Hi @Daminark

Hi chat

Hi @Alessandro

What's up?

I did the probability exam, I'm studying analysis now

Which is an improvement, I guess

Definitely better :)

$X$ is hausdorff and compact . $f:X \to X$ continuous , i need to prove that there is $K \subset X$ closed s.t $f(K) = K$

someone can help?

$X$ is compact so $f(X)$ is compact and therefor closed

So i thought taking $K = f(X)$ or $K = f \ ^ {-1} f(X)$ but none seems to work.

@Liad This is just an idea, but what if you apply $f$ repeatedly?

also , $X$ is normal (because Hausdorff + compact --> normal) so maybe we can use Uryshon's lemma

Can a compact Hausdorff space have an infinite chain of closed sets with proper inclusions?

nvm of course it can

(it can because every chain of closed sets with no empty intersection has a point in common? @TobiasKildetoft )

If $f,g:X\to Y$ are continuous functions with $Y$ Hausdorff then the set of points $\{x\in X: f(x)=g(x)\}$ is closed

Hmm, is the intersection of an infinite chain of closed subsets closed?

I don't think compactness is needed here

@AlessandroCodenotti You need compact to make that set non-empty

@TobiasKildetoft i think no $[-1/n,1/n]$

the intersection is $\{0\}$ which is not closed

@Liad Yes it is closed

@TobiasKildetoft right, but the empty set is closed so that's not a problem

huh right ..

@AlessandroCodenotti I assumed Liad needed to find a non-empty subset with that property

and you are right @TobiasKildetoft

otherwise you don't need Hausdorff either

@AkivaWeinberger I see he mentions Underwood Dudley at one point. That guy is a brilliant crank-chaser

@AlessandroCodenotti what is your $K$ ?

The set of fixed points of $f$

so you need compactness

wait. is it necessarily not empty?

@Liad Wait, the infinite intersection of closed sets is always closed (duh). The question is whether it can be empty in this case

But I need to go now

@TobiasKildetoft alright , thanks.

@Liad hm, not sure

because it is stated that we need $K$ to be not empty @AlessandroCodenotti

FINISHED MY EXAMS !!

Hi chat

Hi@Astyx

ha refreshed

but tired too i guess!

@Astyx im jealous :P

:)

Soo tired

(technically I still have some exams, but they don't really matter)

hm

It is possible to turn the real line into a compact space by adding a single "point at infinity" which we will denote by ∞.

I was thinking how we can turn $\Bbb{R}$ into a compact space by adding $\infty$

That's called alexandroff compactification, you can do it for every space

@Astyx !!

hmm

@AlessandroCodenotti

for real number line it is illustrative :)

I can finally learn differential topology in peace

Learn In Peace

I advocate that

Any continuous function from S1 to R has image containing an non empty open subset of R ?

No

Pick a constant one

You mean constant function?

Yep

in the image then it will be a point which is not open

right

?

@AlessandroCodenotti $K$ could be empty - math.stackexchange.com/questions/891674/…

Right

also a quick intuition about why any continuous function from S1 to R is uniformly continuous

because

S1 is compact

Yeah, that's a consequence of the Heine-Cantor theorem

continuous function on a compact space is uniformly continuous

can't we use that?

@AlessandroCodenotti

oh

yes

its Heine Cantor theorem!

also how we show that there is a point $z$ in S1 such that $f(z) = f(-z)$

@AlessandroCodenotti

Given a flag $\mathcal{F}$ of linear subspaces $0 \subsetneq F_1 \subsetneq F_2 \subsetneq \ldots \subsetneq F_k \subsetneq \mathbb{C}^n$, when is the stabilizer group $\text{Stab}(\mathcal{F}) = \{g \in GL(n,\mathbb{C})$ with $gF_i = F_i$ for all $i\}$ defined by polynomial equations with rational coefficients, i.e. an algebraic $\mathbb{Q}$-group?

