Mathematics

Q: Open subsets of a complete metric space.

1

I've been going over previous exams, and I came across a question that I missed. It is as follows: Let $X$ be a complete metric space. Show that every open subset of $X$ is homeomorphic to a complete metric space. I am having difficulty showing this. Any help would be greatly appreciated.

This seems to answer the open case

not sure about the general case.

Completely metrizable space

In mathematics, a completely metrizable space (metrically topologically complete space) is a topological space (X, T) for which there exists at least one metric d on X such that (X, d) is a complete metric space and d induces the topology T. The term topologically complete space is employed by some authors as a synonym for completely metrizable space, but sometimes also used for other classes of topological spaces, like completely uniformizable spaces or Čech-complete spaces. == Difference between complete metric space and completely metrizable space == The difference between completely metrizable...

12:20 AM

G_\delta makes a lot more sense than F_\sigma

since \Bbb Q is F_\sigma

oh yeah, I wrote nonsense with $G_\sigma$ there

I meant $G_\delta$

Related question, I wonder how one proves that a metrizable space is compact iff every metric inducing the topology is complete

that's an exercise in a German general topology book I have

I think it's a cool reult

Maybe the idea is to embed a non-compact space as a non-closed subset of some complete metric space

Jenna Sloan

If any number multiplied by zero equals zero, then shouldn't zero divided by zero equal every number?

you probably just take some non-bounded positive function f:X \to \Bbb R and add f to the metric

and that gives a non-complete structure

assuming the original metric was complete.

lush

I'm off to sleep

eh

lush

12:33 AM

cya guys :)

that doesn't work

Jenna no

cya

good night @lush

@robjohn: It appears I'll be going to LA (despite my health woes). If I can figure out how to park and find you (:)) I'm happy to meet up for lunch or after lunch for coffee. I'm staying somewhere in the valley (eastward, I think).

BTW, any ideas on this? I constructed a stupidly wrong counterexample, but now I'm not sure if I believe it or not.

12:37 AM

@JennaSloan Any number multipled by zero equals zero is the reason why division by zero is undefined. In the set of real numbers :-)

Q: Proof that the Trace of a Matrix is the sum of its Eigenvalues

66

I have looked extensively for a proof on the internet but all of them were too obscure. I would appreciate if someone could lay out a simple proof for this important result. Thank you.

linear-algebra eigenvalues-eigenvectors trace

@MikeMiller Hey, do you know what exactly Besse means by $\ker P^*$ in Appendix 32? Somehow they jumped from $W^{k,2}$ in 31 to $W^{k,p}$ on 32 which doesn't really make sense to me.

@TedShifrin wicked awesome proof

Especially when $p<1$, I don't expect the $p=2$ result to give much, if anything.

Well, I guess $\ker P^*\subset C^\infty$...

so it doesn't depend on which bundle you define $P$ on to begin with.

\\shrug

//shrug

Jenna Sloan

12:52 AM

Is zero even a real number? Or is it more of a concept, like infinity?

@Pseudohuman is zero real?

Pseudohuman

@JennaSloan 0 is a real number

1:05 AM

The negative numbers are also "real numbers" @JennaSloan

@skullpatrol I haven't seen a negative number in the night sky

I think only the numbers $\{1,2,3,4,5,6,7,8,9,10,42\}$ are real

everything else is made up

If the temperature was below zero you would feel the effect of them :P @0celo7

use the Kelvin scale

Negative Kelvin does exist.

1:09 AM

^

though that more reflects the way temperature is defined in statistical mechanics

Crap

The answer I need is in Hormander

usually, increasing the internal energy of a gas goes hand-in-hand with increasing its entropy

I am throwing basically the entirety of linear elliptic theory at this problem

@0celo7 sounds fun

but you can definitely find examples where putting more energy into a system makes it more ordered, and when that happens you'll get negative temperatures

1:11 AM

that's weird

I have no idea how temperatures work

ironically some of the best answers are in physics-y books because GR people worry about sobolev regularity metrics

or any physics for that matter

you need to use the fact that certain Sobolev spaces are Banach algebras

$W^{s,p}$ is an algebra if $s>n/p$

@0celo7 hooray, we have rings!

@MatheinBoulomenos So I've got rings whose projective limit is $C^\infty$

suddenly I don't like PDE as much

1:13 AM

sounds more fun to me

@MatheinBoulomenos in stat mech $T=\left(\frac{\partial U}{\partial S}\right)_{V}$ where $U$ is the internal energy of the gas, $S$ is the entropy, and $V$ is the volume which is held fixed

yeah, I have no idea what entropy is

something something order

lol

ok, physicist

why would you use $u\otimes v$ when you clearly mean $uv$

context?

1:15 AM

why the hell should $u\otimes v$ be the pointwise product

@Semiclassical nonlinear PDE

Didn't you guys record a copy of your live-stream on YouTube? @Semiclassical

gotta label it somehow, I guess?

