Mathematics - 2019-05-27

5:11 AM

10 red balls (all alike) and 10 blue balls (all alike) are to be arranged in a row. If every arrangement is equally likely then the probability that the balls at the two ends of the arrangement are of the same color is ...

This is my attempt: The denominator will be $\frac{20!}{10!10!}$ while, the numerator is $2\cdot\frac{18!}{8!10!}$, hence probability is $\frac9{19}$. Am I right?

pre-kidney

6:40 AM

So.... I worked hard answering a bounty with a well-thought out and comprehensive answer, well before the bounty expired, and yet the bounty was not awarded. In fact, it looks like a no-mans land math.stackexchange.com/questions/3230195/… with not a single response since I posted the solution. What gives?

The bounty ended 3 hours ago, with my answer as the only one, and with 0 votes (looks like no one saw the answer). It says the grace period lasts for another 20 hours. Will the bounty be auto-awarded to me?

Is there any way I can contact a moderator about this?

Anyone here?

F.White

@pre-kidney It gets auto awarded to you when the grace period ends, and you get half the value of the bounty

pre-kidney

Even the answer has 0 votes and is not accepetd?

@F.White ?

Looks like this is not the case, per the rules;

If you do not award your bounty within 7 days (plus the grace period), the highest voted answer created after the bounty started with a minimum score of 2 will be awarded half the bounty amount (or the full amount, if the answer is also accepted). If two or more eligible answers have the same score (their scores are tied), the oldest answer is chosen. If there's no answer meeting those criteria, no bounty is awarded to anyone.

F.White

Oh, it needs two upvotes

pre-kidney

darn, looks like I won't get that bounty unless 2 kind souls see the question in the next 20 hours...

F.White

I would upvote it, but I don't know enough about the material to judge its correctness/quality etc

pre-kidney

6:54 AM

no worries, wish there was a wau yo bump a thread so it doesnt get buried

*way to

a probabilist's bat signal :)

Isiah Meadows

Speaking of weird questions, just stumbled into this. For something that sounds really simple in theory (the sum of the first N terms of the sequence where you start from 2 and square it each time), I can't find hardly any material on it.

In math notation, it's $\sum_{i=0}^{n-1}2^{2^i}$, which does make it sound slightly more complicated, but it only subtly looks different from $\sum_{i=0}^{n-2}2^i$ on the surface. I know they're not exactly the same and that the different exponent does make a big difference, but I didn't expect literally nothing to show up in basic Google searching. (And I don't know enough about number theory to know where to search for more information.)

Isiah Meadows

7:44 AM

Oops, just realized a mistake here: I meant $\sum_{i=0}^{n-1}2^{2^i}$, not $\sum_{i=0}^{n-2}2^{2^i}$.

Subhasis Biswas

9:59 AM

hello all

@BalarkaSen , can you please verify this. An easy thing for you

0

Q: Show that every ordered set with the well ordering has the least upper bound property

Here is a proof attempt: Let $S_a =\{x\in A:x \leq a\}$ (also known as a section of $A$). We firstly prove that $\forall a \in A$, $S_a$ has a supremum in $A$. Clearly, every $S_a$ is bounded above by $a$. We consider $T =A \setminus S_a$. Now, $T$ being non empty [unless $a = \max\{A\}$], it ...

Here is a proof attempt:

Let $S_a =\{x\in A:x \leq a\}$ (also known as a section of $A$).
We firstly prove that $\forall a \in A$, $S_a$ has a supremum in $A$.

Clearly, every $S_a$ is bounded above by $a$. We consider $T =A \setminus S_a$. Now, $T$ being non empty [unless $a = \max\{A\}$], it has a minimal element $m$.

Now, $a$ is an element of $S_a$ with these properties:

1. $a$ is an upper bound of $S_a$ and $a \in S_a$ (not strict partial order relation).
2. Any element less than $a$ is not an upper bound of $S_a$.

i literally copy-pasted the entire thing here.

Another thing can be done: $D= \{x \in A: x $ is an upper bound of $K$ $\}$ . Taking the minimal element...then

Balarka Sen

10:25 AM

I'm not particularly keen on looking at this, @SubhasisBiswas. You asked it on main, someone will answer I'm sure.

Subhasis Biswas

i did

a little question: does every bounded set in a non-strict partial order relation contains its maximum?

Balarka Sen

10:41 AM

@Subhasis isn't a maximal element an element of the set by definition? of course, it need not always exist

Alessandro Codenotti

10:55 AM

@Subhasis let $A$ be a set bounded above in a well order $(P,<)$, we want to show it has a least upper bound. Consider $\{p\in P\mid p\text{ is an upper bound of }$A$\}$. Since $P$ is a well order this set has a minimal element

Balarka Sen

least upper bound is different from a maximal element tho

Hi @Alessandro

Alessandro Codenotti

Hi @Balarka

Leaky Nun

@AlessandroCodenotti inb4 take the union

Alessandro Codenotti

There's no need to know that every well-order is isomorphic to an ordinal for this proof

Leaky Nun

but there's no kill like overkill

Alessandro Codenotti

11:00 AM

Want to see a nuke proof that there is no surjection $\Bbb N\to\mathcal P(\Bbb N)$?

