Mathematics - 2017-02-03 (page 1 of 3)

Akiva Weinberger

00:48

Cool: $\displaystyle\lim_{x\to\infty}\frac{ae^{ax}+be^{bx}+ce^{cx}}{e^{ax}+e^{bx}+e^{c‌x}}=\max\{a,b,c\}$, and the $\lim_{x\to-\infty}$ is the minimum.

Samuel Yusim

01:34

so my first paper got published the other day

I am pretty hype about this

Akiva Weinberger

01:45

@SamuelYusim What's it about?

Congrats

Samuel Yusim

@Akiva some graph invariants called the metric dimension and the strong metric dimension

we proved some stuff about graphs where they're close together and graphs where they're far apart

Adeek

hey @AkivaWeinberger

Donnie

hello people

I need help

x = 2 + cos t, y = 3 + sin t, 0 ≤ t ≤ 5π/2

I must find the starting point of this

I have an idea how to do it with multiplication but the addition is throwing me off. Are there any sources out there where I may learn this?

Akiva Weinberger

02:02

Are you trying to see what shape the parametrization gives? Or what?

What's the question?

Balarka Sen

02:15

@SamuelYusim Congratulations!

Samuel Yusim

@BalarkaSen thanks!

Donnie

yes I want to see the shape and all its points

Balarka Sen

@Akiva 's kinda obvious tho. if WLOG $a$ is the largest of the three that's the term which will dominate. so proof by handwaving, limit better be $ae^{ax}/e^{ax} = a$.

but it's easy to make it non-handwavy.

Secret

If you want the equation of the plot, you will want to eliminate t by moving the constant to the other side, square the equation so you can use the trigonometric identity $\sin^2 t+ \cos^2 t =1$ to add them

Alternately, you can feed these parametic equations for the range of t given to find the x,y coordinates and then plot these

Akiva Weinberger

@BalarkaSen Right, yeah, it's trivial to see that it works. I was just sharing because I hadn't seen it before.

Balarka Sen

02:19

Fair enough

Secret

You equation appears to be a circle. I will let you figure out where its centre is

euclid

@arctictern Hi tern

arctic tern

hello

Donnie

anyone?

euclid

@arctictern i dont have any question just wanted to say hi to you.our room is frozen.

Secret

02:25

@Donnie Try this source for some examples on how to plot parametric equations

arctic tern

@euclid cool.

Balarka Sen

@Donnie Secret already gave you an answer

arctic tern

(cos t, sin t) parametrizes the unit circle centered at the origin (0,0) (counterclockwise starting from (1,0)), so (3+cos t,3+sin t) parametrizes the unit circle centered at (3,3) (counterclockwise starting from (4,3)).

euclid

@arctictern i hope see you later. take care

arctic tern

later

Secret

02:45

[More infinite set questions] Since rationals are dense in the reals, that is there always exists a rational z such that x < z <y given x < y, is this the same as one rational is separated infintesimally from the next one along the line

thus unlike the integers, each integer differs from the next one by 1, which is not an infintesimal amount?

Another related question: What is the cardinality of an infintesimal interval or a open interval of infintesimal length?

Ramanujan

$\cos x ≈ 1- \dfrac{x^2}{2}$ what it means?

Secret

@Ramanujan Hint: what is the taylor series of cos x?

Ramanujan

Don't know :/

Secret

have you learned taylor series before?

Ramanujan

No,so when is it useful?

Secret

02:57

What you have there is a 2nd order approximation of cos x obtained from truncating the taylor series of cos to $x^2$. That is useful in physics and engineering provided your x is small enough

arctic tern

@Ramanujan \approx means "approximately"

e.g. $\pi\approx3.14$

Secret

taylor series are a type of power series $\sum_{i=1}^n a_ix^i$ which capture the behaviour of some nice functions. It can be used to simplify integration as integrating powers of x is very easy

Our calculators actually use taylor series to calculate trig functions

Ramanujan

Interesting

Secret

You will came across them eventually at some point in your physics, maths and engineering courses

Ramanujan

Sorry

Akiva Weinberger

03:14

@Ramanujan It essentially means that they're near each other when $x$ is near zero. To quantify just how near, notice that:$$\lim_{x\to0}\frac{\cos(x)-(1-x^2/2)}{x^3}=0$$

So the distance between them goes to zero faster than $x^3$ does.

In fact, of all cubics, the function $1-\frac{x^2}2$ is the closest to the function $\cos(x)$ near $x=0$.

(Yes, cubics; the $x^3$ coefficient is $0$.)

TheGreatDuck

@AkivaWeinberger any good at weird floor sums?

