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12:05 AM
How'd your talk go, @Fargle?
 
ok I finished my second lecture... 2 down 4 to go
 
re everyone else
 
but it's 12AM now
 
Apparently there was a competitive aspect to it, @Ted. I got third place (out of about 8-10 speakers in the math part of the day).
 
Was it like an MAA meeting, @Fargle? I've judged student talks at those before.
Congratulations! That's great.
 
12:07 AM
It was similar, although maybe slightly less demanding. Strictly speaking it was an annual meeting of a science society, so math wasn't the focus, but yeah, something like that.
 
Well, that's still nice. Well done.
 
Thank you. I'm rather happy to have done so well (and also to be finished, for now, with numerical things...)
 
You can even mention the talk and the prize in your list of achievements when you apply to grad school.
 
Indeed!
 
Right. Now back to my number theory question and all our diff top and diff geo. watches Fargle flee like the wind
or maybe it's flea ...
 
12:10 AM
Nah, now I can dive headlong (in between my extracurricular fun-and-games nonsense). Though that number theory question's still stumping me, I'll be honest.
Mostly because I haven't done the due diligence of doing it.
 
@TedShifrin is this what typical US food looks like?
 
The key idea is quadratic reciprocity, as you mentioned, although I don't know that stuff. I just know from group theory that if $a$ is not a square mod $p$ and $b$ is not a square mod $p$, then $ab$ must be a square.
 
what is the number theory question?
 
No, that's gluttonous barbecue American food. Far from universal.
Showing that $x^4-10x^2+1$ is reducible mod every $p$, @Mathein. And don't give any more away.
Demonark already gave one slick solution ages ago.
 
okay, I won't say anything
 
12:13 AM
To be fair, I don't know it either. I hope to know it by the time this number theory course finishes.
 
Your assignment, @Mathein, is to prove that "almost every" $f(x)\in \Bbb Z[x]$ is irreducible over $\Bbb Q$ and reducible mod every $p$.
 
@TedShifrin a typical meal outside is at least 10 pounds... but cooking by myself takes a lot of time
 
I doubt it, @Fargle. But I phrased it in terms of group theory so you can prove it with the algebra you know.
 
I'm starting to not like London :P
 
@Leaky: Don't you have friends you can share cooking/eating with?
You can't afford to eat out all the time as a college student.
 
12:14 AM
I don't
 
Well, make some friends!
This is why dormitories are good.
 
I share a flat with 2 other people
their eating schedules are... conflicting
they never cook, at least
 
Ah.
 
one of them buys those ready meals every night
where you just put it to the microwave
and then you're done
yeah, no
 
Yes, not my style of cooking, but it has become more universal.
 
12:17 AM
I mean, in Hong Kong you can at least find cheap restaurants and you can choose
 
@Ted, I think I see. Odd plus odd is even, and if something isn't a square, it's an odd power of whatever generator you're considering for $\Bbb Z/p\Bbb Z^\times$.
 
I'm sure there are cheap Asian and Indian places in London, Leaky, perhaps not near you.
 
yeah no
I live in London London
 
Oh, you don't even need to know that group is cyclic, @Fargle. That's overkill. But OK.
 
so hard luck
 
12:18 AM
I'm sorta surprised, Leaky, but I haven't been in London in a long time.
 
by "London London" I mean central London
 
You've done a serious google search?
 
yes
I even walked like 20 blocks down a street
you can only get food less than 5 pounds in mcdonalds
and that's like 20 blocks away and super unhealthy
 
no, don't eat that crap.
 
sure
 
12:21 AM
googling indian restaurants in central London I found a few that aren't super expensive.
 
where?
 
central londond is big
and it's hardly worth it to go out of walking distance
 
there are tube and buses ...
 
they're expensive
and it takes me 20 minutes to go to the nearest tube station
it would take me less time to cook
and less money
 
12:23 AM
There's nothing wrong with cooking. Make food that'll last 3 or 4 meals.
That's what I usually did in school.
 
@TedShifrin That's overkill? Huh. I'll try to see if I can see that another way.
 
hmm
that's a good idea
so reheat it with microwave each time?
man I've never thought of that
stupid me
 
Sure. Most stuff will do fine if you're smart about it.
I do that with Chinese stir fries every time I make 'em.
 
my rice cooker is small though
 
well, it's easy enough to make more rice. I'm talking about the other stuff.
 
