Mathematics - 2018-10-05 (page 1 of 3)

Secret

00:03

Mancala

00:34

Could someone tell me where to find a result that says: "two surfaces minimal conected that match an open are equal"

Ted Shifrin

@Leaky: Back to the drawing board. I have no idea what it means to say "the function converges uniformly there."

Faust

01:12

Morning everyone

CaptainAmerica16

@Faust It's 9:14 pm where I am. Good evening.

Adam

I wanted to ask everyone and anyone to give me their most fundamental and simplified description of any relations between the natural logarithm and the harmonic series but also a heads up if you are known to be clinically insane. like relative to the degree I am considered to be on average

@Mike Miller no it's not everyone has been really nice. I mean relatively nice when compared to Alfred's fascist\topologist facebook group or the greathouse forum

Ultradark

01:29

@Adam the $\ln$ series is $\sum_2^\infty 1/n(\ln(n))^p $

it's related to the p-series

CaptainAmerica16

@WillHunting I figured it out. I'll type the proof here when you show up again.

Ultradark

@CaptainAmerica16 I found out something kinda cool: $\lim\limits_{n\to \infty; n\in \mathbb R^+}\text{surface area} \{(x,y): y= x^n(1-x)^n,x\in[0,1],\}=\pi$ at least it seems to be true I'm going to try to prove it

it'll be a good exercise to go through

CaptainAmerica16

@Ultradark Ooh, that does look interesting. Keep me updated on how it goes, I haven't done any proofs like this yet.

I realize "like this" is a vague statement. You may not care, but leaky always has something to say...

Ultradark

01:55

okay, so basically the function as $n \to \infty$ results in a flat line (a disk) with radius 1. So of course the area of a circle with radius $1$ is $\pi$!

CaptainAmerica16

@UltraDark What area of math is this? This looks like the properties I learned about in precalc, (but at a deeper level of course).

Ultradark

I think it's probably precalc and calc

CaptainAmerica16

@Ultradark Oh, ok. That's what I thought XD

@Ultradark I think I finally solved the set theory thing. I might do a calc proof next.

Adam

02:11

Would it be wrong to do a social experiment, spamming the internet with clickbait claiming that this function's value at $n=5$ proves that there exist exactly 5 dimensions in each universe of the multiverse. and then later get more ridiculous, like we threw Joel Olsteen into the boomdoogle and he came back as someone that actually makes a positive contribution to society? $$\frac{(n^n)!(((n+1)!)^{(n-1)^2}}{(n!((n+1)^{n-1})!)^n}$$

I mean it's blatant misinformation and a false claim of scientific certainty but isn't that pretty much what the ball park is there, if Morgan Freeman can cash in on dubious implied conclusions surely it wouldn't be a bad thing. like for charity

CaptainAmerica16

@Adam You should do it, if only for yourself...or me. I kind of want you to do it.

Adam

I lied i absolutely would not give it to charity either. and its actually enough motivation if someone commits to doing the cgi for joel being sucked into the plasma vortex

oh well im done

CaptainAmerica16

@Adam I no longer know what you're talking about.

Rithaniel

02:28

You have a mismatched parenthesis there. Or is that the part that proves the 4.5th dimension?

Mike Miller

@Adam That comment linked to something in particular that was very disconcerting. The chat has now been made private.

There's some comments directly below that message giving context.

Adam

02:42

$$\frac{(n^n)!((n+1)!)^{(n-1)^2}}{(n!((n+1)^{n-1})!)^n}$$ @Rithaniel nice catch the float approximation at $n=5$ is something to the order of $10^{-5040}$ it's just a ridiculously small number

Rithaniel

at n=8, it's on the order of $10^{-125047912}$

Which is a little bit smaller.

Just a tiny bit.