0

Q: Does anyone know how to reduce this sum of sums into something simpler in order to find a special value?

I was given this from a friend. They asked me to deduce what the equation is of. I played around with trying to compute alpha for some time. Plugging it into f(x), the function appeared to equal $0$ almost everywhere. I could never find the actual value of alpha. It appears to be an infinite irra...

Akiva Weinberger

I don't know if I want to try that

It looks annoying, and the hour is late

TheGreatDuck

fair enough

Secret

floor functions have some identities you can use, and also the notion of fractional part. Maybe that wil help knock down those $\lfloor \frac{n}{m}\rfloor-\lfloor \frac{n-1}{m}\rfloor$

also do you have an image of the plot of f(x). You said it is zero almost everywhere, is it continuous?

TheGreatDuck

03:31

@Secret I'd have to calculate alpha

note that alpha is an infinite decimal

and trivially speaking, I doubt that function is continuous

this is all assuming a crude 2-3 digit approximation of alpha is even useful here.

TheGreatDuck

03:52

@Secret how do I compute values of a function containing an infinite sum?

Secret

That I have no idea. I am not terribly familar with numerical computations of infnite summations

Especially that infinite sum is so messy, it might not have a closed form (not even as an integral representation) at all

Secret

04:09

7

Q: Distinguishing properties of $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ that lead to differing cardinalities?

I have what many on here would consider an elementary question, but I would very much appreciate responses that use only elementary ideas, if possible, so that I can understand them. I would also appreciate detailed rather than brief responses. By construction, $\mathbb{Q} \subseteq \mathbb{R}$...

real-numbers

0

A: Division of segments into infinitely many parts.

In modern mathematics we have several ways of formalizing infinity. The one that is most relevant to your question was provided by Abraham Robinson; see here. Following his framework, there are both infinitesimals and infinite numbers. Thus if $H$ denotes an infinite number, one can indeed div...

Secret

04:49

But if $f(x)-M=0$, then $g(x)=\frac{1}{f(x)-M}$, which is also absurd??

arctic tern

@Secret huh?

Secret

If M is a value of f, then there exists a x such that $f(x)-M=0$

Then the $g(x)$ is clearly undefined for that x

---
http://math.stackexchange.com/questions/4202/induction-on-real-numbers

Real induction is quite interesting. It sometimes helps to think in terms of intervals so that we will be working in the continuum without referencing to any countable points

SAWblade

05:17

hey all

Secret

0

Q: Order preserving bijection between the integers and the rationals

Is there a bijecttion $f:\mathbb{Z}\rightarrow\mathbb{Q}$ s.t $f(x) < f(y)$ iff $x < y$

elementary-set-theory functions

JessyunBourne

05:37

0

Q: Want to prove $Aut(A_{n})\simeq GL(n, \mathbb{Z})$ where $A_{n}$ is a free abelian group of rank $n$

I'm trying to prove that $Aut(A_{n}) \simeq GL(n, \mathbb{Z})$, where $A_{n}$ is a free abelian group of finite rank $n$. I have already available to me the result that $A_{n} \simeq \mathbb{Z}^{n}$, so essentially what I have to prove is that $Aut(\mathbb{Z}^{n}) \simeq GL(n, \mathbb{Z})$. To ...

abstract-algebra group-theory group-isomorphism free-groups free-abelian-group

Some folks calls it a @SAWblade, I calls it a Sling Blade, mmm hmmm.

SAWblade

:P

SAW stands for self-avoiding walk

JessyunBourne

in Sling Blade voice "All right, then! Mmm hmmm"

SAWblade

i dont get the reference

JessyunBourne

Sling Blade

Sling Blade is a 1996 American drama film set in rural Arkansas, written and directed by Billy Bob Thornton, who also stars in the lead role. It tells the story of a man named Karl Childers who has a developmental disability and is released from a psychiatric hospital, where he has lived since killing his mother and her lover when he was 12 years old, and the friendship he develops with a young boy and his mother. In addition to Thornton, it stars Dwight Yoakam, J. T. Walsh, John Ritter, Lucas Black, Natalie Canerday, James Hampton, and Robert Duvall. The movie was adapted by Thornton from his...

You have to see this movie before you die.

SAWblade

i have no interest in something that takes place in arkansas

JessyunBourne

05:40

LMAO!

It's actually really good!

@arctictern dunno if you're around or anything. But that question I just posted a link to. You probably know everything there is to know about free abelian groups and you're awesome.

arctic tern

buttering me up huh? :)

r9m

05:57

Why would anyone butter up an arctic tern .. only to roast it! :P

Secret

06:09

[Random abstract algebra]
$\exists m,n, ma=na=an=am=a \text{ and }nm=mn=1$
$\forall w, s(w)=w+a$

Lucas

Hi ! Does someone have an idea about math.stackexchange.com/questions/2124516/… ( Transpose = Change of basis ? )

DHMO

@Lucas what is K?

arctic tern

field

Lucas

06:25

Yes. Actually F seems more appropriate.