12:25 AM
I'm not exactly sure yet how to relate this to the problem entirely, though. I think I need to just work on it again for a while.
 
@TedShifrin man I feel so stupid now
 
Yup @Fargle. I gave you hints about different ways of factoring ...
@Leaky: I serve a purpose here.
 
ok I'd need to buy those thin plastic things that keep things from spoiling
those things made of ethylene and what not
 
plastic containers? yeah ... they're cheap
 
Yeah, I remember the hint and I remember some of my factorizations. As I recall, my problem was that I couldn't systematize that.
 
12:26 AM
what do you call them?
no, those flexible ones
 
storage containers
 
where you cover it on top of the food
 
oh, plastic wrap
 
aha
 
I put stuff in containers. Better seal.
 
12:27 AM
alright
ok I now feel like an entire idiot
 
Someone should star that :P
 
you should
I mean, I'm an idiot that is defined everywhere and complex differentiable everywhere
 
I did not star it.
 
but I still need to wash the dishes...
 
Yes, cooking takes some effort. So does eating.
 
12:29 AM
but I can do maths while eating
I can't do maths while washing the dishes
 
Sure you can ...
 
...
@MatheinBoulomenos kannst du?
 
Oh oh, it's Semiclassic.
 
1:32 AM
help anyone?
1
Q: Non unique decomposition of series in a "particular way"?

More AnonymousIntroduction I recently was playing around with some series and a natural question came to mind how many ways can a series be decomposed "a certain way" to a sequence? Background I believe for if the below holds for all $n$ holds then: $$ \sum_{r=1}^n\frac{a_r}{r} = \frac{1}{n}\sum_{r=1}^n b_...

 
2:29 AM
Hi. I got $n$ equations $p_i^2 \sum_{j=1}^n(d_j + p_j^2) = (d_i + p_i^2)^2$ for $i=1,\dots,n$, where $d_i$ are known nonnegative integers and I wish to solve the system for $p_i\in [0,1]$. Any hopes for a closed form solution?
 
 
1 hour later…
3:50 AM
Hello, if i have y = f(x), x = e^t, why is it that when i differentiate dy/dx with respect to x, I get d^2y/dx^2 * dx/dt
 
4:30 AM
Whiteclaws chain rule isnt it
 
4:42 AM
Also the question I mentioned above ... is it me or can some sort of group theory be used?
 
 
1 hour later…
6:03 AM
@LeakyNun If it isn't immediate you should stop learning about schemes and revisit basic topology dude
 
vzn
holy @#%& hes ALIVE! o_O
in The h Bar, Nov 13 at 19:51, by Blue
hBar has gotten boring without 0celo and Balarka
lol 4 stars! go star it yourself man! :P
 
6:19 AM
15 hours ago, by Leaky Nun
Consider $\Bbb R$ but for every rational number we add one copy, so for $q \in \Bbb Q$ we have $q^-$ and $q^+$
Balarka, what's your take on this?
 
I got ontoness, How do I complete the proof of continuity and homeomorphism?
2
Q: Prove that the Cantor set is homeomorphic to $(X,\mathscr T)$.

Unknown xFor each $n\in \mathbb N$, let $X_n=\{0,2\}$ and let $\mathscr T_n $ be the discrete topology on $X_n$. Let $X=\prod_{n=1}^\infty X_n,$ and $\mathscr T$ be the product topology on $X$. Prove that the Cantor set is homeomorphic to $(X,\mathscr T)$. Proof. Let $C$ be the Cantor set, Let $(C,\math...

 
7:06 AM
@Secret @LeakyNun Oh wait this is just the Cantor set
 
cantor???
But cantor does not have the property $q_- < q_+$ for its order topology?
 
7:23 AM
A graph admits an embedding in projective plane means the graph is projective planar right? The graph has a projective arrangement right?
 