Adam

@Mike Miller yes it's pretty standard but on the whole my point was, the format of the website is such that it doesn't slowly become one of those idiot online gated communities that are less and less about the original purpose of it's use as time goes on, I guess simply because of the magnitude of it's use in math, I've only posted once since joining the historical community but I have a sinking feeling it's not going to have the same quality, and has got a lot of nose bleeds in store for me

haha yeah it was just the first significant jump that I observed there is actually a lot more to be said about the patterns in recurring consecutive digits in functions like that involving factorials and positive powers, I mean if you have a mess around a generalized form like say $\mathcal U_{{n,k}}= \left( {n}^{k} \right) !- \left( n! \right) ^{k}$

you will see there is a proportionality between the increase of the sequence itself and the number of repeating $9's$ in each term's base 10 representation, I want to at least properly learn Kummer's Theorem before I take a serious look at crap like this but if you evaluate a few lines you will see what I mean

seems to be a proportionality based on inductive reasoning which is for level 24 sleep deprivation and im only around 6 or 7 atm

Anonymous

03:00

Anonymous

We concluded that $BH || PC$ but then how similarly $BP || HC$?

Secret

F888 JEE

F8888 JEE

F888888 JEE

Anonymous

What does that mean? lol

Holo

@IceInkberry secret ramble a lot

Secret

Wait a sec... how on earth BP perp PC? Must be another right angle I have missed

Anonymous

03:09

Yeah, she(?) has weird dreams too. Everyone knows.

Anonymous

@Secret It isn't.

Holo

BP and PC... please reread this :)

Secret

Ok I misread EC as HC, no wonder...

hmm...

Anonymous

(Also, by the way, we don't have geometry in JEE)

Holo

Good, geometry is extremely annoying

Anonymous

03:18

Yeah, I miss out on trivial things like this one. And it's annoying when I don't understand why I am missing out.

Holo

It is basically Brute-force search

Secret

ah H is orthocentre, thus $BA \perp AC$. Hence $BA \perp HC$. Meanwhile, angle at semicircle said $BP \perp BA$

Thus conclusion follows

ijcai.org/Proceedings/89-2/Papers/121.pdf

Well, technically one can systemise these proofs but it relies on identifying elements of the geometry

Anonymous

@Secret Got it! (It's $BA \perp FC$)

Anonymous

Thanks secret!

Holo

@Secret using vector geometry is nicer, but still annoying as hell

Anonymous

03:24

But isn't that called bashing (the beauty of) the problem?

Secret

I sometimes like to say if one want to know how hard finding the shortest proof is, try any high school geometry problems

Holo

@Secret how hard is to find the longest?

Secret

Geometry in its most minimalistic sense, is really a collection of objects called points and relations between them.

@Holo There is probably no upper bound if you allow redundancy

Holo

@Secret how hard is to find the longest assuming you only show relevant information for the proof each step*?(Better putted?)

Secret

as for long proofs without redundancy, I am not sure. One can in theory set up a chain of proofs so that the main thing is to prove A, but to prove A you need to prove B,C, but to prove B,C you need to prove and so on. I vaguely recall something along the lines of countable model consistency something, but I don't recall the details

one can in theory extend a proof to countably long by using similar infinite regress as it will take a while to get to the axioms

Holo

03:31

But will it take infinite steps to get to the axioms? How can you continue the chain after getting to pure logic proof?

Secret

30

Q: Why can't proofs have infinitely many steps?

I recently saw the proof of the finite axiom of choice from the ZF axioms. The basic idea of the proof is as follows (I'll cover the case where we're choosing from three sets, but the general idea is obvious): Suppose we have $A,B,C$ non-empty, and we would like to show that the Cartesian product...

Well in theory you can do infinitary logic with countably long statements, but then you cannot even write it down

I cannot seemed to find anything that mention about existence to upper bound of proof length

news.cnrs.fr/articles/…

There are also very looooooong proofs made by conputers

Holo

Hmm... I can create longer proof(not write it down tho)

Will Hunting

04:00

@CaptainAmerica16 Or you can just type it and ping me so that I will receive the notification.

Secret

google.com.au/amp/s/rjlipton.wordpress.com/2010/01/20/…

The gist of uncountablity is that you can always expand the object by one indefinitely

despite already have mapped countably many of them

Holo

04:15

I disagree, I think that this intuition is no good because how is it true also for countable infinity

@Secret despite $\omega$ already mapped countably many elements $\epsilom_0$ is still countable

Secret

Ah right

Holo

It is actually a very good question @Secret , what intuition there is to the existence of uncountable cardinals for students who are new to this stuff

Secret

04:31

I wonder in ZFC + CH there is a way to show that if you can induct the diagonal argument exactly $\omega_1$ times then you will ran out of reals and hence terminate the procedure

This is because in many of the proofs, often the bijection is used to do this in a single step thus we sort of "retroactively" knew it has to stop at $\omega_1$

but if we do the induction proper, is it possible to show by induction it must stop there?

hmm...