Secret

unless you are talking about number fields, commonly denoted K

Lucas

Ok. Thank you artic tern :)

DHMO

07:09

$\{S \subset \Bbb N | \forall s \in S: \forall a < s: a \in S\}$ has a cardinality of $\Bbb N$, but can have a cardinality of $2^\Bbb N$ if you change the ordering...

Alessandro Codenotti

07:26

Not sure what that means

DHMO

@AlessandroCodenotti Dedekind cuts

Alessandro Codenotti

I think it's reasonable to expect some properties of downward closed sets to be modified when changing the order since that's the only thing they depend on

DHMO

@AlessandroCodenotti I know I've been asking this for days now, but I still can't get through how a chain of cardinality $2^\Bbb N$ is possible...

Alessandro Codenotti

Did you see a solution and don't understand it or are you still looking for one?

DHMO

@AlessandroCodenotti I saw a solution and understood it.

I just can't intuitively understand how a chain under $\Bbb N$ can have cardinality $2^\Bbb N$.

which is equivalent to how there are $2^\Bbb N$ possible cuttings of $\Bbb Q$.

Secret

07:36

I had the same problem and we are currently discussing about it in the set theory room
The dedekind cuts in Q is quite visual as the subsets are all of the form (a,r) for reals a,r, but the subsets of N is not that easy

Alessandro Codenotti

07:50

Most of this cuts cannot be represented by a finite string of operations on $\Bbb Q$

And I don't think you can prove in a very straightforward manner that $|\Bbb R|=2^{\aleph_0}$ via Dedekind cuts, you use them to construct $\Bbb R$ and then show that it bijects with $P(\Bbb N)$

Secret

Thus for any $x$ and $x'=x-\epsilon$ for any $\epsilon > 0$ $S_x'$ is missing countable number of elements than $S_x$. There is no problem with this as there are only countable disjoint open intervals in reals

Alessandro Codenotti

You can't define $S$ like that @secret, unless $x$ is rational there is no $x$ yet

DHMO

which leaves open the question "how to define $\pi$ in terms of Dedekind cut?".

Secret

@AlessandroCodenotti I thought we can put the cuts on the real line and select the rational subsets from there...

DHMO

@Secret the cuts define the real line. The real line does not exist before you cut.

Alessandro Codenotti

07:57

Which real line? The whole point of Dedekind cuts is to construct $\Bbb R$

Secret

O great, now I have no idea what those subsets look like...

DHMO

@Secret $\{S \subsetneq \Bbb Q | \forall s \in S: \forall a <_\Bbb Q s: a \in S\}$

Alessandro Codenotti

A cut is a partition of $\Bbb Q$ into $2$ disjoint subsets $A$ and $B$ such that $a<b$ for all $a\in A$, $b\in B$ and $A$ has no greatest element

Secret

@AlessandroCodenotti (x,x,x,x,x,...| ...,x,x,x,) something like this (the left bracket is A and clearly there is no maximum, similarly B has no minimum) ?

DHMO

Another question is to define a well-ordering on $\mathcal P(\Bbb N)$...

SteamyRoot

08:00

You cut $\mathbb{Q}$ in two pieces, but the "boundary" between the pieces turns out not to be a rational number.

DHMO

@SteamyRoot the boundary can be rational

@Secret B can have a minimum

SteamyRoot

It can be, but those are the boring ones.

DHMO

@SteamyRoot that depends on your definition of "boring"

Secret

(... x,x,x,x,x, ....| x,x,x,x,x,x, ...) ?

DHMO

@Secret B may not have a minimum

SteamyRoot

08:01

Well, they're useless for constructing the real numbers because you already have $\mathbb{Q}$ in the first place.

Secret

@DHMO @AlessandroCodenotti Argh...
(... x,x,x,x,x, ....| x,x,x,x,x,x, ...) or (... x,x,x,x,x, ....|.... x,x,x,x,x,x, ...) ?

Then one get that continuum at the cuts because at .... all those rationals are bunching up as they converge to the cut?

Alessandro Codenotti

@SteamyRoot I'd like my reals to contain a copy of $\Bbb Q$ :P

DHMO

@Secret no idea what "bunching up" means

not sure why that implies continuum

Secret

At least one sequence at .... converges to whatever number that is at the cut, i.e. .... will be a cauchuy sequence since A is bounded above by the cut

SteamyRoot

@Alessandro You already start with $\mathbb{Q}$ though. You could just extend it with the "irrational" Dedekind cuts; rather than reconstructing $\mathbb{Q}$ all over with the "rational" cuts.