@Secret Cantor set seen as the subspace topology of $\mathbb R$
 
hmm...
Ok so that means since it is isomorphic to the cantor set, it is nowhere dense
 
Let $(a_1,a_2,...,a_n,...)\in B=B_1\times B_2 \times B_3...\times B_N \times B_{N+1}...$. I can choose $N\in \mathbb N$: $B_n=X_n, \forall n\ge N$. Let $x\in C$: $x=\sum \frac{a_i}{3^i}$
Geometrically I could prove that \psi((x-\frac{1}{3^{N+2},x+\frac{1}{3^{N+2})\subset B $
How do I prove abstractly?
 
Cantor set without endpoints @Akiva but you're probably right
Is this order the Dedekind completion of an order of type $2\cdot\eta$ where $\eta$ is the order type of the rationals? If this is true (and it looks true) you're right
 
7:38 AM
@AlessandroCodenotti I took end points with second representation with ternaries 0 and 2 s.
 
Oh, we're talking about another problem also (maybe) involving a Cantor set, sorry
 
okay
 
 
3 hours later…
10:38 AM
@AlessandroCodenotti Ah, true
I recant
(pun)
@Secret Look at the points on the "boundary"
(boundaries of the deleted intervals)
Those give you pairs of points with nothing in between
 
11:08 AM
Hello
I am studying probability, there's a question : given pdf of random variable $X$, find pdf of $Y=sinX$. I have studied a theorem that helps solve these type of problems, it says that the pdf of $y=g(X)$ is given by $f(g^{-1}(y)) |(g^{-1}(y))'|$, given that $g'(x) >$ or $< 0$ for all $x$, but here, $g'(x) = cosx$ and for the given pdf of $X$ this function is not $>$ or $< 0$ for all $x$. what to do now?
 
11:27 AM
Does anyone know if there is any result that says "if two minimal surfaces equal to neighborhood then they are equal"?
 
12:09 PM
@Astyx hi. not sure hwy showing $span(\{f_x\})$ is dense is enough (we dont need its closure). i want to show that Parseval's equality holds for each $f\in L_2(0,1)$
 
12:30 PM
$$f(t)=\frac{\text{sgn}
(Z(t))\left|\sum\limits_{n=1}^{n=k} \frac{1}{n} \zeta(1/2+i \cdot t)\sum\limits_{d|n}\frac{\mu(d)}{d^{(1/2+i \cdot t-1)}}\right|}{\frac{\partial \vartheta (t)}{\partial t}+H_{\text{k}}}$$
 
12:55 PM
Hi folks, I have a mathematical puzzle for you. :-) Here it is:
0
Q: Just a simple puzzle on calculus

Artificial StupidityDon't down vote unless you know that you are actually wrong. $$ f(x)= \begin{cases} \frac{x^2+2kx+\mathrm{l}}{x+1} & x<-1\\ x^2+2 & x\geq -1 \end{cases} $$ Find $k$ such that $f(x)$ is continuous at $x=-1$.

 
1:44 PM
Does someone of you have an idea about my question: math.stackexchange.com/questions/3003026/… ?
 
When does first countable imply second countable?
 
2:04 PM
Remember about a week ago when I said I had an idea to attack the Collatz conjecture?
Well, I procrastinated too much on writing it down, and now I can't remember what it was
 
@AkivaWeinberger it probably won't succeed anyway :P
 
True
But we'll never know for sure, now, will we? :P
 
r9m
Is it a well known fact that $\displaystyle \frac{\sin |\xi|}{|\xi|}$ is not in range of fourier transform on $L^1(\mathbb{R}^n)$ for $n \ge 3$?
@Blue cool profile pic! :) Itachi ..
 
Anonymous
@r9m Heh, thanks ;)
 
2:31 PM
@Liad If the span is dense, you can find a sequence of elements $(f_n)$ converging to any $f$. Then you have $\left\langle f, f_n\right\rangle = \sum_k \langle f_n, v_k\rangle\langle f, v_k\rangle$ (because $f_n = \sum_k \langle f_n, v_k\rangle v_k$)
Left term converges to $||f||^2$
Right one to $\sum_k \langle f, v_k\rangle^2$ (there's probably some work to do here)
 
Fun fact: if the sun spun once every 15 seconds it would break the speed of light
'cause its circumference is 14.5 lightseconds long or something
 
Around itself ?
 
I sure am glad it doesn't spin that fast then
Imagine if light was broken !
 