Holo

What do you mean by "this", also, if the reals are aleph 1 there is still no reason to stop, because $\omega_1+1$ is still the size of aleph 1 we can bijective from the reals minus point to the reals and so adding another point

The reals will run out only if you get to aleph 2(assuming CH)

Secret

Ok maybe I should ask it this way. Assuming CH, if the reals are aleph1, then is it possible to induct the diagonal argument to $\omega_2$ and then see how it cannot continue afterwards and hence terminate there?

like, what happens when we try to take the $\omega_2$ th diagonal, is it being excluded from the procedure because it has now gone too long, or something else?

it is also tricky to ask this properly because $\omega_2$ really has no predecessor so one can go up from below forever and never reached it

Holo

So you are saying; let's take a sequence of reals, call them $a_0$ and define transfinite induction by using the diagonal on them again and again?

Secret

yup

Holo

But how do you define the limit case?

Base and successor steps are easy, problem is, for example, what is $a_\omega$?

Secret

04:46

I think it will be similar to how limit ordinals are defined, the supremum of: diag(a0), diagdiag(a0) diagdiagdiag(a0) ...

Holo

For this, how do you define the order of the sequence?

Secret

We always start with the some (a0) which is countable (because that's how the 1st step of the diagonal proof begin). We then index each instance of applying the diagonal argument as diag, thus the procedure can be iterated and indexed with ordinals

so asumming the order type of a0 is $\omega$, then diag(a0) should have order type $\omega+1$ and so on until the limit ordinal cases are reach, where we then take supremums

Holo

Lol, I find this method kind of funny, as the induction is over ordinals, but there is no reason it won't work

Secret

Induction over ordinals is called transfinite induction, it is a common tool in many proofs using well ordering theorem

Holo

Now the question, does $a_{\omega_1}=a_{\omega_1+1}$?

No, I mean that the ordering is funny

Actually, it is pretty obvious that at $a<\omega_2$ the sequence will be a constant, otherwise you created too much reals @Secret

Secret

05:01

Ah so the diagonal operation will eventually get stuck at a fixed point

and thus concluding that we ran out of reals

Holo

Yes, pretty much, the question is, when will it stop, given $a_0$, although I don't think it is easy to answer, if answerable at all

Secret

Well, while we can kind of simulate all the computable countable ordinals with a suitable encoding in a turing machine and hence physical computers, the trouble is $\omega_1$. Nature has to have some object that is already of order type $\omega_1$ otherwise the answer is uncomputable

And way before we reached $\omega_1$, we would already got stuck somewhere before $\omega_{CK}^1$

Holo

Well, this is computability problem, yes

Corellian

05:23

58 mins ago, by Holo

It is actually a very good question @Secret , what intuition there is to the existence of uncountable cardinals for students who are new to this stuff

I've read a remark where someone mentioned the problem of the limit case for the perimeter of a square as you cut corners off to eventually get a circle, and how the uncountability in the circle's curve has to do with the apparent discrepancy

Not sure if this is really true though

Holo

I am not sure what city corners off means

Corellian

old relevant meme

yeah I didn't word it very clearly my bad

of course, there's no issue with respect to area, so I don't know what they exactly meant by uncountability

Holo

I wouldn't call it the limit "case", I'll just say limit. Also I forgot the name of this but indeed the problem is within the limit(If I remember correctly you can give any number instead of 4 there)

Corellian

think it's a continuity thing, or something stronger. but yeah

Holo

No, I mean this specific "paradox" has a name. Continuity is way more general

Corellian

05:43

hmm not aware of an actual name for this problem or the staircase problem (diagonal of a square) related to it

Karl Kronenfeld

howdy @Ted

Corellian

"quadrature of the circle" but the other way :p

hey @Ted long time no see

Holo

@Corellian this is the worst way to put it in +)