@TobiasKildetoft if you define $F$ to be the family of closed sets $K \subset X$ s.t $f(K) \subset K$ , this family has the finite intersection property so it has a nonempty intersection. do you think we can continue from here?

\frac - > $\frac{\sin(x)}{\cos(x)}$

\dfrac -> $\dfrac{\sin(x)}{\cos(x)}$

difference bold and magnified a bit in dfrac

:)

actually, there's, dfrac, tfrac, and frac

dfrac means: has to look good like a formula

tfrac means: has to look good inside a sentence of text

$\tfrac{\sin(x)}{\cos(x)}$

oh

and frac means: pick whichever looks best depending on the context

nice@steam

There's cfrac as well, for continued fractions, I believe

Well, kinda, yeah. You should use \frac as your default, and use \dfrac or \tfrac if you need to force how it looks for some odd reason.

$\cfrac1{1+\cfrac1{1+\cfrac1{1+\cfrac11}}}$

$\dfrac1{1+\dfrac1{1+\dfrac1{1+\dfrac11}}}$

Hm, cfrac seems to have more spacing or something

So maybe it doesn't stand for "continued," I dunno

cfrac is indeed just a frac combined with spacing

Also has some optional arguments for left or right outlining or so I think...

Is tfrac the same as frac in text, then?

woops

$$1\tfrac12\quad1\frac12$$

And, yes, using frac in text defaults it to tfrac

$\afrac11\bfrac11\cfrac11\dfrac11\efrac11\ffrac11 \gfrac11\hfrac11\\ \ifrac11\jfrac11\kfrac11\lfrac11 \mfrac11$

lol

Haha :P

I think cfrac is the only "special" frac

It does stand for continued, by the way

Defined as

$\nfrac11\ofrac11\pfrac11\qfrac11\rfrac11\sfrac11 \tfrac11\ufrac11\\ \vfrac11\wfrac11\xfrac11\yfrac11 \zfrac11$

Whelp.

Cfrac, dfrac and tfrac it is, then.

#SayNoToFracking

Lol

Don't frac your crack Jack.

@SteamyRoot there is a problem im trying to solve maybe you will be able to help

$f:X\to X$ continuous , $X$ is hausdorff and compact, i need to prove there is a closed set $K \subset X$ s.t $f(K)=K$. $K \ne \emptyset$. if $f$ has a fixed point we can take $K$ to be the set of fixed points, otherwise idk..

I assume $K\ne X$ as well?

if $K = X$ then you need $f$ to be surjective

Oh, true

Umm

Do like the Banach fixed point theorem

Repeatedly take $f$

We don't know if it's a contraction map, or even if it's metrizable

But yeah that's what I was thinking as well

About what?

Well, it's not the BFPT itself

It just uses the same idea

pick $K_0 = X$, $K_{i+1} = f(K_i)$

The set you want will be $\cap K_i$

That doesn't quite work

Wait hold on I misread

Maybe it does

it is not empty because $X$ is compact.

it is closed.

Why does $f(\cap K_i) = \cap K_i$ ?

If $A\subset B$ then we have $f(A)\subset f(B)$, right? Just making sure

hm, yes

@Liad Don't we have $f(A\cap B)=f(A)\cap f(B)$? Or maybe that should be $\subset$ and not $=$

Yeah, it's $\subset$

no we done have it

take disjoint sets that goes to the same place

We have $f(A\cap B)\subset f(A)\cap f(B)$, though, right?

yes

This is putting me to a trance

So $f(\bigcap K_i)\subset\bigcap f(K_i)$

I should be doing work lol

Command?

hm, yes @AkivaWeinberger

@SteamyRoot why the set you mentioned is the right one ? :P

$\bigcap f(K_i)=\bigcap K_i$

because $f(K_i)=K_{i+1}$

@aminliverpool Sometimes putting spaces in to break it up works

LaTeX tends to ignore spaces

@AkivaWeinberger so why does $f(\cap K_i) = \cap f(K_i) $

Hm

:P

You could just work elementwise.