Oh, also, since it's about 4 lightseconds wide, that means the bits of the sun on the edge of the half we can see are 2 lightseconds further away than the bit of the sun in the middle of the half we can see
 
2:35 PM
That's cool
 
so if we look at the sun, there's a sort of relative lag
It's distorted
 
So if the sun were to go out, we would see the center go out before the edges
 
Yeah I guess
Well, there's still the bits we can't see, that are further away
We would slowly see the sun erode over the span of four seconds
And it would be eight minutes after it actually happened
I'm trying to think if we would see the core of the sun when we're two seconds in
 
Anonymous
@Astyx We'd be too dead and cold to see that :P
 
We wouldn't die immediately
Sun goes out, we have eight minutes of ignorant bliss, four seconds of "What the hell is happening," and an indeterminate amount of panic
and then we die
 
Anonymous
2:40 PM
@AkivaWeinberger If we've no gravity to hold the planet, we'd float out in space as a projectile. And that would happen immediately
 
If we really tried I guess we could make a small portion of the population live for a while longer
 
Yeah but we're already floating in space
It would just change from moving in a circle to moving in a line
That said, I'm starting to think that this is a bad idea, and that maybe we shouldn't destroy the sun
@Astyx Geothermal-iness
 
Instead of going around and around in circles we would be going somewhere
 
@Astyx Oh so maybe it is a good idea
Let's do it
 
Anonymous
@AkivaWeinberger Yes, I do agree we wouldn't die immediately. But we'd get pretty cold within those 8 minutes. I need to look up the exact figures though
 
Anonymous
2:44 PM
Maybe you're right. It might not be that cold either. The atmosphere does retain heat
 
Anonymous
And a considerable amount of it
 
We wouldn't get cold within the 8 minutes. The eight minutes are the amount of time it would take for the information to get to us.
During those eight minutes we're still getting the final rays
We wouldn't know anything was wrong
 
Anonymous
@AkivaWeinberger A force suddenly vanishes. No acceleration?
 
There's a lightspeed delay if the force source is sufficiently far away
Gravity travels at the speed of light
We're not attracted to the sun, we're attracted to where the sun was eight minutes ago, I think
@Semiclassical Is that right?^
 
@Astyx yea, it's exactly what i got that is left to prove ^^
 
2:49 PM
As stated, that would depend on the reference frame, so it's probably wrong
 
@Liad So you're good ?
 
Anonymous
@AkivaWeinberger Okay, if we regard GR as partly true you're correct. :P
 
We're probably attracted to where the sun was going eight minutes ago (i.e. where it would be if it didn't accelerate in the last eight minutes)
 
Anonymous
But the sun vanishing instantaneously in itself violates GR
 
Anonymous
So not much use of talking about it
 
2:52 PM
Hm, true
The worst case scenario is for the sun to just go out, but the mass would still be there
 
Anonymous
What do you mean by "go out"?
 
Turn off
Go dark
 
Anonymous
ehhh
 
@Astyx working on it
 
Anonymous
22 mins ago, by Blue
@Astyx We'd be too dead and cold to see that :P
 
Anonymous
3:00 PM
In any case, my this claim was wrong ^
 
Anonymous
8+ mins of no sun isn't enough to kill us ;) (mainly because we have a great atmosphere due to earth's size)
 
Anonymous
11
A: If only the sun's light went out, how long would it take for all living things to die out?

kingledionHow fast will the Earth cool down? This is going to be the dominant factor. If it takes too long to cool down, then maybe all the plants will die without sunlight and creatures will starve to death. But if it cools down fast enough, everyone will freeze. We can apply the Stefan-Boltzman law to ...

 
3:28 PM
in Group Theory, 44 secs ago, by Shaun
Quick question:

Let $\phi: G\to H$ be a surjective homomorphism. Suppose that $H$ is virtually solvable. Is $G$ then virtually solvable?
 
3:56 PM
Well, if $G$ has a free, non-abelian subgroup, then $G$ cannot be virtually solvable, but $G$ is homomorphic to the trivial group, which is virtually solvable.
 
can someone help with a projection quesiton ?
 