Nick

06:20

#FrivolousFriday 5 minute challenge:

Q1. gcd(42823, 6409)

Q2. Find Multiplicative inverse:
(i) 50 mod 71
(ii) 43 mod 64
(iii) 42828 mod 6407

Q3. Solve 50x := 63 mod 71

[solve each question in 60 seconds or less]

ÍgjøgnumMeg

07:41

Hmm.. let $v_\infty(\alpha) = -\deg \alpha$ (with $\alpha \in K(X)$)

then $v_\infty(\alpha + \beta) = -\max\lbrace \deg \alpha, \deg \beta\rbrace$ right?

and this is $-\max\lbrace \deg \alpha, \deg \beta\rbrace = \min \lbrace -\deg \alpha, -\deg \beta\rbrace$

Secret

Never understood the magic of number theory

ax mod b = 1 mod b is generally hard

Discrete logarithm

In the mathematics of the real numbers, the logarithm logb a is a number x such that bx = a, for given numbers a and b. Analogously, in any group G, powers bk can be defined for all integers k, and the discrete logarithm logb a is an integer k such that bk = a. Discrete logarithms are quickly computable in a few special cases. However, no efficient method is known for computing them in general. Several important algorithms in public-key cryptography base their security on the assumption that the discrete logarithm problem over carefully chosen groups has no efficient solution. == Definition... ==

ÍgjøgnumMeg

That's not the discrete logarithm problem tho

Secret

ah right, my bad

ÍgjøgnumMeg

$ax \equiv b \bmod m$ is "easy" in the sense that the euclidean algorithm is quite efficient

but of course $a^x \equiv b \bmod m$ is hard, and that's the discrete logarithm problem

Secret

The FrivolousFriday challenges suggests those are not problems that should be solved with the euclidean algorithm though. Unless b is prime, I don't see any obvious symmetry I can exploit

Nick

07:54

@Secret No, I think you're free to use Euclid and extended Euclid. Just don't use Wolfram.

ÍgjøgnumMeg

@Nick sure, but you don't know how many steps of Euclid there will be before you start so under pressure you might assume there is a simpler way

Secret

08:39

[Random joke]

Size of objects used by types of people:

Ultrainfinitists: Inconsistent size

ZFC: $\kappa <$ Kunen inconsistency

ZF: $\kappa < $ Super Reinhardt

Finitist: $\kappa < \omega$

Ultrafinitist: $\kappa < M$

Valve: $\{0,1,2\}$

Akiva Weinberger

09:21

What is your favorite proof?

Leaky Nun

@AkivaWeinberger holomorphic implies analytic

Tobias Kildetoft

@AkivaWeinberger Anything that uses way more theory than necessary to make the proof itself easy to understand.

For example Cayley-Hamilton using that matrices form an irreducible variety.

Or for that matter any similar statement about matrices which is easy for either invertible or diagonalizable matrices

Akiva Weinberger

That's basically from the fact that, if $f$ and $g$ are (multivariate) polynomials, $f$ is not $0$ everywhere, and $fg$ is $0$ everywhere, then $g$ is $0$ everywhere.

Tobias Kildetoft

@AkivaWeinberger Right

Akiva Weinberger

(Proof: Ring of multivariate polynomials have no zero divisors, so $fg=0$ implies $f=0$ or $g=0$.)

Combined with the fact that, whether or not a matrix is diagonalizable is equivalent to $f=0$ for a certain polynomial in its entries.

(And similarly for invertibility - here, the polynomial is the determinant.)

Tobias Kildetoft

09:30

right, and for the first one it is the discriminant of its characteristic polynomial (so not actually equivalent, but sufficient).

Or more precisely, there is an open set of diagonalizable matrices coming from the preimage of the complement of $0$ of that polynomial.

Akiva Weinberger

Er, $f\ne0$ I meant.

Tobias Kildetoft

I am looking through this proof of the classification of finite abelian groups now that I will be going over on Monday. It is almost too slick, since it shows that a cyclic subgroup of maximal possible order will have a complement, but it took me a while to figure out where the maximality was even used.

Akiva Weinberger

Let me rephrase my first statement also: If, everywhere, either $f=0$ or $g=0$, then either everywhere $f=0$ or everywhere $g=0$.