Send it piece by piece

How long is this formula

@aminliverpool then put newlines

Anyway, what you want to prove is $\cap K_i = f(\cap K_i)$.

@aminliverpool no, literal newlines

@SteamyRoot yes, how?

I would just prove both inclusions elementwise.

right to left is by set - theory

Pick $y \in f(\cap K_i)$. Then $\exists x \in \cap K_i: f(x) = y$. Then $\forall i, x \in K_i$, hence $y \in K_{i+1}$ for all $i$.

But clearly also $y \in X$ so $y \in K_0$ and thus $y \in \cap K_i$.

@aminliverpool yes

@SteamyRoot alright, the other way is harder doesn't it?

$x \in \cap K_i $ implies $f(x) \in \cap f(K_i) = f(\cap K_i)$ ,right? @SteamyRoot

wait im not sure about the last '='

The other way around should be pretty much the same.

Can you write it? im not sure it is the same

You have to realise this intersection isn't just some arbitrary intersection of sets $K_i$

It's the limit of the decreasing sequence $(K_i)_i$

I'm back

Also, it's false

Bout to lose internet again, by

@SteamyRoot we have $\cap f(K_i) = \cap K_{i+1} $ ?

Pick $y \in \cap K_i$. Then $y \in K_{i+1} = f(K_i) \forall i$, so $\exists x: \forall i: x \in K_i$ and $f(x) = y$. But then $x \in \cap K_i$, so $y \in f(\cap K_i)$.

@Liad Yes, by definition of $K_i$. You shouldn't take intersections of $f(K_i)$'s, though.

@Liad This seems quite an interesting question, here is a related post on the main site: How to show that a continuous map on a compact metric space must fix some non-empty set.. I have mentioned this also in General topology chat room.

Wait I messed up

@MartinSleziak alright, thanks.

Technically, you have to invoke the decreasing sequence-ness to make sure such $x$ exists and you don't have a different one in every $K_i$, but meh

@SteamyRoot Shouldn't that be $\forall i\exists x:x\in K_i$?

See previous comment :P

@SteamyRoot ok i will just believe you are right :-) too much time i spent on this question.

How do I solve this question (Hints would suffice)? Research Effort: Learnt(and derived) Sine formula, learnt (and derived) cosine formula, Napier's Analogies, solved some basic questions on the same.

It's a Prove that/ To Prove question.

@Abcd but you don't even specify that $a,b,c$ are side lengths of a triangle with angles $A,B,C$...

For $α$ an arbitrary real number, how many solutions might you expect $z^α =1$ to have?

@ManeeshNarayanan it should be $\lfloor \alpha \rfloor$

$z^{\pi}=1\implies \pi\ln z=0 \implies \ln z=1 \implies z=e$

@Secret the first implication is wrong, the second implication is not even wrong, so the third implication is hopeless

and then overall $e^\pi$ is far from $1$

$e^\pi \approx 20 + \pi$ :P

@ManeeshNarayanan it may be $\lceil \alpha \rceil$ instead

isn't it always holds that for reals, $\ln z^x = x \ln z$?

(I know for complex you need to worry about branches...)

@Secret did he say that $z$ is real?

No

O i see...

Z can be complex

@LeakyNun Frankly, I think the second implication is the most painful one of all, though.

@SteamyRoot I agree, which is why I labelled it as "not even wrong"

well, I initially thought z is also real, so I just use the real log laws

Yes, but $\pi \ln z = 0$ does not imply that $\ln z = 1$.

@Abcd This question seems kind of similar: Proof of the identity: $c\sin \frac{A-B}{2} \equiv (a-b) \cos \frac{C}{2}$

OMG, that's a careless mistake!

I am getting only commands. It is not compiling

Either way, I doubt this is going to be doable for arbitrary real numbers.

@SteamyRoot it is.

I think that careless mistake is because when I tried to cancel out the $\pi$, my thinking is one step ahead thinking about $\ln 1$ which is $0$, and then for some reason that messed up thought process lead to me somehow dropping the ln by mistake

@LeakyNun Enlighten me.

