Mathematics - 2019-06-02 (page 1 of 2)

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5:21 AM

I found the source of the quote again! It's from Mumford's foreword to Parikh's The unreal life of Oscar Zariski: "Everyone knows that physicists are concerned with the laws of the universe and have the audacity sometimes to think they have discovered the choices God made when He created the universe in thus and such a pattern. Mathematicians are even more audacious. What they feel they discover are the laws that God Himself could not avoid having to follow." — Qiaochu Yuan Dec 12 '10 at 2:06

5:57 AM

Omnipotence paradox

The omnipotence paradox is a family of paradoxes that arise with some understandings of the term 'omnipotent'. The paradox arises, for example, if one assumes that an omnipotent being has no limits and is capable of realizing any outcome, even logically contradictory ideas such as creating square circles. A no-limits understanding of omnipotence such as this has been rejected by theologians from Thomas Aquinas to contemporary philosophers of religion, such as Alvin Plantinga. Atheological arguments based on the omnipotence paradox are sometimes described as evidence for atheism, though Christian...

Akiva Weinberger

@Ultradark Fiction or nonfiction

I enjoyed The Martian very much

For nonfiction maybe check out Atul Gawande's books on medicine

There's also How to Invent Everything, a science book with the premise of, if your time machine broke and you're stranded in the past, here's how to reinvent civilization

7:02 AM

[Random]

to be typed 0(x+x)=a and bracket algebra

In a vector space, if $cv=0$ then either $c=0$ or $v=0$ ($v$ is a vector). Proof says "if $c=0$, done, if $c$ not zero, then.... $v=0$". Shouldn't it also have to prove that if $v$ nonzero, then $c=0$?

$cv = 0$ implies $cv_i = 0$ for all entries of the vector

7:56 AM

Guys, can you help me with this question? math.stackexchange.com/questions/3246087/…

8:37 AM

@ÍgjøgnumMeg TY. But shouldn't the textbook have to prove that, too?

9:00 AM

is the integral $\int_0^{\pi/2} \frac{\tan 2x}{1+x^2}dx$ convergent?

guys, how can I create a new chat room or a private chat room?

You click on a person you want to invite and you press "start a new room with this user"

@Jeff No, because that's redundant

but there are no private chat rooms

^

2 hours later…

Subhasis Biswas

10:53 AM

@BalarkaSen, can you please check my work, unless you are busy

...

I'm a little busy right now. If it's short go ahead.

And/or if you have specific questions

Subhasis Biswas

@Ultradark just before this one

What are you trying to prove?

Subhasis Biswas

i wanted to show that with respect to usual metric, a metric space $X$ is disconnected if some $y$ such that $x<y<z$ not in $X$, where $x, z \in X$

What does $<$ even mean in an arbitrary metric space

Subhasis Biswas

10:56 AM

metric of $\mathbb{R}$

that's why I went around it

I cannot apply ordering unless specified

i believe the message is self contained.

If $A \subset \mathbf{R}$ such that there exists $x, y \in A$ with $x < z < y$ for some $z \not \in A$ then consider $U = A \cap (-\infty, z)$ and $V = A \cap (z, \infty)$.

$U$ and $V$ are disjoint open subsets of $A$ with $U \cup V = A$, so $A$ is disconnected.

Subhasis Biswas

@BalarkaSen that's very clear.

and easy

That answers the question, then?

Subhasis Biswas

Supposing that $|p_0-q_1|=r_1$ , $|p_0-q_2|=r_2$ , $|p_0-q_3|=r_3$ such that $r_1<r_2<r_3$ (fixed) and any $p_0, q_1, q_2, q_3$ (consider the values like a set) satisfying the equalities are such that $p_0, q_1, q_3 \in X$ but $q_2 \notin X$. Prove that $X$ is not connected

Now, consider the set $\{x \in X: |x-p_0|< r_2\}=A$ and $\{x \in X: |x-p_0|>r_2 \}=B$. Evidently, $A \cup B =X$. Now, $A \cap B = \emptyset$. And $A, B$ are non-empty. Both $A$ and $B$ are open. So, $X$ must be disconnected.

