Mathematics - 2024-12-11

ModularMindset

00:35

youtube.com/@jackzimmerman2217

in case anyone wants to subscribe or follow my math journey

Jakobian

09:01

I'm starting studying Rudin's example of a Dowker space

X4J

09:59

At propositional calculus, if L is a language with $|L| = k$ and $(\psi_0, \dots, \psi_{l-1})$ is a sequence of sentences s.t for each $i < l-1$ we have $\psi_i$ tautologically implies $\psi_{i+1}$ but not the conserve. What is the maximum $l$?

I figured it is possible with exactly $k$ sentences but I'm not sure how to prove it is the maximum length.

*I probably might be unclear because I lack of the correct terms (I translated that question) so I'd like to clarify if needed

psie

14:42

Define the equivalence relation $x\sim y$ on $[0,1)$ iff $x-y$ is rational and let $N$ be the subset of $[0,1)$ that consists of a member of each equivalence class. If we also define $R=[-1,1]\cap\mathbb Q$, is it true that $[0,1)\subset\bigcup_{r\in R} (N+r)$?

psie

15:11

I think I've figured it out. If $x\in[0,1)$ is rational or irrational and it happens to be the chosen representative in $N$, then clearly it is in $N+0$ and hence in the union. Otherwise, if $x\in [0,1)$ is rational or irrational and belongs to $[y]$ (i.e. the representative is $y$), then it belongs to $N+(x-y)$. And $(x-y)\in [-1,1]\cap\mathbb Q$. Bingo.

psie

15:49

Follow-up question. If $N$ is as before and we consider $E\subset N$, and we let $t\in (E+r)\cap(E+s)$, where $r,s\in \mathbb Q\cap[0,1)$. Then $t-s,t-r\in E\subset N$, but is $t-s=t-r$?

Certainly $t-r=(t-s)+(s-r)$ and $s-r\in\mathbb Q$, so $t-r\sim t-s$, but are they equal?

psie

16:20

Ah, ok. Indeed, $t-s=t-r$, since they both belong $N$ and the same equivalence class, and since $N$ consists only of one member of each equivalence class, ... ! Bingo.

Slate

What's on your mind?, an invitation to discussion on Meta Stack Exchange

ephe

I can solve part (a) but I can't really figure out how to make use of the hint in part (b). How do I even find the value of that function on the boundary? I tried to put $z=Re^{i\theta}$ and take one $R$ out but that didn't do much. Can someone help me out?

Thorgott

17:55

@Ben there should be plenty manifold examples, but I'm blanking right now: math.stackexchange.com/questions/5010195/…

psie

18:35

> Exercise Let $f:[0,1]\to[0,1]$ be the Cantor function, and let $g(x)=f(x)+x$. Show that $g$ is a bijection from $[0,1]$ to $[0,2]$, and $h=g^{-1}$ is continuous from $[0,2]$ to $[0,1]$.

Attempt If $x,y\in[0,1]$ and $x<y$, then $g(x)=f(x)+x\leq f(y)+x<f(y)+y=g(y)$ and hence $g$ is injective. Furthermore, $g$ is continuous. By the intermediate value theorem, we see that $g$ takes on all values between $g(0)=0+0=0$ and $g(1)=1+1=2$. Hence $g$ is surjective.

Regarding $h=g^{-1}$ being continuous, we show $g$ is an open map. Let $I=(a, b)$ be an open interval contained in $[0,1]$. It follows from $g$ strictly increasing and the intermediate value theorem that $g((a, b))=(g(a), g(b))$. Here is where I'd like to conclude; since any open subset of $[0,1]$ is the union of intervals, it follows that $g$ is an open map.

Question Is the last statement true and how can I specify the intervals more explicitly?

Thorgott

yes, it is true that any open subset of $[0,1]$ is a union of intervals. of course, you can also have intervals containing an endpoint $0$ or $1$, but the conclusion is similar for those.

and I don't know what specification of an interval you imagine that is more explicit than giving its endpoints

psie

18:51

@Thorgott ok, thanks. So the open subsets of $[0,1]$ are unions of intervals of the form $(a,b)\subset[0,1]$ and $(a,1]\subset[0,1]$ and $[0,b)\subset[0,1]$. I'll try to bake them in somehow.

Ben Steffan

19:45

@Thorgott I don't see anything immediately either

Thorgott

7

Q: Realizing homomorphisms between fundamental groups

Let $X,Y$ be compact connected manifolds and $\varphi\colon\pi_1(X)\to\pi_1(Y)$ be a homomorphism between their fundamental groups. Under what conditions on $X$, $Y$ and $\varphi$ is it true that $\varphi$ is the homomorphism induced by an appropriate continuous map $f\colon X\to Y$?

