So I'm trying to prove that every nonempty open subset of $S^{n-1}$ contains an open subset $V$ such that $V^c \cong D^{n-1}$, but I'm running into some trouble. I figured that since $V^c$ is compact and $D^{n-1}$ is Hausdorff it suffices to find a bijective continuous map $f : V^c \to D^{n-1}$, but actually constructing such a $f$ is proving hard to do

Going back to my question from earlier can an homeomorphism decrease the Hausdorff dimension?

Like in the case $n=1$ and $n=2$, I can sorta see what such a map should do, but still I don't know how to write down an explicit formula for it

Also can every topological manifold be embedded in $\Bbb R^n$?

@AlessandroCodenotti mathoverflow.net/questions/34658/…

Nice, thanks!

hello

can someone tel me why the integral on X equal the integral on X_0?

@Vrouvrou because the function is zero outside so It won’t add any value to the integral

@AlessandroCodenotti The proof for smooth manifolds can be word-by-word analogized to embed a topological manifolds in Euclidean space. In fact, it's slightly easier

I see

Anyway in general there are homeomorphisms of metric spaces reducing the Hausdorff dimension: the inverse of the Osgood curve

If your topological manifold is compact, you have a finite cover by charts, call it $\{U_i\}_{i = 1}^n$. Choose a partition of unity $\rho_i$ supported on this cover. Then define $f : M \to \Bbb R^{n+1} \times \cdots \times \Bbb R^{n+1}$ by $f(x) = (\rho_1(x) \varphi_1(x), \rho_1(x), \rho_2(x) \varphi_2(x), \rho_2(x), \cdots, \rho_n(x)\varphi_n(x), \rho_n(x))$. This defines an embedding.

So the interesting question should be "if $X$ is an $n$-dimensional topological manifold and $f:X\to Y$ is an homeomorphism (onto its image), can the hausdorff dimension of $f(X)$ be smaller than $n$? Especially for $Y=\Bbb R^m$

(Two envelopes problem) you are given two envelopes A and B, and you are also given the knowledge that one of them contains twice as much money as the other. You open envelope A to see $10. You then calculate the expected value in B: 0.5*20 + 0.5*5 = 12.5 > 10. So should you switch to envelope B?

Where $Y$ is a metric space of course

@AlessandroCodenotti Good point

@LeakyNun The expected gain is 0, so clearly not, I'd say

The point is the conditional expectations of A|B and B|A are the same

then what's wrong with the calculation?

Nothing, you're just not computing the right thing

You're computing E(B|A=10). So what? Why should that matter at all?

if E(B|A=10) = 12.5 > 10 then shouldn't E(A|B=12.5) > 12.5 as well?

True

so you should just keep switching?

I mean, I just gave you an argument for why you shouldn't switch :P

E(B|A=10)>10 does not imply that the situation is unsymmetric for you.

That's simply it

Can't figure out whether the paper I'm reading wants $\text{Tr}(AB)$ or $\text{Tr}(A^H B)$.

ah, finally, found it...and I was still wrong

$\langle A,B\rangle = \text{Tr}(AB^H)$

Hi chat

@AlessandroCodenotti I thought that topological dimension was bounded below by Hausdorff dimension

Does anybody know how I map a doubly slit unit disk onto the unit disk in the complex plane?

You have to use the modular lambda function

@JoeShmo Wait, did you really mean what you wrote? Not the unit disk onto the doubly slit unit disk?

Because the double punctured unit disk has a holomorphic universal cover by the unit disk

i mean if i can conformally map one onto the other, the inverse exists and is conformal

slit, not punctured

Those are not biholomorphic domains

$D_1(0) \setminus \{(-1, -h] \cup [h,1)\}$ for some $h\in(0,1)$

A doubly slit disk is biholomorphic to a doubly punctured disk so it doesn't matter

how do we know? in fact i suspect the opposite is true if it's pointed out in this exercise

but its slit at its boundary

Oh

i.e. the slit isn't there at all

That makes a lot of sense.

Yes, then they are biholomorphic but I don't know how to explicitly construct one

said set is topologically equivalent to a disk $\iff$ there exists a homeomorphism (better yet, a conformal one) that maps one to the other, etc.

Correct, Riemann mapping theorem.

I thought you had slits in the interior, in which case they are not topologically equivalent, of course.

something to do with the Zhukovsky map

right

the Zhukovsky map maps the unit disk onto a slit ellipse

Aha. I don't know this

@BalarkaSen I don't believe doubly slit in the interior is bihol to doubly punctured :)

@MikeMiller Is it false that homeomorphic domains are biholomorphic?

sure ^

continuous but not differentiiable

false on the complex plane, but its probably fairly easy to construct sets where that property holds true

Oh of course that's garbage, disk and complex plane

What's a better example that doesn't break Liouville's theorem?

of what exactly

Two homeomorphic open subsets of C which are not biholomorphic

try taking moduli

Uh, what's that

otherwise its harder to come with something that isn't differentiable on the complex plane

plural of modulus. The modulus is continuous but not differentiabel

That's not at all what I am asking

what are you asking

You're giving examples of homeomorphisms which are not biholomorphisms. That's easy :P

isn't that what you asked..