4:32 PM
Any clue to as of why completeness of the deductive closure of a theory is equivalent with all models of it being elementarily equivalent?
 
What do you mean with deductive closure?
Nevermind I got it, which direction is giving you problems?
 
4:59 PM
what does it mean for an infinite series to be convergent in $L^2$?
 
the same meaning as in any other metric
 
yeah i understand the definition for a sequence of functions to converge in L^2
 
The series $\sum\limits_{i=1}^\infty f_i$ converges to $f$ if $\lim\limits_{n\to\infty}\|f-\sum\limits_{i=1}^n f_i\|_2=0$ just as in every normed space
@S.Crim A series converges if the sequence of partial sums converges then
 
so ive got a sequence of nonnegative numbers where the infinite series of the square of the sequence is finite
now i must prove that the infinite series of (this sequence multiplied by a random variable)
is convergent in L2
How do i start that proof up
 
5:15 PM
@AlessandroCodenotti: I don't really know how to start.
 
What do complete and elementarily equivalent mean to begin with?
 
It means that for all $\mathcal{V}$-structures $\mathcal{M}$ and $\mathcal{N}$ we have that $(\mathcal{M},\mathcal{N}\models T)\Rightarrow\mathcal{M}\equiv\mathcal{N}$.
 
@OskarTegby it's trivial once you understand all the definitions
3
 
Okay.
 
you're missing a quantifier on $T$ ignore me
 
5:19 PM
That's the problem formulation.
 
So how do you define "complete"?
 
in that case
2 mins ago, by Oskar Tegby
It means that for all $\mathcal{V}$-structures $\mathcal{M}$ and $\mathcal{N}$ we have that $(\mathcal{M},\mathcal{N}\models T)\Rightarrow\mathcal{M}\equiv\mathcal{N}$.
no, this is what it means for "all models of it being elementarily equivalent"
the "elementarily equivalent" part is in the $\equiv$ which you didn't define
 
stop sending us screenshots, we know what the terms mean
 
Do you mean that I didn't define what the symbol $\equiv$ means?
3 mins ago, by Alessandro Codenotti
So how do you define "complete"?
 
5:22 PM
I mean you didn't answer Alessandro's question "What do complete and elementarily equivalent mean to begin with?"
 
It seems not...
Isn't that the definition of complete?
 
yes, but if you send us screenshot of the definition then we can't be sure whether you understand the definition
 
Point.
The definition of a complete $\mathcal{V}$-theory means that a $\mathcal{V}$-sentence or its negation has to be a part of the theory, and it can't be the case that both it and its negation belong to the theory.
 
Hi, if a graph is hamiltonian connected we can directly say that the graph is hamiltonian, right?
 
Ok so let's start with complete implies all models are elementarily equivalent
 
5:24 PM
Okay.
 
i.e. a hamiltonian cycle exists?
 
So let's say $M,N\vDash T$, you want to show that every sentence holding in $M$ holds in $N$, right?
 
Correct.
 
I can't really say more without giving a complete solution. If $M\vDash\varphi$ what does tell you about the relationship between $T$ and $\varphi$?
 
$\varphi\in T$?
 
5:26 PM
@AlessandroCodenotti I'll excuse myself to finish my elliptic curve pset...
 
have fun!
@OskarTegby Ok so can it be the case that $N\nvDash\varphi$?
 
We're assuming completeness and that $M,N\models T$. Right? We want to show that $M\equiv N$. Right?
 
Okay. So if $M\models\varphi$, then $\varphi\in T$. Because both $M\models T$ and $N\models T$ we should have that all sentences in $T$ holds under the assignment of $M$ and $N$, respectively. Right?
 
5:33 PM
So, since $M\models \varphi$ where $\varphi\in T$ we have that $\varphi\in T$. As this holds for all sentences in $T$ for both $M$ and $N$ we have that they give the same assignments for them. Does this mean that they are equivalent?
That first sentence didn't make much sense, but whatever.
If $A:S\to\{0,1\}$ maps $M$ and $N$ to the same elements of $\{0,1\}$, then $M$ and $N$ are equivalent. Correct?
We didn't use completeness here.
 
How do you know $\varphi\in T$ without completeness?
 
Because we assumed that $M\models\varphi$, but this wasn't given.
 