This

@BalarkaSen that is very much the case when $X = \mathbb{R}$. I want to prove almost this type of scenario when $(X,d)$ is an infinite metric space with the usual metric of $\mathbb{R}$

Now, using this result, I want to prove that if $\forall x, y \in X$, $d(x,y) \in \mathbb{Q}$, then $X$ must be disconnected

Sure, that's not hard either. Fix $a \in X$ and look at $f : X \to \mathbf{R}_{\geq 0}$ defined by $f(x) = d(x, a)$. This is a continuous function, so $f(X) \subseteq \mathbf{R}_{\geq 0}$ is a connected subset if $X$ is connected - those are intervals, and has the intermediate value property.

It seems you are doing something like this in your message but in a very convoluted way

Subhasis Biswas

11:03 AM

Yes, by fixing $p_0$

Just use continuity of the metric

Subhasis Biswas

@BalarkaSen I will. But is my proof valid? Yours is much better :D

It sounds right but I didn't look carefully. You certainly mean $d(x, y)$ instead of $|x - y|$ which you are using throughout, since you're working in a general metric space now.

Subhasis Biswas

@BalarkaSen hmmm...but what does this "usual metric" even mean here

in Logic, 3 mins ago, by Secret

A
A => B
----------
B

My brain is really not designed to find the truth value of A => B

which is why classical logic, compared to temporal logic, is such a headache to me

11:07 AM

@SubhasisBiswas That's what, you kept saying usual metric, so I thought you were proving it for submetric spaces of $\Bbb R$, in which case it's trivial and I gave you a proof.

otherwise "metric space with usual metric of R" is nonsense

Subhasis Biswas

@AlessandroCodenotti now look at this message for the argument I used to show that if $d(x,y)$ is rational, then the space is diconnected. Last one for noe

now

I don't see where the argument is. It's the same proof; $d : X \times X \to \mathbf{R}_{\geq 0}$ is continuous, so has connected image. Cannot be ontained in $\Bbb Q$, that's totally disconnected.

Subhasis Biswas

okay

so if $X$ is connected, then $f(X)$ should also be connected

If $f : X \to Y$ is continuous, $X$ is connected, $f(X) \subset Y$ is connected, yes.

That's "intermediate value theorem" in topological spaces.

Subhasis Biswas

I didn't think of this fact in case of metric functions. Learned yesterday that metric functions are continuous

and should've used that connectedness is a productive property.

11:14 AM

Right, you shouldn't immediately blurt out a sketch of an argument even if it's a correct one. These are one-liners, try to streamline the argument to something fundamental instead of pages of inequalities, manipulations, etc.

Subhasis Biswas

@BalarkaSen I observe this so often. The arguments are so elegant yet so short

another question

If a product topology is connected, should each individual constituent spaces be connected?

sort of like the converse to the theorem that you asked me to prove

Alessandro Codenotti

That's something you can figure out by yourself

^

Alessandro Codenotti

Remember that the image of a connected space through a continuous function is connected

I have to decide what to read today

Subhasis Biswas

11:35 AM

$f:X_1 \times X_2 \times X_3 ... \times X_k \to X_j$ be such that $f((x_1,x_2,...,x_k))=f(x_j)$

@AlessandroCodenotti , is it gonna work?

You meant $x_j$ on the right

Not $f(x_j)$

But yes

Subhasis Biswas

damn it man. I sure meant that

for example $f(x,y)=x$ which is continuous.

taking $\epsilon = \delta$

$\epsilon-\delta$ arguments only work in metric spaces.

The coordinate projection maps are continuous for arbitrary topological spaces, with the domain given product topology.

Subhasis Biswas

@BalarkaSen okay, (i am going to utter a word which I almost know nothing about) $\epsilon - \delta$ for metrizable spaces?

where the notion of "distance" is valid?

If you don't know what the word means you should not say it

Subhasis Biswas

11:42 AM

@BalarkaSen As far as my knowledge goes, it is a topological space where some $d$ can be defined with the properties of metric spaces

correct me if I am wrong

You missed the crucial point that the metric topology coincides with the given topology.