@SteamyRoot well I don't know what you mean by "doable"

Sometimes I wish there's a field of maths that can track careless mistakes, cause I made way too many of them and I don't often know how

If you don't know what it means, it's weird for you to claim it's that o.O

I just took it to mean solvable

How many solutions exists?

@ManeeshNarayanan infinitely many, if you take the correct branch

it should be countably infinite

if you take the principal branch, it should be $\lfloor \alpha \rfloor$

and I'm not going to torture my mind by thinking about negative numbers

$z ^3 = 1$ only has $3$ solutions

@LeakyNun I thought that was quite understood. Moreover, it's not even given in the question because it's naturally implied

@SteamyRoot arbitrary.

for rational numbers there would be finitely many solutions

But for irrational case ,can you explain

for irrational numbers there would be countably infinitely many solutions

@MartinSleziak Thank you :)

wolframalpha.com/input/?i=solve+x%5Epi+%3D+1

How?

Why the complex counterpart of everything almost always look prettier than their real counterparts

@ManeeshNarayanan wait, I'm typing

Because going along the circle in irrational quantities means you spin around forever, probably.

@SteamyRoot yep

it would be useful to think of the unit circle as $\Bbb R / \langle 2\pi \rangle$

How to compile these tex commands

@ManeeshNarayanan follow the last link in the chatroom description

I wonder if there's some rule on the number of real roots of unity ...

@Secret it's either 1 or 2

for odd integers, it is 1; for even integers, it is 2

Just a question to all the online users: Why do you like Math?

for rational numbers, check the numerator, and follow the above rule

for irrational numbers, don't bother, because $(-1)^\pi$ isn't even defined.

P.S. $(-1)^{2/3} := \left(\sqrt[3]{-1}\right)^2 = (-1)^2 = 1$

notice how I used $:=$.

(NB, analytic here means things that can be systematically broken down or assembled, unlike nonstructural things such as emotions and the feeling of art)

@Secret Okay. Thanks for answering

@LeakyNun What about you?

@Abcd I don't have an answer.

oh

Anyone else?

@LeakyNun Wut

@SteamyRoot see my PS

Why are you redefining complex exponentiation o.O

@SteamyRoot no, Secret was talking about real exponentiation

"Spelling inaccruately inaccurately while accusing someone being inaccurate is a terribly inaccurate and inaccrurate thing to do, and yes I spelt inaccurately accurately on the final sentence. Now bartender, give me a drink with $\aleph_1$ orange juice in it"

@LeakyNun That's not allowed for real exponentiation either.

@SteamyRoot I thought $(-1)^{2/3}$ is allowed

Yes, it's a complex number.

@SteamyRoot no, it's real

$-1 = (-1)^1 = (-1)^{2/2} = ((-1)^2)^{1/2} = 1^{1/2} = 1$

So $(-1)^{\frac{2}{3}}\neq ((-1)^2)^{\frac{1}{3}}\neq ((-1)^{\frac{1}{3}})^{2}$?

@SteamyRoot the second equality is kind of invalid

@Secret they are all $1$ so they are all equal

No, it's not.

@SteamyRoot see my link

Meh. Non-standard conventions.

How should I correctly evaluate $(-1)^{\frac{2}{3}}\neq ((-1)^2)^{\frac{1}{3}}\neq ((-1)^{\frac{1}{3}})^{2}$ if the above convention is not used?

@Secret follow my link

$(-5)^{\frac{\pi}{2}}=(\text{insert suitable operator})\{(-5)^{\frac{3}{2}},(-5)^{\frac{3.1}{2}},(-5)^{\frac{3.14}{2}},...\}$ probably does not even converge

@Secret it's called "discrete" for a reason

wolframalpha.com/input/?i=(-5)%5E%7Bpi%2F2%7D

Defined in the complex though

for the "genuine" exponential for complex numbers, $(-5)^{\pi/2} := \exp(\frac\pi2 \ln -5)$

for the "genuine".

