Mathematics - 2023-02-25 (page 2 of 2)

Silly Goose

21:20

is one way of looking at metric spaces to see metrics as generalizing the notion of order in sets of elements of higher dimension than one

err i guess that probably isn't true. since you can assign an order without additional structure on any set with just the definition of order perhaps :P

冥王 Hades

I hate iced tea

Gwyn

Is it correct to say that $\|x\|$ (Euclidean norm) satisfies the triangle inequality on $\mathbb{R}^n$ because, since $t\le |t|$ for each $t\in\mathbb{R}$, since the root is increasing and for the Cauchy-Schwarz inequality, it is $\|x+y\|=\sqrt{(x+y)\cdot (x+y)}=\sqrt{x\cdot x+2x\cdot y+y\cdot y} \le \sqrt{\|x\|^2+2\|x\| \|y\|+\|y\|^2}=\sqrt{(\|x\|+\|y|)^2}=\|x\|+\|y\|$?
I mean the scalar product in $\mathbb{R}^n$ with $\cdot$, that is $v\cdot w=\sum_{i=1}^n v_i w_i$.

And I define $\|x\|:=\sqrt{x\cdot x}$.

Ted Shifrin

21:37

@D.C.theIII Yes, although I did write some different computational ones as well.

@Silly No, there will be no notion of order in most of these spaces. But you can talk about "closeness" of two points without an ordering.

Xander Henderson

Not to mention the fact that there are one-dimensional metric space which are unordered...

(and which have no reasonable notion of "order")

Ted Shifrin

@Gwyn That is fine. Your argument works in any (even infinite-dimensional) inner product space.

remands @Xander to custody for disorderly conduct

Xander Henderson

((of course, this depends on what you mean by "dimension", which is not an obvious thing by any means, either))

@TedShifrin unorderly conduct, thank you very much!

Ted Shifrin

@Hades Effectively, though, one way to prove that the 3D determinant gives signed volume is to use the scalar triple product, if you know that the magnitude of the cross-product is the area of the parallelogram. It all depends on what order you derive things.

@Xander You were so unordered that you became disordered.

Xander Henderson

@TedShifrin Oh, dear. I'm totally disconnected, too. :/

Ted Shifrin

21:42

We need the term discombobulated for set theory, now.

Gwyn

@TedShifrin thanks!!

monoidaltransform

This is another problem in the same exam question (I have included all the details): Assume $f:V\rightarrow X$ is a continuous map, where $X$ is compact and $g:U\rightarrow X$ is a continuous map, where $U\subseteq closure(U) \subseteq V$ Assume $C$ is a closed set in $X$ contained in $V$ and $\partial{C} \subseteq U$. If $g$ is the identity near $\partial{C}$ and $g$ and $f$ coincide away from a closed set. Show that the map $H(x)=f(x)$ on $V-U$ and $H(x)=g(x)$ on $U$ is continuous.

WHAT IS THIS EXAM ugh

Ted Shifrin

The author is fond of sloppy statements/mistakes.

monoidaltransform

maybe i'm too tired, but there's no reason $H$ is continuous right?

Ted Shifrin

Just forget this guy. How is $C\subset X$ then contained in $V$?

As with the other question, you first have to rewrite the question so that it actually makes sense.

monoidaltransform

21:52

Okay just a second let me try to write it more clearly. There are too manythings going on

I found this in his lecture notes: Assume $X$ is a compact metric space. W is open in X, U is open in X, C is closed in W and we have $\partial{C}\subseteq U \subseteq closure(U)\subseteq W$. Given two continuous map $f:W\rightarrow X$ and $g:U\rightarrow X$, such that $g=i$ on the boundary of $C$, and $g$ and $f$ coincide away from a closed set $K$ in $U$. He then states " It is easy to see that the following map" F(x)= f(x) on $W-U$ and $g(x)$ on $U$ is continuous

Why is it easy to see?

ugh. I wish all teachers were as clear as you, Ted

apparently he is a famous mathematician, but all his students said he was very very hard to understand, along with his thick accent

Ted Shifrin

22:07

I'm not sure how my clarity comes in here :P

I know some very famous mathematicians who were notorious for being sloppy.

monoidaltransform

Like everything you write makes so much sense

and your lectures on youtube too

Ted Shifrin

So why are we saying $C$ is closed in $W$ rather than saying $C$ is closed in $U$?