Like I said, if you only barely know some terminology, don't say it at all. That's journalism, not mathematics.

Subhasis Biswas

@BalarkaSen I am here to make mistakes.

and I am not ashamed to make it.

This isn't a mistake; it's just using a word you are using without knowing what it means!

aka journalism, as popularly known

Subhasis Biswas

@BalarkaSen don't get me started. plis. Don't awaken my inner Arnab Goswami

@BalarkaSen I try to keep this in mind all the time. If I don't know a terminology, I say it to everyone around that I am going to use a word that I know nothing about.

No showing off my collection of big words

I really appreciate the time you are taking to teach me bits of details.

@BalarkaSen how the notion of continuity hold up without $\epsilon$ $\delta$ definition?

Alessandro Codenotti

12:34 PM

@BalarkaSen arxiv.org/abs/1904.07004

12:46 PM

Lol

I like $\infty$-categories now. They're just simplicial sets where every inner horn has a filler

Alessandro Codenotti

Oh, why didn't you say that before? I understand $\infty$ categories now

LOL

It's really cool you should check out Lurie's "What is an $\infty$-category" on the AMS Notices thing

1:03 PM

@BalarkaSen I've heard a result that goes something like the category of $(\infty,1)-$groupoids is equivalent to the category of topological spaces

have you heard of anything like that?

Just $\infty$-groupoids I believe.

Alessandro Codenotti

And spaces up to weak homotopy equivalence I would say?

$\infty$-groupoids are simplicial sets where every horn as a filler, they are also known as Kan complexes

The model category of Kan complexes is equivalent to the model category of CW complexes, for example

Alessandro Codenotti

Actually that doesn't work, I was thinking of sending every space to its fundamental $\infty$-groupoid

That's correct

Alessandro Codenotti

1:08 PM

But if I have two spaces which are weak homotopy equivalent but not in bijection I'll get two distinct groupoids

Like $S^1$ and the pseudocircle

They will be weak homotopy equivalent as Kan complexes!

Alessandro Codenotti

Ah ok, so we need the right notion of equivalence in the category of groupoids too

Makes sense

Yup

You need to pass to the homotopy category if you want an honest to god equivalence of categories. The classical statement is that the homotopy category of topological spaces is equivalent to the homotopy category of simplicial sets

Alessandro Codenotti

Oh ok, that makes sense

(I stopped following the topology II course when they started talking about simplicial sets and nerves :P)

Actually $\infty$-groupoids are a little stronger than Kan complexes probably. Every horn there has a unique filler

Something of that sort

Nah, I think that's right.

@AlessandroCodenotti The correct way to properly define the fundamental $\infty$-groupoid of $X$ is as the collection of the set $X[n] := \text{Map}(\Delta^n, X)$ of singular simplices for $n \geq 0$ with face maps $d_i : X[n+1] \to X[n]$ give by restricting a singular simplex $\Delta^{n+1} \to X$ to the $i$-th face, and degeneracy maps $s_i : X[n] \to X[n+1]$ given by composing a singular simplex $\Delta^n \to X$ with the $i$-th degeneracy map $\Delta^{n+1} \to \Delta^n$.

This is the "singular Kan complex" of $X$. Closely related to the singular chain complex

The reason this is the correct definition is because: $X[0]$ (points on X) should be thought as the objects of $\Pi_{\leq \infty} X$, $X[1]$ (paths between points on X) are then morphisms. But if you have a morphism $x \to y$ and $y \to z$, there is no canonical choice of a morphism $x \to z$ (indeed, the composition of paths doesn't work because you don't have associativity; that's why in the fundamental $1$-groupoid the set of morphisms is paths upto homotopy)

Alessandro Codenotti

1:23 PM

Ah I see

The choice of such a morphism being kept track by the singular $2$-simplex that bounds $f : x \to y$, $g : y \to z$ and $h : x \to z$ (i.e., that 2-simplex is a choice of a homotopy between the composition $g \circ f$ and $h$) So elements of $X[1]$ are morphisms upto $X[2]$

And higher "coherences" of the n-morphisms (elements of $X[n]$) upto (n+1)-morphisms (elements of $X[n+1]$) thereof

Alessandro Codenotti

Makes sense

> In general, the inclusion–exclusion principle [for Euler characteristic] is false. A counterexample is given by taking X to be the real line, M a subset consisting of one point and N the complement of M.