Leaky Nun

go ahead

Secret

^

Alessandro Codenotti

Let $\Bbb C$ denote the Cohen forcing (${^{<\omega}2}$ ordered by reverse inclusion). Let $f:\omega\to\mathcal P(\omega)$ be any function, let $X\prec H(\aleph_1)$ be a countable elementary submodel containing $f$ (which exists by Löwenheim-Skolem) and let $\pi:(X,\in)\to(M,\in)$ be its Mostowski collapse, so that $M$ is transitive.

In particular $M$ is a transitive model of $\mathsf{ZFC}^-$ (because $H(\aleph_1)$ is and elementarity) containing $\mathrm{ran}(f)$ ($\pi(\omega)=\omega$ and $\pi(f)=f$). Also $\Bbb C\in M$ because of absoluteness reasons. By the Rasiowa-Sikorski lemma there is a filter $G$ which is $\Bbb C$-generic over $M$.

Now it's a standard argument that $\bigcup G$ is not in $M$ (that's the new real added by Cohen's forcing), but $\bigcup G$ is a function $\omega\to 2$, so it is the indicator function of a subset of $\omega$ that cannot be in $M$, hence $\mathcal P(\omega)\not\in M$ and so $f$ cannot be surjective, being contained in $M$.

Balarka Sen

set theorists are crazy

Alessandro Codenotti

I could say the same about topologists or geometers :P

Secret

11:13 AM

there are 99% of model theory terminology that I do not know in that proof, hmm...

Balarka Sen

~~model theory~~ model category

Secret

the only part I can understood in the above is there is some real number produced by the forcing procedure, and is some subset of $\omega$, cannot be found inside some countable object whose nature I do not understand, thus resulting in the conclusion that $f$ is not surjective

Leaky meanwhile, might have enough model categoric background to fully comprehend the above proof

Leaky Nun

@AlessandroCodenotti what is $H(\aleph_1)$?

Alessandro Codenotti

$H(\kappa)=\{x\mid |\mathrm{trcl}(x)|<\kappa\}$

Leaky Nun

so $V_{\aleph_1}$?

Alessandro Codenotti

11:19 AM

If $\kappa$ is regular $H(\kappa)$ models $\mathsf{ZFC}^-$ and it is the same as the set of all sets of cardinality hereditarily less than $\kappa$

No, $V_{\omega_1}$ is way bigger than $H(\aleph_1)$

Secret

a "Hereditary $\kappa$ set"?

Alessandro Codenotti

$H(\aleph_1)$ is the set of hereditarily countable sets

Leaky Nun

ok

@AlessandroCodenotti why $\pi(f)=f$?

@AlessandroCodenotti why does $\aleph_1$ exist?

Alessandro Codenotti

@LeakyNun If $a\in X$ and $\mathrm{trcl}(a)\in X$ then $a$ should be fixed by the collapse, hmmm I'm not 100% sure about this right now

@LeakyNun Why shouldn't it? We have $\mathsf{ZFC}$ in $V$, in which this whole construction is done

Leaky Nun

@AlessandroCodenotti yeah but how do you prove that $\aleph_1$ exist? I'm asking about the standard argument

Alessandro Codenotti

11:29 AM

By powerset and choice there is a cardinal above $\aleph_0$, since they're well ordered there is a minimal one

I'm sure there are more "minimal" approaches, but this is very short

Leaky Nun

@AlessandroCodenotti why does powerset give you a cardinal above $\aleph_0$?

Alessandro Codenotti

Actually there's no need for choice here, with $\mathsf{ZF}$ you can already prove Hartogs theorem that there is a least well-orderable cardinal above every well-orderable cardinal

(And powerset is needed because you want to consider all possible well-orderings of $\omega$, which is a subsets of $\mathcal P(\omega\times\omega)$)

Leaky Nun

11:45 AM

@AlessandroCodenotti why is it above?

Alessandro Codenotti

Ah actually I was misremembering the exact argument. Consider the set $W$ of well-orderings of subsets of $\omega$ (rather than well-orderings of $\omega$). By replacement $\{\mathrm{ot}(w)\mid w\in W\}$ is a set, and it's not hard to prove that this set is actually an ordinal, call it $\alpha$. Suppose there were an injection $\alpha\to\omega$, then there would be a subset of $\omega$ of order type $\alpha$, hence $\alpha\in\alpha$

This gives an ordinal not injecting into $\omega$

With choice it's easy to conclude from here, I don't remember how to go about this in $\mathsf{ZF}$, but I'm pretty sure it can be done. Anyway we have choice in the argument above so all is fine

Leaky Nun

@AlessandroCodenotti ok circularity avoided lol

Secret

12:01 PM

(unrelated) actually, the mere existence of $\omega$ by axiom of infinity is circular enough

You either have some form of axiom of infinity or the universal set (New Foundations) to produce $\omega$ from stratification

as far the foundations I am aware of

$\omega_1$ is somewhat similar, but power set saved its butt

Secret

12:15 PM

...

hmm...

