Probably nonexistent.
 
You know that $\sin$ is typically defined as a series
 
5:19 AM
Yes it is
Maclaurin
 
So series are okay.
What's wrong with $f^{-1}$ also being defined by a series, @Rem?
 
Nothing wrong But it just doesnt make one happy@Soham:p
 
You need to learn to love special functions. For things like what Chris's sis does, there are often no conventionally "pretty" solutions.
Look at this‌​, though. Apparently $\sin x \tan x$ has a nice inverse.
 
Look at that
THe plot is so weird
 
I posted that, minus the x, a while back.
No, it's not. It just looks weird because there are two intersecting graphs.
 
5:22 AM
Yes
 
Anyway, I should probably go and do a bit of Munkres.
 
Good luck questions are hard
Some of them
 
5:43 AM
What are you learning about in topology right now @Soham ? I just finished a topology course myself. It's among my favorite math courses I've taken so far.
 

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