:o

Alessandro Codenotti

I'm reading a bit of the HoTT book today, soon I'll also believe in categories :P

@AlessandroCodenotti of cardinality at least $\aleph_{17}$?

Alessandro Codenotti

1:31 PM

I don't believe in small categories directly

@LeakyNun You have to take subsets which have good neighborhoods and the intersection of their neighborhoods

That's where Mayer-Vietoris works

That is, $X = A \cup B$ where $A, B$ are neighborhood deformation retracts with such neighborhoods say $U$ and $V$ such that $U \cup V = X$ and then you can say $\chi(X) = \chi(A) + \chi(B) - \chi(U \cap V)$

@Alessandro Cool stuff

I have an observation about Mayer-Vietoris actually. If $X$ is a topological space you can associate a $\text{Ch}_\bullet$-valued presheaf to it which assigns to each open subset $U \subset X$ the singular cochain complex $C^\bullet(U)$.

This is nearly a sheaf. For any pair of open subsets $U, V$ you have a pullback diagram $C^\bullet(U) \to C^\bullet(U \cap V) \leftarrow C^\bullet(V)$. The pullback is exactly $C^\bullet(U + V)$, the chain complex made out of sum of chains of $U$ and chains of $V$. This is a subcomplex of $C^\bullet(U \cup V)$, but Mayer-Vietoris is precisely the statement that the inclusion $C^\bullet(U + V) \hookrightarrow C^\bullet(U \cup V)$ is a chain homotopy equivalence.

So it's like a "sheaf upto homotopy", because gluing+identity is the same statement as saying that the limit of the pullback diagram $\mathscr{F}(U) \to \mathscr{F}(U \cap V) \leftarrow \mathscr{F}(V)$ is $\mathscr{F}(U \cup V)$.

I should say $C^\bullet(U \cup V) \to C^\bullet(U + V)$, dual of the inclusion $C_\bullet(U + V) \hookrightarrow C_\bullet(U \cup V)$ at the level of chains.

This is one of the things in the singular setup that's just clear in de Rham setup, because Mayer-Vietoris there is just the fact that the sheaf $\Omega^\bullet$ of differential forms on a smooth manifold is fine, ie., has partition of unity.

And well, the fact that it's a sheaf comes from partition of unity as well.

Akiva Weinberger

1:57 PM

Hey, given two polynomials, what's the name of the polynomial whose roots are the products of roots of the first two?

Like, $P=\prod(x-p_i)$, $~Q=\prod(x-q_i)$, $~R=\prod(x-p_iq_j)$

Alessandro Codenotti

Are you sure it as a particular name?

The coefficients of R can be written in terms of the coefficients of P and Q, right? Because keeping q_i constant, they are symmetric in p_i, so write it as a polynomial on the elementary symmetric polynomials of p_i. That guy is also symmetric in q_i, so do that once more.

Akiva Weinberger

@AlessandroCodenotti No

But it feels like it probably does

Unrelated: just noticed that ${F_{n+1}}^2+{F_{n-1}}^2=3{F_n}^2\pm2$

Ex: $9+64=3(25)-2$

Ex: $25+169=3(64)+2$

the new US visa requirement for information about social media accounts is quite egregious

livefromalounge.boardingarea.com/2019/06/01/… (with screenshot)

bbc.co.uk/news/world-us-canada-48486672

Akiva Weinberger

What

That's pretty horrible

2:06 PM

@BalarkaSen I guess more precisely, if $S_n \times S_m$ is acting on $A = \Bbb Z[x_1, \cdots, x_n, y_1, \cdots, y_m]$ the invariant subring $A^{S_n \times S_m} = \left ( \Bbb Z[x_1, \cdots, x_n][y_1, \cdots, y_m]^{S_m} \right)^{S_n} = \Bbb Z[x_1, \cdots, x_n][e_1, \cdots, e_m]^{S_n} = \Bbb Z[e_1, \cdots, e_m][x_1, \cdots, x_n]^{S_n} = \\ \Bbb Z[e_1, \cdots, e_m][f_1, \cdots, f_n] = \Bbb Z[e_1, \cdots, e_m, f_1, \cdots, f_n]$