DHMO

08:04

@Secret some sequences do. some don't.

@SteamyRoot that would make operations difficult to perform.

For example, we say that $A \le B \iff A \subseteq B$.

Alessandro Codenotti

Yeah but then it's just a pain when I have to treat rationals and irrationals differently when defining the operations and the order on $\Bbb R$ (and you surely remember that it is painful enough already if you went through the whole construction in detail)

Secret

@DHMO Ok I guess that kinda clarified it. I used to have a misconception that countable means they are uniformly spaced apart, just like the integers

If there exist at least one sequence that can bunch up like this, then there is no problem seeing why we cannot get a continuum from the cuts despite the set being countable

SteamyRoot

Well, I guess it makes things more annoying :P

Secret

So similarly for the naturals with inverse lexicographic order, you have in general a nonuniformly distributed countably many integers in each subset in the chain. This provide the extra degrees of freedom to allow continuum many cuts, if this logic holds...

Another question: We knew that rationals are dense in the reals. Does that mean a given rational q is infinitesimally separated from the next one?

DHMO

@Secret No. That means there's no "next one".

> In topology and related areas of mathematics, a subset A of a topological space X is called dense (in X) if every point x in X either belongs to A or is a limit point of A.

> In mathematics, a limit point of a set S in a topological space X is a point x (which is in X, but not necessarily in S) that can be "approximated" by points of S in the sense that every neighbourhood of x with respect to the topology on X also contains a point of S other than x itself.

> Informally, [if A is dense in X, then] for every point in X, the point is either in A or arbitrarily "close" to a member of A

Secret

08:16

@DHMO (If I understood correctly based on dedekind cuts, not all limit points are rationals) You said the rationals are not infintesimally separated, then how far away is the next nearest rational from a given one, or is that nonmeasurable?

DHMO

@Secret That means there's no "next one".

Secret

Wait, rationals are countable, but there is no next one??????

DHMO

@Secret you're being silly. what's the next rational number after 1?

Secret

I thought there exists a 1+1/M for some sufficiently large M, such that even we cannot write that down explciitly, it still exists

DHMO

@Secret what about 1+1/(M+1)?

Secret

08:20

because this M is sufficiently large, M+anything=M? (does not seemed like a good fix)

but then that will mean it is infintesimal, a contradiction

DHMO

if M = M+1 then M isn't an integer.

and I'm not willing to go to the hyperreals

Secret

Ok so that's what I get wrong about rationals: Thinking they have some well defined spacing and there's a next one

DHMO

They're definitely counter-intuitive.

The rationals do not form an open subset of $\Bbb R$.

The rationals are not discrete: there's no open set containing a rational that does not contain another.

However, the integers are discrete: there's plenty of open sets containing an integer that does not contain another.

Secret

The rationals are not discrete someone please star this

8

DHMO

Nor are the rationals closed...

Though the integers are definitely closed.

Secret

08:24

So they are both countable sets, but with very different properties...

DHMO

The rationals are the limit points of the irrationals, and vice versa.

@Secret indeed, drastically different.

Secret

Now I see how the p adics are the inverted version of them

Now that that is sorted. The next question will be: Why we cannot get chains longer than $\mathfrak{c}$ from countable sets. Is it because there are only $\mathfrak{c}$ subsets?

or some more counter intuitive reason?

DHMO

@Secret Each chain contains distinct elements of $\mathcal P(\Bbb N)$.

therefore, they can be injected into $\mathcal P(\Bbb N)$.

Therefore, their cardinality must be at most $\mathcal P(\Bbb N)$.

Secret

ah yes, you cannot inject a larger cardinality into a smaller one

DHMO

Now, my next question: assume that $\mathfrak c = \aleph_2$. Can I construct a chain of length $\aleph_1$?

Secret

08:29

why not, there are enough subsets to do the job

DHMO

I mean, is there a constructive proof?

Secret

and you can always dedekind cut at countable number of points to produce $\aleph_1$ number of them, I guess...?

DHMO

@AlessandroCodenotti ^

Secret

Another more meta question: Suppose I have the set of all cardinals (which forms a transfinite sequence hence countable), and I perform a dedekind cut on it, can I get a chain of length $\mathfrak{c}$?

DHMO

@Secret what the hell is "the set of all cardinals"?

Secret

08:36

A set that contains all these things

this set is countable since in previous discussion we cannot have a continuum

DHMO

@Secret "The aleph numbers are indexed by ordinal numbers."