Hartogs number

In mathematics, specifically in axiomatic set theory, a Hartogs number is a particular kind of cardinal number. It was shown by Friedrich Hartogs in 1915, from ZF alone (that is, without using the axiom of choice), that there is a least well-ordered cardinal greater than a given well-ordered cardinal. To define the Hartogs number of a set it is not necessary that the set be well-orderable: If X is any set, then the Hartogs number of X is the least ordinal α such that there is no injection from α into X. If X cannot be well-ordered, then we can no longer say that this α is the least well-ordered...

wait a minute, why we cannot use hartog's theorem to produce $\omega$?

ok nvm, you only produce successor ordinals

going from finite to $\omega$ is a much bigger gap than from $\omega$ to $\omega_1$

user193319

Dumb question that is bothering me: how do you prove that $1$ isn't even. Clearly $1$ is odd: $1 = 2 \cdot 0 + 1$; i.e., there exists a $k \in \Bbb{Z}$ such that $1 = 2k+1$, so by definition it is odd. But I need to show that a number can't be both even and odd. However, when I tried to prove this, the question reduced to showing that $1$ isn't even...

How do I avoid this circularity?

Secret

All even numbers $r$ satisfy $r = 2k$ for some $k$, so you want $2\not| 1$

user193319

Okay, how do you show that $2$ doesn't divide $1$?

Secret

and I suspect you might need Euler division algorithm in some form to solve that equation

user193319

Sounds circular.

There should be a very easy proof.

Martin Sleziak

12:23 PM

@user193319 What are you allowed to use?

Secret

0

A: Is it true: 2 does not divide 1?

How deep do you want to go? First we need to define integers. And then we need to define $1$ and $2$ and what "$a$ divides $b$" mean. "$a$ divides $b$" means that there is a number $k$ so that $a*k = b$. This is usually taken in context and it is usually assumed that $k$ is an integer. (It's t...

Martin Sleziak

If $2k=1$ then $k=\frac12$, which is not integer. Would this proof suffice?

Secret

order theory stuff $0 < k <1$

davidlowryduda

@user193319 This is very foundational. Whether or not there is an easy proof will depend entirely on what you are using as your definitions/axioms.

user193319

Hmm...this isn't a problem from a book, so I didn't have any particular axiomatic system in mind...

The problem arises because I was trying to show that two vectors in a certain quotient vector space of $\Bbb{F}^\infty$ are linearly independent.

Martin Sleziak

12:29 PM

Or, as Secret suggested, for $k\le 0$ you have $2k\le0$. For $k\ge1$ you have $2k\ge2$. So there is no $k$ such that $2k=1$.

skullpatrol

short answer: broken numbers :-)

or a part of a number

Odestheory12

1:04 PM

Hi there. I have 11 balls labeled from 1 to 11. Any combination of the first 4 balls makes you win. What is the probability of winning if we choose randomly 4 balls?

Is it $\dfrac{\frac{11!}{(11-7)!}}{11!}$?

I dont think so since the result is really low.

$\frac{11!}{(11-7)!}$ is the total winning combinations we can get.

Secret

$$\sum_{I=0}^{n}2^{f(i)}$$

Eli Bashwinger

1:27 PM

Here is a truly irritating question:

0

Q: Sums of Commutators in a Group Ring

Let $G$ be some group, and let $\Bbb{C}G$ denote the complex group ring over $G$. Let $x,y \in \Bbb{C}G$, and define $[x,y] := xy-yx$ to be a (ring) commutator in $\Bbb{C}G$. Let $K$ be collection of all sums of commutators. I am reading a paper in which the authors claim that any element in $K$ ...

group-theory operator-algebras group-rings

user131753

@SubhasisBiswas: I would like to tell you about a small typo (just before "Clearly $\sup S_a=a$". in 3. there it should be $a<m$ and not $a\le m$.

user131753

Also a little more clarification regarding the generalization would be good.

user131753

2:36 PM

I was thinking about the following generalization of the derivative of function in the context of arbitrary rings.

user131753

Let $R$ be ring and $\tau_1,\tau_2$ be any two topologies on $R$. A function $f:(R,\tau_1)\to (R,\tau_2)$ will then said to be $(\tau_1,\tau_2)$-differentiable on $R$ at $a\in R$ iff there exists a continuous function $g:(R,\tau_1)\to (R,\tau_2)$ such that, $$f(x)-f(a)=g(x)(x-a)$$for all $x\in R$.

user131753

Is anyone already familiar with this kind of notion a derivative for arbitrary rings?