I applied for a visa before this took place but I still haven't gone to the US embassy

I don't know if they will ask me for my facebook account there

Akiva Weinberger

This is government doxxing

Although it seems kinda unenforceable

$e$'s are elementary symmetric on $y$'s and $f$'s are elementary symmetric on $x$'s.

Akiva Weinberger

This was prompted by a question asking how to show that, if $a_n$ and $b_n$ satisfy a linear relation, so does $a_nb_n$

which is clearly true, since you can think of $\{a_ib_j\}$ as living on a finite-dimensional vector space

and you're asking for $\{a_ib_i:0\le i\le N\}$ to be linearly dependent for some $N$

but finding the actual linear relation seems to be the same as finding this polynomial

{1, 2, i} {sqrt(2), sqrt(3), 2sqrt(3)}

Akiva Weinberger

2:11 PM

I mean infinite sequences $a_n$ and $b_n$

Linear *recurrence relation

oh...

Akiva Weinberger

Like the Fibonacci numbers do: $F_{n+2}-F_{n+1}-F_n=0$

pass to the generating function

$a_n$ satisfies a linear recurrence relation iff $\sum a_n x^n$ is a rational function

then... idk lol

Akiva Weinberger

$(F_1x+F_0)/(x^2-x-1)$

Hm: I wonder if you can find $\sum x^n/(n!)^2$

put everything in jordan normal form, done

2:13 PM

I wonder what invariant theory on formal power series rings look like

Seems like the worst question to ask

$a_n = \sum p_i q_i^n$, $b_n = \sum r_i s_i^n$, etc

(this is assuming the matrix is diagonalisable)

otherwise you should have $a_n = \sum p_i n^{q_i} r_i^n$ right

Akiva Weinberger

This is where the ${F_{n+1}}^2+{F_{n-1}}^2=3{F_n}^2+(-1)^n2$ observation came from

my expression is still clearly closed under multiplication and whatnot

probably can be converted to a proof

Akiva Weinberger

(To get rid of the $\pm2$, shift the recurrence over by one and add it to the original)

@BalarkaSen formal power series ring is just a completion of the polynomial ring

Akiva Weinberger

2:17 PM

What's $\int e^{a/x+x}dx$

Oh, W|A can't do it

It's hopeless then

it involves the exponential integral

Akiva Weinberger

I was trying to see if I could figure out what $\displaystyle\sum_{n=0}^\infty\frac{x^n}{(n!)^2}$ was

@BalarkaSen here I have the poset category of the powerset of a finite set

\O/

I feel like it is very related to, eh, higher homotopy

2:25 PM

How so

because, say, functors from $I_1$ to $C$ correspond to morphisms in $C$

$I_n$ is the poset category of the powerset of a set with cardinality $n$

functors from $I_2$ to $C$ correspond to... morphisms between morphisms?

it's like a homotopy between paths

natural transformations between two functors from $I_2$ to $C$ gives a functor from $I_3$ to $C$

Remind me what the poset category is again?

the objects are the elements of the poset

the morphism $x \to y$ is whether $x \le y$

Gotcha. The power series is partially ordered by inclusion, got it.

so...

a homotopy between a homotopy is a functor from $I_3$ to $C$

and this can be iterated

what is this situation similar to?

2:29 PM

Isn't it better to study functors from $[n] \to C$

that's a simplicial set right

These model "simplicial $n$-homotopies"

Your situation is roughly similar

@LeakyNun Yeah.