There's a cool recipe for how you could theoretically come up with examples here, but even in simple cases, it's too annoying

like, I wanted to start with the $2$-complex given by attaching a $2$-cell to $S^1$ via a degree $3$ map and then attach a $3$-cell that messes up your ability to realize the non-trivial automorphism of $\mathbb{Z}/3\mathbb{Z}$, but I couldn't carry it out

$\pi_2$ of that $2$-complex should be $\mathbb{Z}^2$, but to be honest, I cannot see the generators explicitly

Ben Steffan

my immediate idea would be to consider maps of lens space $L(p, q) \to L(p, q')$ and hope that a map inducing an iso. on $\pi_1$ must already be a homotopy equivalence, but this is probably naive

psie

21:23

Consider $f+g$ when $f,g:E\to\mathbb R$ are both measurable. The way its proved in my book is as follows;

Jakobian

What is proved!

psie

> ...$f+g$ is the composition of the two functions $x\mapsto (f(x),g(x))$ and $(a,b)\mapsto a+b$ which are both measurable (the second one because it is continuous, using also the equality $\mathcal B(\mathbb R)\otimes \mathcal B(\mathbb R)=\mathcal B(\mathbb R^2)$).

What do they mean by "using also the equality $\mathcal B(\mathbb R)\otimes \mathcal B(\mathbb R)=\mathcal B(\mathbb R^2)$?" Where exactly is it used? I understand why $x\mapsto (f(x),g(x))$ is measurable and why $(a,b)\mapsto a+b$ is continuous, but I don't see what it has to do with $\mathcal B(\mathbb R)\otimes \mathcal B(\mathbb R)=\mathcal B(\mathbb R^2)$.

Jakobian

@psie so the two sigma-algebras agree

the theorem is that if $f_i:(X, \mathcal{A})\to (Y_i, \mathcal{B}_i)$ are measurable then so is $f = (f_1, f_2):(X, \mathcal{A})\to (Y_1\times Y_2, \mathcal{B}_1\otimes \mathcal{B}_2)$

so with products you take product sigma-algebra

but continuity $f:X\to Y$ allows you to say its $f:(X, \mathcal{B}(X))\to (Y, \mathcal{B}(Y))$ measurable

so for continuity you take Borel sigma-algebras

so now in the domain of addition, and codomain of $x\mapsto (f(x), g(x))$ you have different sigma-algebras

seemingly different - I mean

but they do agree and that's what we use

write more details next time

psie

ah ok, yeah, if they disagreed, we'd be in trouble

Jakobian

to yourself I mean - not to me

it will help you see where this is used

because what I really did here is I wrote details

precisely so you can see everything

if you wrote this on paper it would be same result

it's like Xander says sometimes, there's no math without pen and paper

psie

21:30

@Jakobian but then there'd be no chat :(

Jakobian

simple problems can be resolved on their own

user123234

0

Q: How can I show that the map $t\mapsto \int_0^t |f'(B_s)|^2~ds$ is strictly increasing for a planar Brownian motion $B$?

I'm working on the conformal invariance theorem for planar Brownian motion. I want to prove the following statement: Let $D\subset \mathbb{C}$ be a domain and $f:D\rightarrow \mathbb{C}$ a non-constant holomorphic map. Let $z\in D$ and $B$ be a planar Brownian motion starting from $z$, then $t\...

probability probability-theory stochastic-processes stochastic-calculus brownian-motion

can someone help me?

mo-_-

22:10

Hi

mo-_-

22:25

Determine the sum of the roots of the equation:
$(z - 1)(z - 3)(z + 5)(z + 7) = 297$

Does anyone see the fastest way to get the solution?

$(z^2 - 4z + 3)(z^2 + 12z + 35) = z^4 + 12z^3 + 35z^2 - 4z^3 - 48z^2 - 140z + 3z^2 + 36z + 105$

$z^4 + 8z^3 - 10z^2 - 104z - 192 = 0$

Im sure this is not the best way...

mo-_-

22:48

One way would be to apply the vietas formula...

Jakobian

@mo-_- there is no such thing as "roots of equation" to my knowledge

a polynomial can have roots, not an equation

@mo-_- solution to what. Are you looking for sum of (complex?) solutions of the equation $(z-1)(z-3)(z+5)(z+7) = 297$?

mo-_-

@Jakobian Yes

Jakobian

23:20

@mo-_- sub $u = z+2$ and $v = u^2$ so then you get $(u-3)(u-5)(u+3)(u+5) = 297$, which becomes $(v-9)(v-25) = 297$

That is $v^2-34v-72 = 0$

So $u^4-34u^2-72 = 0$

its not hard to see that this has $4$ distinct solutions and so does the original

The sum of them $u_1+u_2+u_3+u_4 = 0$ clearly

so $z_1+z_2+z_3+z_4 = (u_1+u_2+u_3+u_4)-2\cdot 4 = -8$

this is the sum of the complex solutions to your equation

mo-_-

Oh ok thanks!

But if I didn't have to deal with complexes, I could also write

(z-1)(z-3)(z+5)(z+7) = 0

So z = 1, z = 3, z = -5, z= -7, and the sum Is -8

Using the property that the constant term does not affect the sum of the roots (consequence of Viète's formulas)

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