Secret

2:51 PM

Why do you need to different topologies on a ring, I don't think you need a different topology for a derivative as all it needs is the ability to express a small difference as a linear function?

Formal derivative

In mathematics, the formal derivative is an operation on elements of a polynomial ring or a ring of formal power series that mimics the form of the derivative from calculus. Though they appear similar, the algebraic advantage of a formal derivative is that it does not rely on the notion of a limit, which is in general impossible to define for a ring. Many of the properties of the derivative are true of the formal derivative, but some, especially those that make numerical statements, are not. Formal differentiation is used in algebra to test for multiple roots of a polynomial. == Definition... ==

Otherwise, there's something called a formal derivative

user21820

3:14 PM

@user193319 As a beginner, you should take this as your axioms (under "Equivalent Axiomatizations" in case the link doesn't get you there). This is known as PA−, and adding induction yields first-order PA. One of them is "∀x∈N ( x>0 ⇒ x≥1 )", which is crucial to proving what you want. PA− is also known as the discrete ordered semi-ring axioms, but you can ignore that for now.

Focus on being able to prove stuff rigorously from the axioms. If you are careful, you will notice that the axioms in that linked wikipedia article are actually incomplete, because they didn't define "≥". Normally, we define "x≥y" to mean "x>y ∨ x=y", so that should be all you need.

Subhasis Biswas

4:48 PM

@AlessandroCodenotti want to see a noob proof?

Let us assume that $f$ is a surjection from $A$ to $P(A)$. $f(a) \in P(A)$. Two possibilities are there, 1. $a \in f(a)$. 2. $a \notin f(a)$. Let $S= \{a \in A: a \notin f(a)\}$. $S$ is a subset of $A$ and $S \in P(A)$. $f$ being a surjection, there must exist $a_0 \in A$, such that $f(a_0) = S$. Either $a_0 \in S$ or $a_0 \notin S$.

$a_0 \in S \implies a_0 \notin f(a_0) \implies a_0 \notin S$

$a_0 \notin S \implies a_0\in f(a_0) \implies a_0 \in S$.

Both consequences are absurd, so there is no such surjection

This proof has the flavor of Russell's paradox (as far as my brain can comprehend).

Alessandro Codenotti

Sure, that's the usual proof

It is also a far better proof than the one I gave above, the only point was to use as many big tools as possible to prove an elementary result :P

Subhasis Biswas

@AlessandroCodenotti a personal question. How old are you?

Alessandro Codenotti

I'll turn 24 this year

Subhasis Biswas

@AlessandroCodenotti yes, literally a nuke proof. (i don't understand it though)

okay, i have a question

user131753

@Secret I don't really need it. However there is no need from a formal point of view to have same topologies.

user131753

5:04 PM

Actually I am trying to formulate a notion of "topological derivative".

Subhasis Biswas

Suppose $f: \mathbb{R} \to \mathbb{R}$ is a differentiable function. Then prove that if $f'(x) \leq r<1 \ \ \forall x\in \mathbb{R} $ then $f$ has a unique fixed point.

(don't answer it yet, let me try a bit).

Secret

How can one have a continuous function between two different topologies when the open sets are not even the same?

Regarding topological derivative, you might be interested in this:

Subhasis Biswas

@Secret two different topological structures?

Alessandro Codenotti

@Secret How can you have continuous functions $\Bbb R\to\Bbb R^2$?

Subhasis Biswas

well, as far as I know, an open set wrt one topology may not be an open set wrt another topology.

please correct me

Alessandro Codenotti

5:06 PM

Sure but that's no issue

Thorgott

@SubhasisBiswas I think that should be $|f^{\prime}(x)|\le r<1$, no?

Alessandro Codenotti

"Preimage of open set is open" works regardless of the topologies involved

Secret

Ah ok I see

Subhasis Biswas

@Thorgott nope, nothing's been mentioned. Don't work it out for me rn

Secret

The only notion of topological derivative I am aware of is this:

Topological derivative

The topological derivative is, conceptually, a derivative of a shape functional with respect to infinitesimal changes in its topology, such as adding an infinitesimal hole or crack. When used in higher dimensions than one, the term topological gradient is also used to name the first-order term of the topological asymptotic expansion, dealing only with infinitesimal singular domain perturbations. It has applications in shape optimization, topology optimization, image processing and mechanical modeling. == Definition == Let Ω {\displaystyle \Omega } be an...

(I hate my ctrl C it sometimes failed to respond)

user131753

5:10 PM

I only used the term "topological derivative" to mean a topological analogue of the notion of derivative for arbitrary topological space.

Secret

ok, in that case, I felt your definition is too restrictive. Why $g$ must be continuous? Looking back in real function examples, the derivative of a function is not necessary a continuous function

Thus unless you are aiming for a topological analogue of a $\mathcal{C}^{\infty}$ function, I think $g$ can be any homomorphism

user131753

@Secret Yes. You are right. I only wanted to mean $g$ is continuous at $a$. But it was too late to edit.

user131753

Essentially I had in mind the this definition.