It's the nerve of $C$

Basically, functors $[n] \to C$ are chains of objects $C_1 \to C_2 \to \cdots \to C_n$ in $C$

These are $n$-simplices in your simplicial set, and the face maps are given by "composing the $i$-th and $i+1$-th morphisms" and the degeneracy maps are "introducing the identity map $C_i \to C_i$ at the $i$-th place"

This is a Kan complex where not only every horn is fillable, but every horn has a unique filler

So I guess I don't think it's homotopy theoretically very interesting.

Oh no, it's not a Kan complex. Every inner horn has a unique filler.

If $C$ is a groupoid then it's a Kan complex

For example if the target category $G$ is a group, the geometric realization of this gives the classifying space $BG$

user131753

2:44 PM

Does anyone know of any particular generalization of the notion of metric space such that every topological space is "generalized" metrizable?

@LeakyNun In your case $I_n$ has $2^n$ objects. So I think the corresponding object $X[n] = \text{Fun}(I_n, C)$ is a "cubical set"

probably

Ronnie Brown has some work on this, try googling around to see if this is similar.

@user170039 a generalized metric space $X$ is a collection of subsets $\tau \subseteq P(X)$ that is closed under finite intersection and arbitrary union

user131753

@LeakyNun What is the notion of metric here?

2:47 PM

no idea

@user170039 The collection of subsets :P

user131753

Very informative!

3:14 PM

am i still allowed to post in the forum with workings out, asking what i did wrong in a homeworks tyle question

3:55 PM

Hello!!

I am trying to plot something in matlab but I get the message "error: invalid use of script in index expression". What does this mean?

The code I wrote is the following:

function [ ] = trapezoid(N)
h=1/N;
A=[-5 -2;-2 -100];
y=[1;1];
t2=[0];
y2=[y];
for (i=1:N)
      y=(eye(2)+h*A+h^2/2*A^2)*y;
      t2=[t2, i*h];
      y2=[y2, y];
end
plot(t2, y2(1,:).^2+y2(2,:).^2, 'y');

anyone knows if there's an appropriate room for general R stuff?

there's one in stack overflow

I don't see one

R

Under new administration.

it's hard to search for a room whose title is one letter

had to page through the first few pages of results to find it for ya

oh, thanks :)

it's pretty dead

4:05 PM

Can anyone explain the criterion for differentiabilty of multivariable function? What does good approximation implies? I can take literally any plane, I guess and that condition would be satisfied.

the definition is that there is a linear operator $D$ such that $f(x+h) = f(x) + Df_x(h) + \mathcal O(\Vert h \Vert^2)$, it's a theorem that there is only one such "best" approximation

let me try to find the theorem precisely stated rather than going from memory

if $A$ and $B$ both satisfy it, then subtracting both sides gives $0 = 0 + (A-B)(h) + \mathcal O(\|h\|^2)$

so $(B-A)(h) = \mathcal O(\|h\|^2)$

which clearly implies $B-A=0$

but actually I think it should be $\mathcal o(\|h\|)$ rather than $\mathcal O(\|h\|^2)$

yes I was thinking the same thing actually, I'm not sure it's an iff between the two Leaky

$f(x+h) = f(x) + Df_x(h) + o(\Vert h \Vert)$ as $\Vert h \Vert \to 0$

proofwiki.org/wiki/Definition:Differentiable_Mapping/… here's it without little o notation if you prefer

I'm guessing they picked $r$ for remainder or remaining error or something

it's precisely because there is not an iff between the two that I mentioned it

I agree

I meant to write $D_xf$ not $Df_x$

I've also seen $Df(x)h$

4:34 PM

@GFauxPas But how this relates differentiability? Do the line of thought goes like this: For a sharp turn there are no such best candidates hence the function is non-differentiable?

Alessandro Codenotti

Do you understand how linear approximations relate to differentiability in the single variable case?

you can write the single variable case as $f(x+h) = f(x) + f'(x)h + o(h)$

@AlessandroCodenotti No, In the case single variable we weren't taught that aspect.

as $h \to 0$ it's the same thing as saying $\dfrac{f(x+h)-f(x)}h \to f'(x)$

Ok, then?

4:38 PM

wait, that doesnt look correct

I thought you were going to type something.