No lol

pretty sure it's a couple comments up

@BalarkaSen Yup, already visible for annuli, where the radius determines the bihol type.

@MikeMiller Ah, got it.

(As long as the complement doesn't have a point-component, aka, it should have boundary circles)

Maybe think of them in terms of the universal covers, rectangles equipped with four "corner points"

Yeah, actually, I know examples of closed surfaces which have that as well. Quotients of C by lattices of different proportions

@Joe You completely misunderstood the question. "Homeomorphic open subsets" => There is a homeomorphism between the open subsets. Doesn't necessarily mean I'm asking that same homeomorphism to be the biholomorphism.

im not debating this. got a lot of work

@BalarkaSen There are very good Riemann mapping theorems for multiply connected domains in the plane in fact

@MikeMiller Right, the proportion of the sides of the rectangle determines the biholomorphism type

You can always map to an annulus minus some number of circular slits, IIRC

Though I don't know about uniqueness

Hey guys, studying for my final right now. Was wondering if anyone could take a quick gander at this proof

Prove that $l^p$ is dense in $l^q$ given $p < q$.

For $l^p$ to be dense in $l^q$, it means that any arbitrary sequence $x \in l^q$ is an adherent point of $l^p$, i.e. for any arbitrary radius open $r$ ball centered around any element $x \in l^q$, there exists a sequence $y \in l^p$ s.t. $y \in B_r(x)$, or $d(x,y) < r$.

@OneRaynyDay Show that there is a continuous inclusion from one to the other then show that the set of finite sequences is a dense subset of each

So I think the Zhukovsky map will just do

apply, and then map back the result, but insist that the inverse lies in the unit circle

@MikeMiller I don't think we've gotten to continuity yet unfortunately. We've ended at connected metric space topology

My approach was to construct some $y$ that converges in $l^p$ for any sequence $x \in l^q - l^p$

@OneRaynyDay Don't worry about that sentence

gotcha

Pay attention to the claim that finite sequences are always dense

Yep

That's the key

So I was gonna construct $y$ as follows: For any sequence $x \in l^q$, it must be the case that $\sum_{n\geq N} |x_n|^q < r$ for any arbitrary $r$.

where $N = N(r)$

Then construct $y_n = x_n$ for all $n < N$, and $y_n = 0$ for $n \geq N$

Thumb up

Then since $y$ is a sequence with finitely many non-zero elements it must converge in any $l^p$ space

gotcha, thanks!

btw, jw how's your research been @MikeMiller ?

I've been trying to finish my necessary classes for CS so I've just begun to take some math courses

Trying to get a paper out within 2 weeks

nice :) does ucla usually publish on arxiv preprint before sending into journals or what's the process?

This is a long shot, but any Java people around?

Everyone does, that's just the math process

@Lozansky yes

@MikeMiller I'm not familiar with the math process; that also coincidentally happens to be for CS as well

cool, good luck man :)

@LeakyNun Alright, so I made a chess game

And I want to prevent the player from making a move that puts them in check

Now, I have a method that returns a boolean if you're in check

But obviously, first you move and then it looks at the state of the board

What I want is to simulate the move, then see if that move puts you in check, and if it does then you can't make that move

But for some reason, my "mock" model doesn't really work

ok and then?

if you act on temp, then you acted on this.

right, you should make a copy

I thought that made a copy?

no, temp and this are pointing to the same class instance

a shallow one, yeah :-)

Ugh, how do I make a copy then?

copy constructor

google it

deep copy

But I need the state as it is a the time I'm calling the method

if you make a deep copy, you wilil have a deep copy of the state as well

its comparable to taking everything you have on one piece of paper, and copying it over to another piece of paper. Do you now have the same information on the new piece of paper as you did on the old one, once you've copied it over?

Rather than having two pointers to the same data, it will be two pointers, each to different locations in memory, and the new location gets the same data as the old one.

Hmm interesting

@JoeShmo Yes, but different references?

yup. so what?

alternatively, and that already gets into the particulars of your design, you could just deep copy the internal data structure that you are using to represent the board (2D array is it?)

Yeah 2d array

but i think a deep copy of the entire object is easier to manage.

there's surprisingly a lot of CS people in this chatroom lol

Ah, so this is really only a problem because I have mutable fields

well.... i think you've gone too deep down the rabbit hole

i mean you have to have mutable fields

its a problem because you're mutating them when you're simulating your moves, but you're not keeping track of what you changed. Therefore, you can't undo your changes, once you're done with your simulation

How do I show that there is a hyperplane in $A^n$ which contains $n$ points $p_1, \cdots, p_n$?

as a good exercise in reverse engineering, try inspecting the code that renders and represents the chess board on www.chess.com

Alright, so from what I gather the solution is "Java Object Serialization"?

and add me @Hathatul so we can play

you have to implement a copy constructor

don't serialize anything