$M\vDash\varphi$ does not mean $\varphi\in T$
 
Okay.
 
Let $T$ be theory of groups, and $M=\Bbb Z$, then $\Bbb Z\vDash\forall x\forall y(x+y=y+x)$ but that surely isn't part of the theory of groups
 
5:40 PM
Right.
I guess that we need completeness then.
 
Indeed
Ah, $\varphi\in T$ is false with your definitions, but it is in the deductive closure of $T$
I'm too used to call "theory" what you're calling"deductive closure of the theory", sorry
 
Okay.
Doesn't $M\models\varphi$ imply that $\varphi\in T$ as $M\models T$ and completeness gives us that if $\varphi\notin T$ then $\neg\varphi\in T$?
 
Yes and if we had $\neg\varphi\in T$ then we would have $M\vDash\neg\varphi$ which cannot happen
 
Rightrightright. I get it.
Okay. So, the other way now.
 
Right, so you have that all models are elementarily equivalent and you want to show that the theory must be complete
 
5:49 PM
Well, the deductive closure needs to be complete.
I don't know if that makes a difference.
 
sure sure, the deductive closure
It's not really that useful to keep the theory and the deductive closure distinct in general
 
Okay.
It's the book's idea.
 
user131753
@OskarTegby Which book?
 
So suppose that all models are elementarily equivalent but the theory is not complete, what does that mean?
 
That there exists a sentence $\varphi$ such that it nor its negation belong to $T$.
That's at least what not being complete means, right?
 
5:55 PM
right
So what do you know about $T\cup\{\varphi\}$ and $T\cup\{\neg\varphi\}$?
 
They don't equal $T$.
Something more too.
 
Something more interesting
 
hmm..
 
Problem: If $\epsilon > 0$, $c \ge 0$, $B(x,\epsilon) \subseteq \Bbb{R}^p$, and $\phi : B(x,\epsilon) \to \Bbb{R}^q$ is a continuously differentiable function with $||D\phi(x)|| \le c$ for all $x \in B(x,\epsilon)$, show that $||\phi(x_1)-\phi(x_2)|| \le c ||x_1 -x_2||$ whenever $||x_1-x_2|| < \epsilon$.
I've been thinking about this problem for quite some time and I haven't made any progress. How should I approach it?
 
I don't see it. Any hint?
Are we trying to reach a contradiction?
 
6:01 PM
If $\varphi$ is independent from $T$ then both $T\cup\{\varphi\}$ and $T\cup\{\neg\varphi\}$ are consistent
 
Yeah.
Are we trying to show that then $M$ and $N$ aren't equivalent?
Showing the contrapositive.
 
Who are $M$ and $N$?
 
Just the $\mathcal{V}$-structures in the description of elementarily equivalent.
I'm just wondering how we're trying to show the result here. Is it by contradiction?
 
Contradiction works
Assume that all models are elementarily equivalent, so you get a sentence $\varphi$ such that $T\nvdash\varphi$ and $T\nvdash\neg\varphi$. Think about the consistency thing I told you above
 
I don't see where we want to go with consistency.
 
6:06 PM
What do you know about consistent theories?
Hint: either Gödel or Henkin might give you an hand
 
Just the statement you gave me.
 
Excuse me, how can I easily simplify ugly fractions like 1/2 / (sqrt3/2)?
 
Gödel's incompleteness theorem? No..
 
I see that is equal to sqrt 6 / 6, but I don't know the reasoning behind that
 
no the other one
 
6:09 PM
The completeness theorem?
Ahh! Both can't be consistent. Right?
 
one of its forms, yes
@OskarTegby Why not?
 
@JakeS This kind of fraction is usually simplified by mutliplying the numberator and denominator by some common factor.
 
I see
 
Isn't that just a result?
 
If $T$ is the theory of groups and $\varphi$ is $\forall x\forall y(x+y=y+x)$ then surely are both consistent
 
6:12 PM
I'm using the Gram Schmidt process to normalize a set of eigenvectors, and these ugly numbers are scary. lol
 
Right.
 
So they can both be consistent in general. But not in this case, why?
 
We have the elementary equivalence.
 
@JakeS They are just tedious computations. You just need to know how to simplify those fractions, but it is not really difficult mathematics.
 