Alternatively, if $g=0$ whenever $f\ne0$, and $f\ne0$ for at least one point, then $g=0$ everywhere.

(These are, again, both equivalent to the fact that the ring of multivariate polynomials has no zero divisors, which is easy to prove.)

@TobiasKildetoft $\prod\Bbb Z_{a_i}$, yeah?

How does the proof go?

Tobias Kildetoft

@AkivaWeinberger Given an element $g$ with maximal order, it finds an $h$ such that $\langle g\rangle\cap\langle h\rangle = \{e\}$.

Then it quotients by $\langle h\rangle$ and repeats since the image of $g$ will again have maximal order

Akiva Weinberger

$h\ne e$ I assume

Tobias Kildetoft

09:44

Right

Akiva Weinberger

And if $g$ doesn't have maximal order, you could have like $\Bbb Z_4$ and $g=2$

Tobias Kildetoft

At some point, the quotient will be generated by the image of $g$ under the composed homomorphism, and then the original group is the direct product of $\langle g\rangle$ and the kernel of the homomorphism

Right

And the use of the maximality was buried in the construction of $h$

Akiva Weinberger

Yeah, 'cause there is no $h$ if $g=2\in\Bbb Z_4$

How do you construct $h$?

Tobias Kildetoft

Because what one does is take any $h\not\in \langle g\rangle$, so $\langle g\rangle\cap\langle h\rangle = \langle h^l\rangle$

Akiva Weinberger

Mhm

And clearly $h^l\ne g$

Tobias Kildetoft

09:47

Then shows that there is some $g_1\in \langle g\rangle$ such that $g_1^l = h^l$

in which case we set $h_1 = hg_1^{-1}$ and this new $h_1$ works

And in order to find this $g_1$ we needed to use that the order of $h$ divides the order of $g$.

Akiva Weinberger

Wait, so for example

$\Bbb Z_3\times\Bbb Z_9$

$g=(0,1)$

We can take $h=(1,1)$

and $3h=(0,3)$

so that's $hl$

And I guess we can take $g_1=(0,1)$ (or $(0,4)$ also theoretically but why would you do that)

and then $h_1=(1,0)$ (or $(1,-4)=(1,5)$)

And now $\langle h\rangle\cap\langle g\rangle=\{e\}$

@TobiasKildetoft 'Cause it divides the order of $h^l$, which divides $g$?

Wait no

Tobias Kildetoft

@AkivaWeinberger No, that is the wrong way around

Akiva Weinberger

Right

In my example $h$ and $g$ have the same order anyway

Why does it divide it, then?

Tobias Kildetoft

because the order of $g$ was chosen to be maximal, and the group is abelian

Akiva Weinberger

So?

Leaky Nun

09:55

@AkivaWeinberger to expand, the proof constructs the series expansion from the cauchy integral formula, and for the corollary you have explicit formulas for the n-th order derivative of f(p) based on the values of f(z) for z living in a circle with center p

Akiva Weinberger

Mhm

Tobias Kildetoft

@AkivaWeinberger All element orders divide the largest one in an abelian group (good exercise, you should show this)

Akiva Weinberger

Love that formula

Leaky Nun

the beauty is that the higher order derivatives of f at p are all determined by the value of f on a circle around p

and how you basically get Liouville and maximum modulus for free

Tobias Kildetoft

Though for this proof, we have already covered the Sylow theorems, which immediately reduces the problem to abelian $p$-groups, where it is trivial that all element orders divide the largest one.

Akiva Weinberger

09:56

Technically, they're determined by the value of $f$ on any set with a condensation point, I think @LeakyNun

Leaky Nun

indeed

if the domain is connected

Akiva Weinberger

Right

Why is that, actually? I forget

Oh wait

Is it 'cause you have enough information to find all the derivatives at that condensation point

and we just showed that it's analytic, so that's all you need?

Leaky Nun

@TedShifrin maybe you misunderstood my proof that $\lim_{r\to1^-} \sum r^{n!} = \infty$. Allow me to phrase it more clearly. If the limit does not go to infinity, then it must be a number $L$, because the function $r \mapsto \sum r^{n!}$ is increasing. And then we would have a continuous function $f : [0,1] \to \Bbb R$ that sends $r$ to $\sum r^{n!}$ if $r<1$ and $L$ if $r=1$. Now, let $f_n := r \mapsto \sum_{k=0}^n r^{n!}$, and we see that $f_n \to f$ pointwise.