@AjayMishra you can prove it by yourself

Alessandro Codenotti

@AjayMishra Nothing about the slope of the tangent line at a point and the value of the derivative at that point?

@LeakyNun I've just started studying this, how can I? In the case Calc I, we were taught a function $f(x)$ is differentiable at $x=a$ if $\lim_{h \to 0} \cfrac{f(x+h) - f(x)}{h} $ exists.

if you've "just started studying this" then maybe you should stay away from multivariable calculus for now

4:50 PM

@AlessandroCodenotti Derivative must exist at that point, that's all.

I mean, I've just started studying calc III (MVC).

@TedShifrin I am teaching calculus I am just trying to find a good pictorial way to teach why product rule works what do you think of $\frac{(x + \Delta x)(y + \Delta y) - xy}{h}$

I think one can represent this in some kinda of rectangle and explain it that way.

5:17 PM

[Random]

0x=a

x+y=0

00=0(x+y)=0x+0y

Nuke 1

00=0 no longer a theorem. Define 00=0

00=0=0(x+y)=0x+0y

0(x+y+z)=0x+0y+0z=0+0z=0z

Hmm... a rng with zero terms retained seemed to lead to interesting directions...

0x=(x+y)x=xx+yx

Need to study them systematically later

Now to move on to: Bracket Algebra

Let <0,1,(),B> be a bracket algebra, where () is a n-ary operator with n any natural number. Then we have the following axioms:

10=01=0

(000)=1

(100)=(010)=(001)

(0011)=11

Now given a string 10001101010100010, we have:

10001101010100010=0000000000=1110=0

10001101010100010=101101010100010=00000000

It will be more interesting if we use inference rules instead of identities. Then we we have something like:

(00101) => 00
(0010) => 001011
(001)=>010
01=>10=>0

Then we can encode some logical operations using just a string of binaries and the n ary operator

Subhasis Biswas

6:26 PM

Wish me luck guys. Embarking on a journey called "2nd CHAPTER OF BABY RUDIN".

How does one determine whether a statement is inaccessible to proof?

Hmmm, does anyone know if the GRE mathematics test is paper-only?

Alessandro Codenotti

7:14 PM

@SirCumference what do you mean?

@AlessandroCodenotti As an example, Godel showed that the continuum hypothesis was unprovable using ZFC

I'm wondering if anyone can explain briefly how something like that could be shown

Alessandro Codenotti

He didn't, Cohen did

Gödel showed that the negation of CH is unprovable, by constructing a model in which CH holds

Cohen showed that CH is unprovable by constructing a model in which the negation of CH holds

The first was done with inner models, the second was done via forcing

@Sir For an easier scenario which is not hardcore logic/set theory, the fact that Euclid's fifth postulate cannot be deduced from the first four is done using first creating a model where all five holds (Euclid did that, it's called Euclidean geometry) and creating a model where the first four holds and the fifth doesn't (Gauss did that, with the invention of hyperbolic geometry)

@BalarkaSen So more generally, we would need to show that a given postulate cannot be deduced from any of the axioms of ZFC, to demonstrate that something is unprovable?

"Cannot be deduced" and "unprovable" are synonymous...

7:24 PM

Welp, I didn't think there was a clear number of axioms used in ZFC

It's the construction of models where a statement holds and doesn't (respectively) which constitutes a proof of unprovability, was my point

Alessandro Codenotti

@SirCumference there's infinitey many

@BalarkaSen Ok, I see

Alessandro Codenotti

For a simple example look at the gruop axioms. How do you show that the statement $\forall x\forall y(x+y=y+x)$ is not provable from the group axioms? By constructing a group where it fails

That's the same in set theory, the only difference is that the techniques required to construct models are much more sophisticated

Aah, got it

Thanks

Alessandro Codenotti

7:28 PM

Another way to show that a statement is unprovable is showing that assuming it leads to a contradiction (assuming the theory you started with is consistent, otherwise everything is provable). That's the case with "there exist an inaccessible cardinal" and ZFC for example

@AlessandroCodenotti aSsUmiNG ZFC iS cOnSiStEnT

Grothendieck universes lmao

Alessandro Codenotti

Every set belongs to a Grothendieck universe is equiconsistent with there is a proper class of inaccessible cardinals