The ordinal numbers do not form a set.

Alessandro Codenotti

@DHMO construct a chain of length $\aleph_2$ and extract a subchain of length $\aleph_1$

DHMO

@AlessandroCodenotti can it always be done?

Alessandro Codenotti

We already constructed a chain of length $\mathfrak{c}$ in $P(\Bbb N)$/

Secret

I think we can turn that into an induction proof there and it will hold for all $\aleph_{\alpha}$

where the base case is $\aleph_0$ and $\mathfrak{c}$

DHMO

08:43

@Secret transfinite inductions is more troublesome than it seems

Secret

it can be done as long there is some notion of partial order, if I recall, and even better if there is a well order

Secret

Also I suspect the following might be what some of the subsets in $\mathbb{N}$ will look like under inversed lexicographic order

There should be no problem for naturals as they are discrete

DHMO

@Secret I don't understand the left part of your picture

@Secret they are no longer discrete if we impose a metric based on the new ordering

Secret

But if we plot them on the line of integers in the usual way (not saying that we don't use the new order), then we should expect those dedekind cuts subsets that are closer to zero will have missing integers being very spaced out

given how the dedekind cut of zero is the whole set in this new ordering

because there will be countably many numerals with many digits that are between 0 and 1 compared to 20 and 400

where you no longer can get single digit numerals between them, thus they will look more spaced out in the usual way of plotting them

Kane

09:05

$P(X=k)= \frac{\theta ^{k}(1-\theta)}{1-\theta ^{2n+1}}$
$0\leq k\leq2n$
$0< \theta <1$

Can anyone help me find out the probability that X is odd?

DHMO

@Kane geometric series

Kane

@DHMO hmmm. Cool, how can you tell it's a geometric series?

DHMO

@Kane the ratio is constant

Kane

@DHMO thanks I'll look up the geometric series. I've never really used it before

DHMO

@Secret indeed

Secret

09:14

Also that spacing can be nonuniform (consider a subset that contains 1,45,946,947,...) so you have that extra degree of freedom to get the continuum since every entry in the set has a chance to be in a subset or not (essentially mikeonly's argument) giving us the $2^{\aleph_0}$ number of choices of subsets to get $\mathfrak{c}$ in total

If the spacing from one subset to the next can only be uniform, then this will be impossible as the spacing will then be controlled by the naturals, of which there are onyl finite many of them

A similar argument applies to the rationals, where the nonuniformity can be illustrated by picking some sequence that converge to a limit point. A convergent sequence clearly cannot have uniform spacing (to be checked...)

So in conclusion, nonuniformity of the partitioning for each subset and its complement, and the fact there are countable number of elements in the original set, provide the degrees of freedom to reach countinuity

Claim: Therefore the reason why this argument fails for finite sets is because we only have finite number of elements and $2^n$ finite of choices for each element to be in the subset or not, hence the final number of finite subsets must be finite

correction: of which there are onyl countably infnite many of them

(Hopefully) final question for now: Since $\mathbb{R}$ is dedekind and cauchy complete, thus every dedekind cut and cauchy sequence will converge to some reals (thus no "holes") is it even possible to extend the reals further into a set of cardinality $2^{\mathfrak{c}}$?

DHMO

09:29

@Secret do you have any idea?

Secret

Naively, it will seems I can always made cuts at any point in reals. The issue is, however that the resulting bounded subsets will always converge to a real, thus there is a maximum and minimum for each partition

We can however, always produce a $2^{\mathfrak{c}}$ chain this way, and it will have the ordering of the reals

DHMO

@Secret The chain is still $\mathfrak c$

Secret

8

Q: totally ordered chain in the powerset with big cardinality

Let $B$ be some set. The problem is to find a set $A\subset\mathcal{P}(B)$ of subsets of $B$ which is totally ordered by inclusion and such that there exists a bijection $A\leftrightarrow \mathcal{P}(B)$. This is an easy exercise if $B$ is countable where one can explicitly construct such a set ...

set-theory

You can get a chain of $2^{\mathfrak{c}}$. but maybe not with dedekind cuts.

Alessandro Codenotti

Funny problem to think about: suppose you have a family of countable sets with the finite intersection property, can this family be uncountable?

DHMO

@AlessandroCodenotti Yes. The chain basically does that.

Alessandro Codenotti

09:38

The elements of the chain don't have pairwise finite intersection

DHMO

@AlessandroCodenotti what is the finite intersection property?

Alessandro Codenotti

My bad, I didn't mean the finite intersection property

Ok let me rephrase that correctly: suppose that you have a family of countable sets such that the intersection of any 2 sets in this family is finite, can such a family be uncountable?