Thorgott

@SubhasisBiswas Ok, I just confirmed that the statement is wrong if you are not replacing $f^{\prime}(x)\le r<1$ with $|f^{\prime}(x)|\le r<1$. I can provide a counter-example if you want.

Subhasis Biswas

we consider the function $\phi(x) =f(x) -x$. For existence of a fixed point $c$, $\phi(c)=0$. Now, $(\phi(x)-\phi(c))/(x-c) = \phi'(w)$ (by MVT). For uniqueness, $\phi'(w) \neq 0 \implies f(w) -1 \neq 0 $. Now, the derivative must maintain a strict positive/negative sign throughout (Darboux theorem). So, $f'(w) <1$.

Secret

5:18 PM

That's a sound approach. The only thing I will be careful is when the ring is non commutative and/or nonasosciative, because then you will likely ran into problems

Subhasis Biswas

@Thorgott not if $r$ is positive

?

@Thorgott a counter example would be great

user131753

@Secret Which operation would be non-associative?

Subhasis Biswas

@Secret aren't both operations on ring have to be associative?

Secret

link.springer.com/article/10.1007%2FBF01255614

Ok I was thinking too broad whenever I heard the term "ring"

Subhasis Biswas

you linked us to a Lie ring?

Secret

5:23 PM

Yes , a ring is associative and commutative in +

You knew my thinking have an affinity to not so nice structures...

Subhasis Biswas

@Thorgott, now how to manage this $r$ in my proof?

$f(x)<1$ is a comparatively relaxed condition.

and pretty much sufficient.

Thorgott

I realized my old counter-example was flawed, but I still believe this shouldn't be true. Relaxing the condition to $f^{\prime}(x)<1$ (I think that's what you mean) will however certainly not suffice.

Subhasis Biswas

"Now, the derivative must maintain a strict positive..." I mean throughout in $[c,\infty)$

@Thorgott WHY?

we are eliminating all the possibilities that there is going to be another $d$ such that $f(d)=d$.

Let alone that possibility. We cannot even have $\phi(a) = \phi(b)$ here.

user131753

5:47 PM

The definition has an easy (and ugly generalization) stated as follows,

Let $X$ and $Y$ be arbitrary topological spaces. Then a continuous function $f$ is said to be differentiable with respect to a function $g: X^2\times X^2 \to X$ at $a\in X$ if for any sequence $((f(x_n),f(a)), (x_n,a))\in X^2\times X^2$ converging to $((f(a),f(a)), (a,a))\in X^2\times X^2$ with $(x_n)_{n\in\mathbb{N}}$ converging to $a$, $g((f(x_n),f(a)), (x_n,a))\in X$ converges to $g((f(a),f(a)), (a,a))\in X^2\times X^2$.

Thorgott

Turns out I was misguided in my thought. The result is indeed true for $f^{\prime}(x)\le r<1$.

Subhasis Biswas

@user170039 :O

user131753

In the case when $X=\mathbb{R}$ our $g$ is the function expressed in mathworld.wolfram.com/CaratheodoryDerivative.html.

user131753

@SubhasisBiswas Is there anything wrong with the definition?

Subhasis Biswas

@user170039 this is a huge generalisation

user131753

5:50 PM

@SubhasisBiswas But unfortunately it is very ugly.

Subhasis Biswas

In case of the real number line, the derivative reduces to our very familiar case. Right?

user131753

@SubhasisBiswas Yep. By this.

Subhasis Biswas

@Thorgott so, what's up with this r?

Thorgott

The $r$ is a crucial bound. If you merely supposed $f^{\prime}(x)<1$, the result is wrong. E.g. the function $f(x)=x+e^{-x}$ is differentiable with derivative bounded by $1$, but has no fixed point.

Subhasis Biswas

@Thorgott well, in $\mathbb{R}_e$, it is true :p

Thorgott

5:55 PM

What's that? The affinely extended real number line?

Subhasis Biswas

@Thorgott yes, I also "felt" that it is essential. $1$ is too weak of a bound.

@Thorgott yes.

@Secret homo or homeo?

user131753

@SubhasisBiswas Just take $X=\mathbb{R}$ and assuming $f$ to be differentiable at $a\in \mathbb{R}$ define $\Phi_a: \mathbb{R}^2\times \mathbb{R}^2\to \mathbb{R}$ defined as follows, $$\Phi_a((f(x),f(a)),(x,a)):=\begin{cases}\dfrac{f(x)-f(a)}{x-y} &\text{if}~x\ne a\\ f'(a) &\text{else}\end{cases}$$

Secret

@SubhasisBiswas you cannot really have a homeomorphism between two different topologies which are not isomorphic?

Thorgott

Homeomorphisms are the isomorphisms of topological spaces, so that's tautologous, no?