Akiva Weinberger

ZFC proves CH

@BalarkaSen Oh btw will you ever return to the h bar :O

Akiva Weinberger

7:33 PM

It also proves not CH

Alessandro Codenotti

Which is strictly stronger than ZFC in consistency strength but I'm not sure how strong it is exactly

0celo came back, but some people like Bernardo, Eulb etc. haven't

Akiva Weinberger

At least, ZFC doesn't prove that ZFC doesn't prove CH

@Sir I sneaked in there a few days ago

@AkivaWeinberger so ZFC is inconsistent lmao

Akiva Weinberger

7:36 PM

Yeah

Prove me wrong

What if you are ZFC

Director: M Night Shyamalan

End credit rolls

@BalarkaSen Welp, how's life btw

You're a math major right?

That's correct. It's not bad

user image

Ok, that one's new

I'm out of the thonk meme loop

7:41 PM

@BalarkaSen what properties must $M$ and $N$ have so that every continuous map $f:M\to N$ is homotopic to a smooth map $g:M \to N$?

@BalarkaSen the real axioms were the friends you made along the way

8

@LeakyNun $M, N$ smooth.

@BalarkaSen what's the theorem called?

Whitney approximation theorem

Do you not need compactness as well?

7:43 PM

I don't think, but I only know how to write it down when $M$ is compact.

Some care is needed if it's noncompact I assume.

At least the version we proved assumed both manifolds compact, $M$ so that you can use Stone-Weierstrass and $N$ for having a global $\epsilon$ for the tubular neighborhood. Probably Hirsch has the maximally general version tho :P

lol wiki has Cellular Approx but not Whitney Approx

6

Q: Applications of Whitney's Approximation Theorem

I am reading the book Introduction to Smooth Manifolds by John Lee. In his book he proves a theorem called the Whitney's Approximation theorem which essentially states that any continuous map can be approximated by a smooth map. In the end he gives an application where he proves that any homoto...

differential-topology applications

math.toronto.edu/vtk/1300Fall2015/lecture-oct27.pdf (theorem 2.0.7, without proof)

@Daminark You don't need the full blast of Stone-Weierstrass. If $M$ and $N$ are both compact, take embeddings in $\Bbb R^m$ and $\Bbb R^n$ with tubular neighborhoods $U_M$ and $U_N$, extend to a continuous map $F : U_M \to U_N$, convolve with a smooth function cut-off function $\varphi_\epsilon$ supported on an $\epsilon$-neighborhood of $M \subset U_M$. Then $F * \varphi_\epsilon : U_M \to U_N$ is smooth. Project it back down using the tubular functions.

Akiva Weinberger

You either live a fan of markings on letters, or diacritic

If $\epsilon$ is very small this will be close to $F$ in the $C^0$-norm, so homotopic to $F$. I guess that's where you really need compactness of $M$.

It's not true that if two maps from noncompact manifolds are close in $C^0$-norm then they are homotopic

See eg

7:52 PM

Fucking hell I was glancing through Hirsch chapter 2 and wow I need to sit down for a second

"Globalization theorem"

That's his peak

Unknown MathMan

Can anyone tell me how can I show - "General transformation matrix T transforms a pair of intersecting straight lines to pair of intersecting straight lines" ?

If $M$ and $N$ are noncompact, I think what you need are proper embeddings in $\Bbb R^m$ and $\Bbb R^n$, then a tubular neighborhood with varying $\epsilon$ (this can be done, but again more technical)

Yeah I've heard of the varying epsilon, I guess if there is a way to do the approximation within $\epsilon(x)$ then I can see how even that weaker version of tubular neighborhood suffices

I am sure the same proof pushes through now to show every $C^0$-map $M \to N$ for smooth noncompact $M, N$ is $C^0$-close to a $C^\infty$-map.

@Daminark You just need to convolve with a smooth function supported on a varying epsilon neighborhood of $M$ in the tubular neighborhood (also varying in width) in R^m

7:58 PM

Ah I see

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