@skullpatrol I don't know if they do. I think maybe they don't, which is annoying

I bet you didn't rush the stage

The man has future grant proposals to think about :P

1:47 AM

Any discrete fans here?

maybe

I'm a continuous fan, myself.

I'm trying to show that if $G$ is Hamiltonian with $n$ vertices then $G\square K_{1,m}$ is Hamiltonian $\Rightarrow n\geq m$, but I think I just got it actually

2:05 AM

If $n<m$, and there were a hamiltonian cycle, for every clone of $G$ in that graph, the cycle would have to reach it. This involves using at least two edges which intersect the "inner" copy of $G$: an edge which leaves and an edge which returns.

But this implies that there are, at least, $2m$ edges which are incedent to vertices in $G$

hence there is a vertex that is incident to three edges of the Hamiltonian cycle, contradiction

what's an example of an infinite subset of $R^{2}$ with no accumulation points?
I'm thinking the set of natural numbers, but that's only in R
$R^2$ is like a 2D plane. I know I sound silly x_X!

natural numbers works

Why not $\mathbb{Z} \times \mathbb{Z}$?

of course, the natural numbers, realized as $\mathbb{N} \times \{0\}$ also gets the job done

@Prototank That looks like graph theory, which is something that I don't know very well; I probably cannot help you.

I'm not a fan

I am taking it for credits... sometimes it is interesting and relevant to what I care about (LDT)

off the top of my head, I don't recall what a Hamiltonian path is (it visits every vertex?), and I have no idea what your $\square$ operation is

2:11 AM

you got it

$\square$ is the analogue of direct product for topological spaces

topological spaces are so hard X_X!

so for each vertex in $A$ clone $B$

^ what I think when people talk about being fans

understandable

I am going to go see if I can find a couple of ice cubes

a troika of olives

some vermouth

and a bit of gin

2:14 AM

Xander what do you study?

since I know where to find a glass and shaker...

I'm a gin person myself

@Prototank I do fractal geometry

I never tried gin

never?

2:15 AM

That sounds like a newer field

am I wrong?

Kasmir Khaan

@MatheinBoulomenos mathein :D

sup

Kasmir Khaan

can I invite you to a room for a sec ? :D

okay

Kasmir Khaan

yeeey :D

Q: A collection of most of the properties about a linear operator and its trace.

2:17 AM

0

${\bf Problem }$ If $A$ is a square matrix, we define the trace of A, $tr(A)$, to be the sum of the diagonal entries of $A$. Let V be a finite dimensional vector space over $\Bbb C$ , with $\dim(V ) = n \geq 1$, and let $T \in L(V )$. We define the trace of T by $tr(T) = tr([T]_{\alpha}^{\alpha}...

linear-algebra

fractal geometry? well, I suppose that it has deep roots, but didn't really get of the ground until Mandelbrot started publishing in the late 70s or early 80s

anyone mind trying to help me land the finishing blow on this proof?

son of a b**** turned into a bloody story trying to prove it

every time i turned around i needed to prove some new result to move forward

What is the Pythagorean theorem of fractal geometry?

So... fun fact: the Dirichlet kernel $D_N$ is, like, bounded below by $C \log(N)$, where $C$ is a constant

@skullpatrol I'm not sure what you mean by "the Pythagorean theorem of fractal geometry"

if you mean "most fundamental result", I would suggest that Hutchinson's 1981 paper has a couple of theorems which might qualify

Yup

2:22 AM

for instance, the existence and uniqueness of the attractor of a contractive iterated function system might qualify

Pythagoras tree (fractal)

The Pythagoras tree is a plane fractal constructed from squares. Invented by the Dutch mathematics teacher Albert E. Bosman in 1942, it is named after the ancient Greek mathematician Pythagoras because each triple of touching squares encloses a right triangle, in a configuration traditionally used to depict the Pythagorean theorem. If the largest square has a size of L × L, the entire Pythagoras tree fits snugly inside a box of size 6L × 4L. The finer details of the tree resemble the Lévy C curve. == Construction == The construction of the Pythagoras tree begins with a square. Upon this square...

Not that^

(though that is actually a fairly easy consequence of the Banach fixed point theorem, once you know that the space of compact metric spaces is a complete metric space with respect to the Gromov-Hausdorff distance)

the Moran equation giving the Hausdorff dimension of the attractor of a self-similar iterated function system satisfying the open set condition is pretty fundamental, too

hmm, interesting

2:41 AM

What is a good example of A sequence of closed sets $\{F_{n} \}$ such that their union is not closed. Like I need closed intervals maybe like $[ \frac{-1}{n},n} ]$ and $ [ \frac{1}{n},n} ]$ and then the union is not closed

sigh

$[\frac{-1}{n},n ]$
$[\frac{1}{n},n]$

$F_n := \left[ \frac{1}{n}, 1 - \frac{1}{n} \right]$

each set is closed, the union is $(0,1)$

in your example, the union is $(0,\infty)$

which is also not closed

at least, if you take $F_n = [1/n, n]$

with $[-1/n,n]$, the union is $[-1,\infty)$

which is closed