I keep wondering if $C$ and $K$ are supposed to be the same set. And I don't begin to see how compactness of $X$ is relevant.

monoidaltransform

So, $C$ is closed and contained in $W$, but its boundary is contained in $U$.

Ted Shifrin

Oh, I see. Yeah, that makes sense.

monoidaltransform

I think to find such U, you need compactness of X, which implies local compactness of V.

D.C. the III

22:12

@TedShifrin Cool. THanks...I'm asking because I'm doing MAT3510 now and figure I'm going to need some more practice with just tinkering with regions I'm integrating over.

Ted Shifrin

If $f$ and $g$ agree on $U-K$, then I don't begin to see the relevance of $C$ here. So if I define $F$ as you said then what goes wrong with the standard gluing of continuous functions?

@DCthe Absolutely. You need to do lots of computational stuff here. But there are plenty of problems in the book :) It would be wonderful to make all my thousand hours of WeBWork work available, but I don't have a server lying around in my apartment.

monoidaltransform

It goes wrong because $W-U$ need not be open

in $W$

Ted Shifrin

Of course it won't be open.

But $f$ and $g$ agree on a neighborhood of $\partial U\subset W$.

user1294729

@TedShifrin sorry to disturb again. I wanted to proof your proposition 2.2. in the script on my own to understand everything a bit better. So I took a smooth regular curve $\gamma$ in $\Bbb{R}^3$. I denote by $\beta(s)=\gamma(\phi(s)$ a unit-speed representation of $\gamma$. Then I first remark from our last discussion that $\beta'(s)=\gamma'(\phi(s))\phi(s)$ and $\beta''(s)=\gamma''(\phi(s))\phi'(s)^2+\gamma'(\phi(s))\phi''(s)$.
Then if I could show that $$K(\phi(s)=\|\beta'(s)\times \beta''(s)\|~~~~~~~~(1)$$ then I'm done. Because from the definition of $\beta'(s)$ I can conclude that $\p…

Ted Shifrin

I still think $C$ is a complete red herring here.

D.C. the III

22:15

When I start getting financially compensated for all my endeavour and I could give back I'll make sure to buy you a data storehouse

monoidaltransform

So $F$ is indeed continuous in this case? provided $C$ is irrelevant?

Ted Shifrin

@user1294729 But $K = \|\beta''(s)\|$ here. The usual cross product stuff is useful for formulas for $\gamma$ without any reparametrization.

$\beta'$ is a unit vector orthogonal to $\beta''$, so there's nothing much to show.

user1294729

But can't I show that $K(\phi(s)=\|\beta'(s)\times \beta''(s)\|$?

Ted Shifrin

But for general $\gamma$, $\gamma'$ and $\gamma''$ will not be orthogonal, etc.

It's true, because of what I said just above. But how does that help you?

user1294729

hmm sorry I don't get your argument

Ted Shifrin

22:20

I don't get why you think this formula is helpful.

The magnitude of that cross product is immediately $\|\beta''(s)\|$, which is the curvature. So done. But where does this get you?

monoidaltransform

Not too sure. Maybe because you can adapt $W$ to write it as a relevant union for which $f$ and $g$ agree on intersections?

and one of those intersections is $\partial{U}$?

Ted Shifrin

I'm just suggesting taking an "inner" tubular neighborhood of $\partial U$ in $U$. If it's "thin" enough, then $f$ and $g$ agree on it, because it avoids $K$. Then you can glue.

user1294729

@TedShifrin but I'm confused with $K(\phi(s))$, what does this mean is $K(\phi(s))=\|\beta''(\phi(s))\|?

Ted Shifrin

You have defined curvature to be the magnitude of the second derivative provided the curve is arclength-parametrized.

user1294729

yes

Ted Shifrin

22:22

$\beta$ is arclength-parametrized.

monoidaltransform

Sorry, by inner tubular neighborhood, you mean $\{x \in U$ $:$ $d(x,y)<\epsilon$ for some $y\in \partial{U}$ \}$?

Ted Shifrin

Yeah, @monoidal.

user1294729

@TedShifrin yes

Ted Shifrin

So that formula holds by definition, @user1294729.

user1294729

but the formula goes with $K(s)$ and not K(\phi(s))

Ted Shifrin

22:24

The $\phi(s)$ is there because the curvature function was defined originally on the $t$-parametrized curve ... which is annoying.

user1294729

So you mean any $t$ can be written as $\phi(s)$ since $\phi$ is bijective?