No?
 
6:14 PM
I know. In this case I've just realized they've been chosen to cancel each other later on in my computations anyway
I think I'll have another question in a moment, however...
 
Lots of professors set questions where the numbers will cancel out nicely later on, but in real problems the numbers may not be so nice.
 
All models are elementarily equivalent so for all $\mathcal{V}$-structures $M$ and $N$ we have that $M,N\models T$ implies that $M\equiv N$. Here, we have that if $T$ isn't complete then there exist an independent sentence $\varphi$ such that $T\cup\{\varphi\}$ and $T\cup\{\neg\varphi\}$ are consistent. I don't see how it contradicts the elementary equivalence.
 
Have you seen the model existence theorem form of the completeness theorem?
 
Nope.
 
Hmm that's an issue then
Because I wanted to use that consistent theories have a model here
 
6:27 PM
Do you have a reference?
I'm sure that it's been shown in the course, and that it's in the book.
 
I'm sure it's in every logic book
 
I'll look at it. I have to run. Thanks a lot! I'll talk to you later.
 
7:13 PM
Maybe I should try to read this
"A Classification of Immersions of the Two Sphere" by Stephen Smale
…Never mind.
 
7:33 PM
@user193319 Some sort of Mean Value something
 
@AkivaWeinberger Stephen Smale is an interesting guy. I wonder why he got bad grades in school. There must be some secret reason...
 
Did he? Huh.
> Smale entered the University of Michigan in 1948.[3][4] Initially, he was a good student, placing into an honors calculus sequence taught by Bob Thrall and earning himself A's. However, his sophomore and junior years were marred with mediocre grades, mostly Bs, Cs and even an F in nuclear physics.
> However, with some luck, Smale was accepted as a graduate student at the University of Michigan's mathematics department. Yet again, Smale performed poorly in his first years, earning a C average as a graduate student. It was only when the department chair, Hildebrandt, threatened to kick Smale out that he began to work hard.[5] Smale finally earned his Ph.D. in 1957, under Raoul Bott.
- Wikipedia
Interesting
Maybe there were issues at home
Or maybe it's the initial good grades - sometimes a kid grows up getting easy A's and as a result never learns good study habits or work ethic, causing them to crash when they finally get to a point where they can't keep on coasting
The rest of that section of Wikipedia describes various great mathematical achievements he made, so it seems he eventually pulled himself together
 
Quick question, how is the degree of a pointed map $f : (S^1, 1) \to (S^1, 1)$ defined in terms of $\pi_1(S^1, 1)$?
 
$\pi_1(S^1,1) = \mathbb{Z}$, right?
 
7:48 PM
Yo @Dami, and yeah
 
> Earlier in his career, Smale was involved in controversy over remarks he made regarding his work habits while proving the higher-dimensional Poincaré conjecture. He said that his best work had been done "on the beaches of Rio".[6] This led to the withholding of his grant money from the NSF.
 
Now, continuous maps between pointed spaces induce homomorphisms of their fundamental groups. So you have $f_*:\pi_1(S^1,1)\to\pi_1(S^1,1)$. How do homomorphisms $\mathbb{Z}\to \mathbb{Z}$ look like?
 
So there's only two homomorphisms from $\mathbb{Z} \to \mathbb{Z}$ one taking $1 \mapsto 1$ and the other $1 \mapsto -1$
 
A relevant thing which I haven't read yet: Chaos: Finding a Horseshoe on the Beaches of Rio
@Perturbative Homomorphisms don't need to be surjective
They're not like homeomorphisms (confusingly)
 
7:52 PM
Ohh wait yeah I'm thinking iso
 
This is also a thing I'll watch
(@WillHunting)
 
Well any homomorphism from $\phi: \mathbb{Z} \to \mathbb{Z}$ is uniquely determined by $\phi(1)$
 
Exactly, so in particular it's of the form $\phi(x) = nx$ for some $n$
 
Okay I think I may see it now
 
Yeah they're all multiplications by a thingy
 
7:59 PM
So that guy is called the degree
 
I think $\operatorname{Hom}(\Bbb Z,\Bbb Z)\cong\Bbb Z$ is how the cool kids write it
 
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