(A family of sets has the finite intersection property if they have pairwise nonempty intersection, I misremembered, that's not relevant here)

Secret

NB. Attempt to naively dedekind cut on the reals will produce the surreals, which sadly is no bigger than the reals

Kane

@DHMO so is this the ratio:
$\frac{1-\theta}{1-\theta ^{2n+1}} = \frac{1}{1-\theta ^{2n}}$

DHMO

@Kane i don't know, verify it

Alessandro Codenotti

09:48

@Secret the surreal numbers are so big they form a proper class rather than a set, I'd stay away from them

Secret

ok

@AlessandroCodenotti Btw we know from the MO link that construction of chains $2^{\mathfrak{c}}$ is possible from a size $\mathfrak{c}$ set. So naively we should be able to index the chain of subset that formed with some index set of size $2^\mathfrak{c}$. All the proof uses dedekind cuts, but reals are already complete, so does that mean this new set cannot have the reals embed in it?

user243301

I would like to refer this, if some user want to read and study this nice reference this weekend. The author was a specialist in Number Theory and Combinatorics studying and solving several problems. This is an example that you can read from ScienceDirect: Cilleruelo, Squares in $(1^2+1)\cdots(n^2+1)$, Journal of Number Theory, Volume 128 Issue 8, 2008. Good weekend all users.

Kane

@DHMO yeah I think it is. Then Im left with $\theta ^{k}$

And apparently I can get this: $\frac{a}{1-r}$

user8469759

Hello guys

Can you suggest me any good reference for non elementary functions?

Kane

The answer to the question is: $\frac{\theta(1-\theta ^{2n})}{(1-\theta ^{2n+1})(1+\theta)}$
I'm not really seeing how they got there

user8469759

09:58

Like bessel functions, ellipitc integrals etc...

DHMO

@user8469759 mathworld

20 mins ago, by Alessandro Codenotti

Ok let me rephrase that correctly: suppose that you have a family of countable sets such that the intersection of any 2 sets in this family is finite, can such a family be uncountable?

@Secret ^

user8469759

I was thinking something more academic maybe

that would allow me to study these functions, with a minimal list of what are such functions

because I'm not really familiar with them

(i was aware of wolfram, but I need something for newbies)

DHMO

@user8469759 a list is here:

owlcation.com/stem/…

user8469759

is there no book about it?

DHMO

well I don't read books so you might want to consult others

Secret

10:05

@DHMO that's a different question from mine? (but interesting nevertheless

DHMO

10:25

@Secret @AlessandroCodenotti I finally have an answer to why it is possible to construct a chain of $\mathfrak c$: because it isn't a chain.

it's just a family of sets totally ordered by inclusion

Secret

isn't a chain just a total order?

DHMO

@Secret it isn't a "chain" in terms of "arranged one after the other"

Secret

You can always rearrange the elements in a set, a set is not a tuple

DHMO

@Secret what do you mean?

Secret

{1,2,3,..}={2,3,1,...} etc.

DHMO

10:30

so?

Secret

elements in a set are not ordered by position

so you can always rearrange them so that they follow the order that is imposed on them

and thus the subsets will be a chain under inclusion

Alessandro Codenotti

@DHMO that's the definition of a chain (when the partial order is given by inclusion)

DHMO

@AlessandroCodenotti I know. What I mean is that it is not a literal chain.

Kane

$P(X=k)=\frac{\theta ^{k}(1-\theta)}{1-\theta ^{2n+1}}$
$0\leq k \leq2n$

$P(X=odd)=P(X=1)+P(X=3)+...+P(X=2n-1)$
$P(X=odd)=\frac{\theta(1-\theta)}{1-\theta ^{2n+1}}+\frac{\theta ^{3}(1-\theta)}{1-\theta ^{2n+1}}...+\frac{\theta^{2n-1}(1-\theta)}{1-\theta ^{2n+1}}$
$P(X=odd)=\frac{(1-\theta)}{1-\theta ^{2n+1}}(\theta+\theta ^{3}+\theta ^{5}+...+\theta ^{2n-1})$

Can anybody see any mistakes I'm making?

not really sure how to continue from here

The term in the brackets is a geometric series right?

SteamyRoot

10:59

Yup

$\theta + \theta^3 + \cdots + \theta^{2n-1} = \theta \frac{\theta^{2n} - 1}{\theta^2-1}$

Kane

@SteamyRoot oh daym, how you get that?

SteamyRoot

Ummm... because, like you said, it's a geometric series

Kane

but doesn't the geometric series go down to $\frac{a}{1-r}$?

SteamyRoot

If it's infinite, yes.