Subhasis Biswas

@Thorgott i looked it up :p

topological equivalent of isomorphism

Secret

6:01 PM

@user170039 yeah, that makes sense

Thorgott

They are the isomorphisms in the category of topological spaces.

user131753

@Secret Although there is a typo. There should be no $Y$.

user131753

Further generalization of the definition can be done follows:

user131753

Let $X,Y$ and $Z$ be arbitrary topological spaces. Then a continuous function $f:X\to Y$ is said to be differentiable with respect to a function $g: Y^2\times X^2 \to Z$ at $a\in X$ with values in $Z$ if for any sequence $((f(x_n),f(a)), (x_n,a))\in Y^2\times X^2$ converging to $((f(a),f(a)), (a,a))\in Y^2\times X^2$ with $(x_n)_{n\in\mathbb{N}}$ converging to $a$, $g((f(x_n),f(a)), (x_n,a))\in Z$ converges to $g((f(a),f(a)), (a,a))\in Z$.

user131753

@SubhasisBiswas, @Secret, how about this one?

Thorgott

6:07 PM

Does this notion of differentiability have a corresponding notion of derivative?

Secret

@user170039 Very cool, I can make a further generalisation for you by using nets instead of sequences

It seems you have accomplished something I tried to do some weeks ago:

user131753

@Secret Yep. That was the next step. But you anticipated it.

Secret

(I work more naturally with nets because I love to work with crazy $\omega_1$ objects)

Subhasis Biswas

@user170039, I am just lurking. I know almost nothing about topology rn. I am trying to follow what's actually going on

user131753

@Thorgott Till now I have not thought about it. I will think about it in detail when I have more time later.

Secret

6:09 PM

May 10 at 7:58, by Secret

Generalised topological derivative: Let $T$ be a linearly ordered family of topologies (from the trivial topology to the discrete topology, parametrised by a real number $\lambda$) generated by a map $g : T \to T$ that is "continuous almost everywhere" in the following sense:

that is my failure a few weeks ago

in particular,

May 10 at 9:01, by Secret

In general, one can notice that analogous to the real number case, the linear approximation of a topology is the trivial topology plus sets that were added as we increase $\lambda$ and the change in number and type of sets added is assumed to depend only at the infintesimal difference of the topologies at the endpoint of interest

You definition avoided the need of a family of topologies that is linearly ordered

Subhasis Biswas

@Thorgott how about this. $\phi'(w)$ must maintain a strict positive/negative (either one) throughout $\mathbb{R}$ and in order to eliminate the possibility that $\lim_{w \to \infty} \phi'(w) =0$, we bound $\phi'(w)$ by a $k(<0)$, i.e. $\phi'(w) \leq k<0 \implies f'(w)-1 \leq k<0 \implies f'(w) \leq 1+k < 1 \implies f'(w) \leq r <1$

@Thorgott you mean a certain output function?

Thorgott

Why are you trying to show that $f^{\prime}(w)\le r<1$? Isn't that the hypothesis.

Subhasis Biswas

@Thorgott follow my previous arguments

@Thorgott my message after your message

Thorgott

Your argument is confusing me. The existence is what you are supposed to show, right?

Subhasis Biswas

@Thorgott oops. You are right. first comes the existence. then comes the uniqueness

right?

Thorgott

6:20 PM

Well, you can show that if a fixed point exists, then it is unique before showing that one does indeed exist, but if you don't make that explicit, it makes your argument harder to follow.

Subhasis Biswas

okay. Let me try

Tobias Kildetoft

6:51 PM

So I finally got the program to do what I wanted it to. And then just after making the pull request, I realized that I had slightly misunderstood what it was meant to do.

Alessandro Codenotti

How long is it going to take to fix this slight misunderstanding?

Also what kind of software are you working on if you can share?

Tobias Kildetoft

should not take more than half an hour I think

just misunderstood which parts of a cost calculation were supposed to count which parts

the overall system is one to be used by the branch of the Danish government that deals with forests and logging. It keeps track of all the pieces of forest and what sort of wood they contain and stuff about amounts and such

the part I am working on right now is functionality to automate generating an invoice based on a transport order

i.e. given a bunch of information about the wood that has been transported and by who, it has to put together the correct numbers to get the price of the transportation

Alessandro Codenotti

Oh, sounds cool actually

Tobias Kildetoft

quite straight forward stuff, it is just a matter of learning which parts of this huge system do what

The backbone of the system is a server running Dynamics CRM

which I have had to learn from scratch

Alessandro Codenotti

I have no idea what that is

Tobias Kildetoft

7:06 PM

CRM is Customer Relations Management, so basically a database of customers and other stuff with a ton of built-in functionality to handle whatever you might want to do with customers

Alessandro Codenotti

Except punching them in the face when they are annoying

ÍgjøgnumMeg

I fantasize about that alot

Tobias Kildetoft

But learning this is going to be useful for the main project I am going to be working on soon anyway, since that too is built on Dynamics CRM. That one is a system for a related branch of government which deals with hunters and hunting licences and everything to do with that

