Mathematics - 2020-04-17 (page 1 of 2)

Mona Jalal

00:20

@skullpatrol it is quite bad but I am in Cambridge, MA and is very bad here too

is anyone here proficient in "learning under uncertainty" topic?

nbro

00:46

2

Q: How does size of the dataset depend on VC dimension?

This might be a little broad question, but I have been watching Caltech youtube videos on Machine Learning, and in this video prof. is trying to explain how we should interpret VC dimension in terms of what it means in layman terms, and why do we need it in practice. The first part I think I un...

user456014

01:45

So much of those convergence tests, Abel, Cauchy, Dirichlet, Hardy, Leibniz, D'alembert, and that's not all! How can I remember them all?

user456014

02:44

@MonaJalal learning under uncertainty? what does that mean?

Mona Jalal

03:12

@in

@InFlames when you have noise in the communication channels or when you want to design a robust algorithm

Ted Shifrin

03:23

@InFlames Like with other things in mathematics. Lots of practice (and thinking about what is more likely to work where, i.e., understanding).

skullpatrol

Is that what understanding is about in the infamous Von Neumann quote? @TedShifrin welcome back btw

user456014

@TedShifrin If you believe me, and you do, there is a book that seems to be almost exclusively about homeomorphisms in analysis. My cup of tea.

user456014

Gonna make some tea now.

maths student

06:52

Suppose A is orthogonal matrix and B is matrix similar to A. Then is it true that B is orthogonal?

Martin Sleziak

07:11

@mathsstudent I do not think this is true. I left a short comment in the linear algebra room.

Jaakko Seppälä

Let $X$ be a topological space and $A\subset X$. Why do we have $\operatorname{int]\bar A\subset \operatorname{int}A$?

Martin Sleziak

You want {int} rather than {int]. (Currently it does not render.)

Jaakko Seppälä

Yes. Sorry about the typo.

Martin Sleziak

Maybe looking at the assumption that $x\in\operatorname{int} \overline A \setminus \operatorname{int} A$ and trying to derive contradiction?

@JaakkoSeppälä Is that actually true? What about $A=\mathbb Q$ in $\mathbb R$?

We have $\operatorname{int} A = \operatorname{int} \mathbb Q = \emptyset$.

And $\operatorname{int} \overline A= \operatorname{int} \mathbb R = \mathbb R$.

Jaakko Seppälä

The problem was given by me some student. She said the question was given in some university exam. I was unable to prove it.

Martin Sleziak

07:25

The same would work with and co-dense set (this is the condition $\operatorname{int} A = \emptyset$) which is at the same time dense (i.e., $\overline A = X$.)

@JaakkoSeppälä Weren't there some additional assumptions on $A$?

Perhaps this could work if $A$ is open?

Jaakko Seppälä

There were no additional assumptions.

At least in the exam paper. Maybe if there was given in the course something like "unless otherwise stated, sets are open".

Martin Sleziak

I have posted this also in the general topology room - maybe there is some chance that somebody notices the problem there. (In case somebody has a few more comments on this.)

Anonymous

Hi. Does anyone here know a good textbook or lecture notes that cover how to compute stuff like $H^2(C_2, \Bbb Z_2)$ in the context of central extensions (i.e., ways to centrally extend $C_2$ by $C_2$)? (I'm having a hard time understanding the lecture note my professor gave on this topic.)

Balarka Sen

See any book which does group cohomology a little bit, eg Dummit-Foote. Cyclic groups have a periodic resolution; you can compute from there that $H^2(\Bbb Z_2; \Bbb Z_2) = \Bbb Z_2$.

There are exactly two central $\Bbb Z_2$-extensions by $\Bbb Z_2$; the cyclic group of order 4 and the Klein 4-group. This can of course be proved without group cohomology

Anonymous

@BalarkaSen Thanks, I will check Dummit-Foote

maths student

08:02

@MartinSleziak any example for underlying field real for above question you have answered?

Leaky Nun

@mathsstudent the generic example one comes up with will be a counter-example

as in, pick random numbers and you'll have a counter-example

Martin Sleziak

@mathsstudent I don't really understand what you're trying to say. Anyway, we can continue in the linear algebra room where I have suggested a counterexample.

CaptainAmerica16

@TedShifrin I've thought about it for a bit. Since $f(E \cap F) = f(E) \cap f(F)$ doesn't work when the elements are different ($x$) and ($e$), I have to prove it works when they are the same. That's the one-to-one bit :D I'm going to figure out how to set that up.

if f(x) = f(e), though

Trey

08:45

can someone explain me the difference between integer rings and galois fields?

Edward Evans

10:06

Any German Speaker. In the room?

Other than me

Edward Evans

10:22

@Thorgott ja hoi! Hätte noch eine frage zur komischen Sprache in Neukirch lol

Thorgott

Klar, schieß los

Edward Evans

Also, ich hab Punkte $x$ die einerseits in einem beschränkten "Bereich" eines $\Bbb R$-Vektorraums, und andererseits in einer diskreten Untergruppe (also in einem Gitter) des $\Bbb R$-Vektorraums. Neukirch folgert hieraus, dass es nur endlich viele solche Punkte $x$ gibt. Ich nehme an, dass mit "Bereich" implizit auch abgeschlossen gemeint wird, sonst ergibt das für mich keinen Sinn :D

Weil er glaub ich damit sagen will dass die Punkte in einer diskreten kompakten Teilmenge vom Vektorraum liegen

Thorgott

10:51

Ist ein Gitte nicht eh abgeschlossen?

Edward Evans

11:08

Ja ich glaub schon lol

oops

glS

11:21

ehi uh... is it just me or did mathjax stop working on the site?

Thorgott

Sei $\Lambda\subseteq\mathbb{R}^n$ ein Gitter. Behauptung: Es gibt ein $\varepsilon>0$, so dass $\lVert v\rVert\ge\varepsilon$ für alle $v\in\Lambda$. Wäre dies nicht der Fall, gäbe es eine Folge $(v_n)_n$ in $\Lambda$ mit $v_n\rightarrow0$, aber $v_n\neq0$ für alle $n\in\mathbb{N}$. Dann enthält jede Umgebung von $0\in\Lambda$ ein $0\neq v_n\in\Lambda$, also ist $\Lambda$ nicht diskret; Widerspruch.
Da $\Lambda$ eine Untergruppe ist, gilt $\lVert v-w\rVert\ge\varepsilon$ für alle $v,w\in\Lambda$ mit $v\neq w$. Folglich sind die einzigen konvergenten Folgen in $\Lambda$ die eventuell konsta…

Edward Evans

Nice, gibt auch einen anderen beweis in der englischen Version von Neukirch anscheinend lol

maths student

12:09

0

Q: Cyclic subspace and one dimensional range

$\therefore$.. Let $T$ be a linear operator on a finite dimensional vector space $V$ with $\operatorname{dim} V$; $\geq 2$, and let $R(T)$ be the range of $T$. If $\operatorname{dim} R(T)=1$, prove that there exists a scalar $k$ such that $T^{2018}=k^{2017} T$ What I am thinking since range is...

linear-algebra matrices linear-transformations

user456014

How is continuity defined between arbitrary topological spaces?

Alessandro Codenotti

preimage of open sets is open

user456014

But all the sets are open in some topology over some set?

Alessandro Codenotti

Sure but you have fixed topologies on domain and codomain when you want to check the continuity of a function

user456014

And when topologies vary one and the same function can be for some topologies continuous and for some discontinuous?

Alessandro Codenotti