Kane

ahhhhhh, i see

@SteamyRoot thanks steamyroot, legend

Vrouvrou

11:50

hello

someone have an idea about this : math.stackexchange.com/questions/2126200/…

DHMO

2 hours ago, by Alessandro Codenotti

Ok let me rephrase that correctly: suppose that you have a family of countable sets such that the intersection of any 2 sets in this family is finite, can such a family be uncountable?

@AlessandroCodenotti any idea?

Secret

12:17

Do you mean $\aleph_0^2$ cause $\omega^2\neq \omega$ as an ordinal?

Otherwise, the easiest way is to e.g. partition $\mathbb{N}$ into $\aleph_0$ parts by $\mathbb{N}-p\mathbb{N}$ for each prime $p\in\mathbb{N}$. It is then easy to see that there are $\aleph_0$, well ordered subsets of size $\aleph_0$. Hence there are total of $\aleph_0^2=\aleph_0$ elements by cardinal arithmetic

Alessandro Codenotti

12:43

@DHMO I know the answer

DHMO

@AlessandroCodenotti I guess math overflow has the answer

6

Q: Uncountable family of infinite subsets with pairwise finite intersections

I am searching for a constructive proof of the following fact: If $X$ is an infinite set, there exists an uncountable family $(X_\alpha)_{\alpha \in A}$ of infinite subsets of $X$ such that $X_\alpha \cap X_\beta$ is finite whenever $\alpha \neq \beta$. The way I know how to prove this statement...

set-theory lo.logic

@Secret nice...

Alessandro Codenotti

Yeah that works

What if instead of finite intersection I ask for every intersection to be bounded in cardinality by a fixed $n\in \Bbb N$?

SteamyRoot

@Secret The sets $\{\mathbb{N}-p\mathbb{N} \mid p \text{ prime}\}$ don't form a partition, do they?

Ramanujan

$$\binom {9}3=\boxed {84}$$

What kind of representation is it?

Secret

NB: The logic behind the above $\aleph_0^=\aleph_0$ proof (plus my initially wrong conception that rationals are discrete is the reason why 2 days ago when Paul give us the uncountable chain question, I cannot see there is an uncountable chain

@SteamyRoot They each form a partition with their complements, thus each of these sets partition the nautrals into two sets

DHMO

12:49

@AlessandroCodenotti I don't really understand that; can you explain that to me?

SteamyRoot

@Ramanujan what do you mean with "what kind of representation"?

Secret

@Ramanujan what is the operation here, 9 choose 3?

Ramanujan

How $\binom {9}3=84$?

DHMO

@Ramanujan binomial coefficient

SteamyRoot

Because $9! / (3! 6!) = 84$ ?

Ramanujan

12:51

@SteamyRoot why 6! ?

DHMO

@Ramanujan go study the definition of binomial coefficient before you ask

SteamyRoot

@Secret Right, but that means you have countably many partitions of $\mathbb{N}$ in $2$ parts.

DHMO

@Secret can you demonstrate your partition concretely?

Secret

Lets say we pick: $\{\mathbb{N}-2\mathbb{N} \}$. This give us the odd numbers. It and the even numbers partition the naturals

and they are both countable

Another example subset is $\{\mathbb{N}-3\mathbb{N}\}$. Thus its complement, the integers not divisble by 3, together with it forms a partition of the naturals

Since there are countably many primes, there are countably many such subsets

so the collection $\{\{\mathbb{N}-p\mathbb{N}|\text{ p prime }\}\}$ has a total of $\aleph_0^2=\aleph_0$ elements

(actually Iam thinking about elements in each set in the collection, I am not sure how to phrase that properly)

Akiva Weinberger

Its disjoint union has $\aleph_0^2$ elements

Secret

12:59

Note, however, not all sets in this collection are pairwise disjoint. For example the set $10\mathbb{N}$ can be found in the subset p=2 and p=5

Akiva Weinberger

Right. That's why the disjoint union is different from the union.

(Disjoint union of $\{A_n\}$, I believe, is $\bigcup\left(A_n\times\{n\}\right)$, up to bijections.)

Secret

Actually, I do find it surprising that the box counting method that we learn in high school when we were taught permutations and combinations work quite well up into the infinite cardinals

All my current intuition about infinite sets rely on box counting of some form, updated with the weird rules of cardinals

SteamyRoot

If you want to really "partition" $\mathbb{N}$ into countably many countable subsets, you could've used the prime powers.

Akiva Weinberger

Or bijected with $\Bbb N^2$ and used the rows?