ÍgjøgnumMeg

Also hey @Alessandro and @Tobias

Alessandro Codenotti

Hi @ÍgjøgnumMeg

Secret

7:08 PM

database management is the most boring thing on this planet

ÍgjøgnumMeg

^

Although, it's nice when a database does what it's supposed to

Secret

true

ÍgjøgnumMeg

and it's satisfying to create one that aids people around you

but right, the actual process is boring

Secret

The proofreading is the most annoying part though

I have 3 data entry instance in my 3 past part time jobs

ÍgjøgnumMeg

I've just submitted a call at work to pull down all the information from all of our databases related to about 100 MAC addresses

Tobias Kildetoft

7:10 PM

For the next project, the most intriguing part is that we have been asked to figure out a way to introduce an electronic version of the hunting license, which is quite neat, but also has a ton of technical stuff that needs to be handled, since the hunting licence is also the licence to own and carry a shotgun

ÍgjøgnumMeg

with the intent of like.. chaining up and torturing the network to give us information about those 100 PCs

@Tobias that sounds very cool

@Tobias is the "direction" then to use some kind of near field communication to enable police to scan people?

Tobias Kildetoft

@ÍgjøgnumMeg Not sure yet. One of the main issues is to not assume that the person carrying it will have internet access.

So it needs to be verifiable without that and yet still secure. Probably some sort of signing will be used

ÍgjøgnumMeg

that's cool

hopefully I can sell-out in Crypto if my proposed academic career goes down the plug hole

Tobias Kildetoft

but all of those decisions will in the end be taken by someone with more experience (and who can take the blame :) )

ÍgjøgnumMeg

@Tobias if I'm doing something important at work I always make my boss sign off so I can't be blamed

hahaha

My neeew job will be in Geeermany

skullpatrol

7:30 PM

near Heidelberg? :P

Subhasis Biswas

Okay, I think I have discovered an elementary theorem :p

there is a good chance that it is wrong, tho

Theorem: A function $g:\mathbb{R} \to \mathbb{R}$ is differentiable on $\mathbb{R}$. Then $g(x)$ has one and only one real root when $g'(x) \leq k <0$. (not sure about the converse though).

Here is a proof

Leaky Nun

7:46 PM

@SubhasisBiswas sounds right to me

Subhasis Biswas

Being continuous, $g(x)$ cannot be both positive and negative on $\mathbb{R}$. So, firstly we assume that $g(x) <0$ for every $x \in \mathbb{R}$.

We take some $a>0$. (for convenience, take $a=1$). Now, $g(1)/1<0$ and $g(1)/k >0$. So, $\displaystyle\frac{g(-g(1)/k)-g(1)}{-g(1)/k-1} \leq k <0 \implies -k \leq \displaystyle\frac{g(-g(1)/k)-g(1)}{g(1)/k+1} \implies 0<-k \leq g(-g(1)/k) $ [Note that the denominator must be positive, since $-g(1)/k <1 \implies g(-g(1)/k)>g(1)$, because of $g'(x)<0$. Therefore, $g(x)>0$ for some $x \in \mathbb{R}$ . A contradiction.

[note that the *numerator must be positive

@Thorgott, now, we apply this theorem to our $\phi (x) = f(x)-x$.

@LeakyNun, check this out.

Thorgott

A continuous function can be both positive and negative on $\mathbb{R}$. I think you are trying to do a proof by contradiction, but in that case, please say so.

Leaky Nun

yeah you should say that you're doing a proof by contradiction

also "for convenience, take $a=1$" should read "WLOG assume $a=1$"

Subhasis Biswas

I agree.

Leaky Nun

that sounds more professional

Subhasis Biswas

7:54 PM

@LeakyNun good note

Leaky Nun

I'm afraid your proof is a little bit too complicated

Subhasis Biswas

Now, that we have ensured that $g(x)$ is indeed $0$ for some $x$, the strict-monotonicity [$g'(x)<0$] of the function guarantees that $g(x)$ is one-to-one.

@LeakyNun not much. The entire point of the proof is to find out a point where $g(x)$ is sure to be positive

I am learning so much from only one question. The initial task was to show that if $f'(x) \leq r <1$, then it has a unique fixed point. [$f : \mathbb{R} \to \mathbb{R}$, differentiable on its domain of definition]

that question led me to this question.

Leaky Nun

surely it is enough to "integrate both sides of the inequality" to give you $g(x) - g(0) \le kx$ for $x \ge 0$ and $g(x) - g(0) \ge kx$ for $x \le 0$; so if $g(0) > 0$ then for $x \gg 0$ we have $g(x) \le g(0) + kx \ll 0$, and if $g(0) < 0$ then for $x \ll 0$ we have $g(x) \ge g(0) + kx \gg 0$

making my sketch rigorous is left as an exercise

Thorgott

8:11 PM

Another alternative is to note that if $g$ had no zeroes, it would be completely negative of completely positive, but as it is strictly decreasing, it would have a horizontal asymptote, which means the derivative would get arbitrary close to zero; contradiction.