Seriously, what :P

The complex one is the standard definition of exponentiation

You can use an alternative one for reals or so if you want special stuff

But don't go mixing definitions based on what number you're exponentiating.

Does the complex function $x^y$ have poles (besides $0^0$)?

notice that it would have to be multi-valued

@Secret I'm not in the mood of analyzing 6-dimensional graphs

@Secret "$0^0$"

No.

Even over the reals, that's not well-defined.

@SteamyRoot no, he means near $(x,y)=(0,0)$ is a pole

No, I mean are their poles or singularities besides (0,0)

(I shouldn't have included the word "near" there)

cause I vaguely recall some negative base do weird things...

but I am not sure if all negative bases are tamed by the complex numbers

In this post math.stackexchange.com/questions/2358486/… we are interested in the sequence $(a_n)$ with the property that $a_n \cdot a_1 = 7$ and $a_{n+1}=\frac{a_n^2+3}{2(a_n-1)}$. Wouldn't the first condition imply that $a_n$ is a constant sequence?

It's not going to be a pole.

Maybe an essential singularity, but definitely not a pole.

@user193319 no, given the sequence $a_n$. (full-stop)

$a_1=7$ and ...

@LeakyNun Ah! I thought the period was \cdot Thanks for the clarification!

@TedShifrin Do you know why the arithmetic genus has the name it does? Why not the Todd genus (it is the integral over the Todd class, right?)?

Let $S$ be the sphere that is tangent to the $xz$-plane and whose center is $C=(0,1,0)$. Why does this imply that the radius of the $S$ is $1$?

@user193319 have you looked at a picture of this situation? the point of tangency is (0,0,0), which is 1 unit away

@Danu that sounds ahistorical

the point of tangency is the closest point in the plane to the center

@arctictern I would want to minimize the function $f(x,z) = (0-x)^2 + (1-0)^2 + (0-z^2) = x^2 + z^2 + 1$? Note that $f(0,0) = 1$, but if there were to exist $x$ and $z$ such that $f(x,z) < 1$, then $x^2 + z^2 + 1 < 1$ or $x^2 + z^2 < 0$, which is a contradiction. Would that work as a proof that $(0,0,0)$ is the point of tangency?

@user193319 I think you can use some geometry

you can instead minimize $x^2+z^2$

and forget what I said about geometry

@LeakyNun Wouldn't minimizing x^2 + z^2 + 1$, which I believe I did above, be essentially the same thing?

@user193319 yes

@user193319 you need to mention continuity to establish tangency

@LeakyNun Continuity of what? I am not sure I follow.

@user193319 of $f$

actually, you need differentiability in addition to continuity

I don't follow...I am assuming that $(x,0,z)$ is a point of tangency, and according to what @arctictern said, this point is such that $(0-x)^2 + (1-0)^2 + (0-z^2) = x^2 + z^2 + 1$ is minimum. Where is differentiability and continuity needed?

Functions can have mins/max regardless of whether they are continuous, right?

yes but that doesn't make it tangent

minimum + differentiable => tangent

But I am assuming that it is tangent.

@LeakyNun When am I allowed to cancel the numerator and denominator and when I am I forbidden to do so?

@Abcd when that quantity you cancel is non-zero, then you are allowed

otherwise, you are forbidden

@user193319 whatever

@LeakyNun What if I am given in terms of x or sin(x)? Like sin(x)/sin(x)

@Abcd same rule

@LeakyNun No, no LeakyNun. I mean What if I am unaware of the value of $x$ or $sin(x)$ then am I allowed to cancel?

then I am unaware if you are allowed to cancel

@LeakyNun Ok. We find instantaneous velocity using limits (differentiation) right? What if my function is not defined at the approaching value? Then wouldn't my answer of instantaneous velocity be wrong?

@Abcd no idea

Ok, can I ask that as a question on stackexchange @LeakyNun?

@Abcd could you give a concrete example ?