Secret

I tend to prefer direct constructions, because that tell us what those subsets look like and how the elements are distributed

the bijection helps doing the proofs in an easier way, but at the same time, it helps me to work out what the direct construction look like

SteamyRoot

13:07

Once you go past countable, you'll end up in plenty of situations where you can't really give meaning to what something "looks like", though...

Secret

yeah... that's one reason in the beginning of all the discussion, that is 4 days ago, I asked Martin and others on how to think abstractly about bijection constructions given a set of any cardinality

I need the abstract thinking to simply all the details into what really is the key to the mathematical structure, but I also use pictures and intuition to guide my way through. These go hand in hand. When pictures are no longer accurate, the correct thinking on the abstract side is important

Secret

13:20

Infinite sets interests me not just because infinity is weird and all weird things are potentially interesting. I actually have two application in mind from all this learning in the past 4.5 days:
1. Understand how to work with sets of cardinality $> \mathfrak{c}$, such as $\mathcal{C}(\mathbb{R})$ and $\mathcal{C}(\mathbb{C})$. This is important in helping me on the real analysis studies and also in variational calculus, so I had some semi intuitive idea on what the target function that will solve the variational equation will be

DHMO also gave me good resources to read on reals and rationals and p adics. So in the future I should made less mistakes in my proofs involving them

17

Q: Is the set of real numbers the largest possible totally ordered set?

Because I find any totally ordered set can be "lined up" in a straight line, I'm guessing that the set of all the real numbers is the biggest totally ordered set possible. In the sense that any other totally ordered set is isomorphic (order-preserving) to a subset of real numbers. Is that right?

real-analysis order-theory

It does seems the fact that the reals contains a countable subset prevent it from being able to expand further without the contradiction that the rationals will be uncountable

Martin Sleziak

@DHMO There are several posts about the same problem also here on math.SE. See Countable set having uncountably many infinite subsets or Uncountably many sets of natural numbers with finite intersections and many other posts linked there. They are called almost disjoint families. There are several very elegant constructions.

DHMO

@MartinSleziak thanks, @AlessandroCodenotti ^

Martin Sleziak

I will also link to Stefan Geschke's text Almost disjoint and independent families. He is active both here and on MO.

Akiva Weinberger

13:37

$\Bbb R+\omega_\alpha$ for large enough $\alpha$ (depending on the CH) should work. $+$ here is the addition operation on total orderings, where $a+b$ is essentially all of the elements of $a$ followed by all of the elements of $b$. @Secret

(I think I've seen $\lambda$ or some Greek letter like that to represent the order type of the reals, but I don't see a reason not to just use $\Bbb R$ instead)

Secret

@AkivaWeinberger By expand, I mean the new set is strictly larger than $\mathbb{R}$. That is impossible if it can be embed into $\mathbb{R}$.But anyway, do you mean a<a+b for any a and b?

Akiva Weinberger

$a$ and $b$ are order types, not elements

Secret

I see

Akiva Weinberger

$\Bbb R+\omega_\alpha$ should have strictly cardinality larger than $\Bbb R$ for large enough $\alpha$, right?

@Secret I was answering the question in the MSE post you shared above.

Secret

yeah, I initially did not realise $\omega_{\alpha}$ will give a set of size $\aleph_{\alpha}$ itself. That should work.

It seems apparent I am too comfortable with elements. I need to get more comfortable with proper classes

@AkivaWeinberger Actually I do have a small question about proper classes. Proper class are defined to be classes that are not sets. I know there are examples such as the ordinals. How are they not a set?

Akiva Weinberger

13:50

Though this brings up a question: In ZF, if $\mathfrak c$ and $\omega_1$ are incomparable, must we have $\mathfrak c+\omega_1\ne\mathfrak c$?

Secret

(Answer can assume ZFC)

Akiva Weinberger

Oh, another example of a large ordered set: Any subset of the surreals larger than $\mathfrak c$

(since the surreals include the reals and are a proper class)

@Secret I'm not sure what you mean. They're not a set since if they were, we would get contradictions

The main difference between sets and proper classes is that sets can be elements of other sets and classes, and proper classes can't be elements of anything @Secret

Secret

I see, I always thought there are other properties that distinguish them

Btw, the set $(W,\preceq)$ given by Michael Greinecker is also an interesting one. It basically behave like the reals (of course the ordering is different from the reals), but there are no countable elements in them. It is also interesting in that it is well ordered. I am not sure how spaced out two elements are in that set, as it is obviously uncountable

If we take a subset of size $\mathfrak{c}$ of that, both will basically look like lines pictorically. (Because I am not referencing to the reals, I can always rearrange the indices for each element so that it resembles that of reals) So I guess the only way to tell them apart is via injections

I suspect we can do similar things on the $\mathbb{R}+\omega_{\alpha}$ set

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