It's a rather geometric argument, but, of course, can be turned into a rigorous proof.

Leaky Nun

@Thorgott surely g -> 0 doesn't mean g' -> 0

Subhasis Biswas

@LeakyNun, take $g \to \pm \infty$

Thorgott

I'm not sure I understand what you're getting at

Leaky Nun

@Thorgott the fact that existence of a horizontal asymptote doesn't mean that the derivative tends to zero; but I might have misunderstood "get arbitrary close to"

Thorgott

Oh yeah, I was being imprecise. However, there is a sequence $\xi_n\rightarrow\pm\infty$ such that $g^{\prime}(\xi_n)\rightarrow0$, which suffices for a contradiction.

I'm curious now though. What's an example of a differentiable function with horizontal asymptote whose derivative does not tend to zero?

Leaky Nun

8:21 PM

@Thorgott desmos.com/calculator/asw895mxap

Thorgott

Oh god. I figured it would be one of those $\sin$ examples with crazy oscillatory behavior. Thanks.

Subhasis Biswas

@Thorgott i meant $x \to \infty$

Okay. You guys are going through my work. Am I improving?

any suggestions ?

Thorgott

I'm not sure I follow the part where you switch from the fraction to a normal term.

Ted Shifrin

8:42 PM

General comment re asymptotes. Many people are under the misimpression that a graph cannot cross its asymptote. Very wrong. in fact, $f(x) = (\sin x)/x$ has the $x$-axis as a horizontal asymptote, and it crosses it infinitely often. One need not "approach without touching," although people think so.

@Thorgott: Try $\dfrac{\sin(x^3)}x$.

Alessandro Codenotti

Hi @Ted

Ted Shifrin

Hi, demonic @Alessandro. How you doing?

ÍgjøgnumMeg

Hiya @Ted

Alessandro Codenotti

Quite well thanks

Ted Shifrin

Ausgezeichnet!

hi @ÍgjøgnumMeg

ÍgjøgnumMeg

8:45 PM

Grüß dich :)

Thorgott

Always the sine function ruining your intuitions

Ted Shifrin

Only if it's bad intuitions to start with :P

Rapid oscillations should be part of your intuition :)

Thorgott

I should probably assimilate them into my intuition. They do always make for great counter-examples.

Leaky Nun

@Ted bonsoir

Ted Shifrin

It's middle of the afternoon here, Leaky :)

Leaky Nun

8:53 PM

@AlessandroCodenotti I just finished defining the counting measure in Lean

@Ted le temps passe lentement pour toi

Subhasis Biswas

'not necessary that the curve does not cross the asymptote': because one is concerned at infinity large value of $x$. What happens in the finite region should not be of concern

@AlessandroCodenotti, would that be a valid interpretation ?

Thorgott

Well, in Ted's example, there are arbitrarily large values for which the function crosses the asymptote, so it's not really a "finite region".

Subhasis Biswas

@Thorgott ummm... well the "margin of oscillation" (by that, I mean the amount by which the curve crosses the asymptote at a time, sort of like amplitude ) $\to 0$?

Thorgott

9:09 PM

Yes, that is true.

pre-kidney

9:20 PM

Hi Guys, so I worked hard on answering a bounty only to be met by radio silence afterwards. Literally no other solutions, my solution appears to be correct and well-explained, yet no response whatsoever. I heard that you need at least 2 votes for the bounty to be auto-awarded, and the grace period ends in 5 hours.

Posting the link here in case it gets more eyeballs here: math.stackexchange.com/questions/3230195/…

It's a nice problem, it is a shame that it ended up getting so little response. Any mods here know if there is a way to bump a post show it shows up in people's queues? Seems like this one fell through the cracks somehow

@LeakyNun any tips for what can be done in this situation?

user193319

Okay. The complex projective space $\Bbb{CP}^n$ can be defined either as the quotient space $\Bbb{C}^{n+1} \setminus \{0\}/ \sim$, where $\sim$ equivalence relation on $\Bbb{C}^{n+1} \setminus \{0\}$ which identifies vectors which differ by a complex scalar multiple; or it can be defined as the orbit space of $\Bbb{C}^\times$ acting on $\Bbb{C}^{n+1} \setminus \{0\}$ by scalar multiplication. With the former perspective, it's clear how to topologize $\Bbb{CP}^n$: give it the quotient topology.

But what about the latter perspective? Is there a natural way of topologizing the orbit space when a topological group is acting on a topological space?

Whatever this topology is, it should coincide with the quotient topology.

pre-kidney

Yes, see ncatlab.org/nlab/show/properly+discontinuous+action for example

user193319

Ah, so the quotient map is actually a covering map.

pre-kidney

By the way @user193319 if you can take a peek at the link I posted, I am one upvote away...

user193319

Looks like a good answer. You have my (up)vote.

pre-kidney

9:36 PM

Thank you, good sir or madam