Ted Shifrin

It is the curvature at $\beta(s)$, which is the point $\gamma(phi(s))$ on the "original" curve.

Remember where you got $\phi$ from in the first place.

monoidaltransform

Okay and then you're invoking the standard gluing of continuous functions result?

Ted Shifrin

@monoidal Yeah.

As I say, all the nonsense about $C$ is totally irrelevant.

monoidaltransform

But on what cover of $W$? (Small enough tubular neighborhood of $\partial{U}$ in $W$) $\cup$ $U$ $\cup$ $....?.....$?

Ted Shifrin

22:27

Huh?

user1294729

@TedShifrin what do you mean by that?

Ted Shifrin

I mean exactly what I said. Where did $\phi$ come from?

user1294729

so $\phi(t)=\int_a^t \|\gamma'(u)\|du$

monoidaltransform

Sorry, the standard gluing lemma is if $f:X\rightarrow Y$ and $X=\bigcup_i U_i$ where
$f|_{U_i}$ agrees with $f|_{U_j}$ on $U_i\cap U_j$ for all $i,j$ and $f|_{U_i}$ is continuous for all $i$ then $f$ is continuous

where each $U_i$ is open

Ted Shifrin

@monoidal Just use two open sets.

@user1294729 Then why are you writing $\phi(s)$? I thought you had $t=\phi(s)$, with $\phi$ the inverse function of $s(t) = \int_a^t \|\gamma'(u)\|du$.

user1294729

22:32

Ah yes sorry $\phi=s^{-1}$

respectively we denoted the integral by $L$ since $s$ is another letter

so $\phi=L^{-1}$

Ted Shifrin

As I've said several times, it's easier to work with $s(t)$ than with its inverse function.

Anyhow, $L$ is the arclength $s$.

user1294729

Yes but then $K(\phi(s))$ needs to be rewritten as $K(L^{-1}(s))$

to prove my equality

Ted Shifrin

Well, that is your $\phi(s)$. I still say there's nothing to prove.

You want to be computing the curvature just in terms of $\gamma(t)$ in the first place.

That's the point of prop 2.2.

user1294729

I still don't see why there is nothing to prove. You explained it to me but it does not make completely sense to be because since $\gamma$ has not unit-speed we use the reparametrization $\beta(s)=\gamma(\phi(s))$ which has unit-speed. But then $K(s)=\|\beta(s)\|=\|\gamma(\phi(s))\|$ but I don't see where the cross product $\|\beta'(s)\times \beta''(s)\|$ needs to appear.

Ted Shifrin

NO. $K$ is $\|\beta''(s)\|$. What you've written is wrong.

I think you should talk to your instructor. I'm not helping here and we're just going around and around over and over again.

user1294729

22:41

ah yes $K(s)=\|\beta''(s)\|=\|\gamma''(\phi(s))\phi'(s)^2+\gamma'(\phi(s))\phi''(s)\|$ but then there is still no cross product

Ted Shifrin

For the last time, the relevance of the cross product is to do it with the original parametrization. Write $\gamma''$ as a linear combination of $\beta'$ and $\beta''$ and use the fact that those vectors are orthogonal, etc.

user1294729

but the cross product is not with the original parapetrization but with $\beta', \beta''$?

Ted Shifrin

I give up. Go talk to your instructor.

user1294729

I will do it hopefully he can help me

monoidaltransform

22:59

Hey Ted. Sorry, i'm almost there but still sort of stuck. So our map is $F:W\rightarrow X$ given by $F(x)=f(x)$ on $W-U$ and $g(x)$ on $U$. So continuity on $U$ is clear. So we want to show continuity on $W-U$ and you said, take a tubular neighborhood $TU$ of $\partial{U}$ that is contained in $U\K$. So in this case, $f=g$ on $TU$. Perfect. But why does that imply continuity on $W-U$?

@TedShifrin

oh nevermind.

Thanks so much

you're the best!

monoidaltransform

23:45

@TedShifrin This doesn't make sense though, because $\partial{U}$ and $U$ are disjoint

Ted Shifrin

So what? The union of $W-U$ and that neighborhood is one open set, $U$ is the other.

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