Since the domain is compact, $f_n \to f$ uniformly as well, so $\lim_{r\to1^-} \sum r^{n!} = \sum \lim_{r\to1^-} r^{n!} = \sum 1 = \infty$.

(@AkivaWeinberger please proofread this if you want to)

@AkivaWeinberger yes. that's basically the identity theorem.

Nick

Is any RSETian here? Please do take a video of the TEDxRSET launch.

Akiva Weinberger

10:12

Wait, why does $f_n\to f$ uniformly? @LeakyNun

Is that because $f_n$ is increasing?

I think you can have instances where a sequence of functions approach a limit pointwise on a compact set but not uniformly

but maybe that's not the case if it's monotonic

Leaky Nun

because each $f_n$ is continuous

$C^0[0,1]$ is complete

mercio

@LeakyNun you can do the same argument for $\sum z^n$, but this one can be analytically continued a lot, so you have to do some more work in order to show that $\sum z^{n!}$ can'tbe analytically continued anywhere through the unit circle

Leaky Nun

@mercio that wasn't what I'm trying to clarify

I argued that for the $\sum z^{n!}$ case, one can reduce to the $\sum r^{n!}$ case

I can repeat that argument if you want.

mercio

I don't see how restricting the analysis to the real axis is a good thing, so yes maybe you can repeat the argument

Leaky Nun

it suffices to show that $\lim_{r\to1^-} \sum (r\exp(i\pi\theta))^{n!}$ does not exist for rational $\theta$

mercio

10:17

why

Leaky Nun

because the rationals are dense in the reals

if you're asking me why (limit does not exist --> cannot continue in that direction), I'll just repeat my $C^0$ argument

@mercio is there anything more I should clarify?

mercio

yes

Leaky Nun

namely?

Akiva Weinberger

@LeakyNun Still

mercio

are you saying that the limit won't exist for irrational theta because the limit doesn't exist for rational thetas ?

Akiva Weinberger

10:20

You can have $f_n$ and $f$ continuous, $f_n\to f$ pointwise, and the domain compact, without $f_n\to f$ uniformly

mercio

is that how you intend to use the density of rationals ?

Leaky Nun

if you can continue along a direction $\theta$, then the function can be extended on an open set containing $\exp(i\pi\theta)$, which would contain some $\exp(i\pi\theta')$ for $\theta'$ rational

mercio

ah then yes

Leaky Nun

@AkivaWeinberger and $f_n$ is increasing w.r.t. $n$

Akiva Weinberger

OK, so you do need monotonic.

Otherwise, let $f_n$ to be $0$ on $[0,\frac1{n+1}]$ and on $[\frac1n,0]$, but a spike on $[\frac1{n+1},\frac1n]$ (which linearly goes from $0$ to $1$ in the first half of the interval and back down on the second half)

Leaky Nun

10:22

right

5 mins ago, by Leaky Nun

it suffices to show that $\lim_{r\to1^-} \sum (r\exp(i\pi\theta))^{n!}$ does not exist for rational $\theta$

notice that for $n$ sufficiently large, $(r\exp(i\pi\theta))^{n!} = r^{n!}$

so we reduce to the real case

mercio

@LeakyNun yeah but I was waiting for you to say this

Leaky Nun

@mercio sorry

mercio

I don't think you said it before

Leaky Nun

my fault

mercio

and it IS the important part of the proof

Leaky Nun

10:24

I noticed

I just thought it was too obvious lol

@mercio so does my proof check out?

mercio

yes

Leaky Nun

yay

Lucas Henrique

11:36

Hi @chat

Define a set A as infinite $\iff \exists B(B \subset A \land |A| = |B|)$

We say that a set is finite if, and only if, it’s not infinite.

How do a prove that any finite set has a bijection with a natural number?

Secret

This is ZFC?

(as otherwise a lot of thing can happen with that definition)

(The following proof is assuming you are working in ZFC)

Lucas Henrique

Yes.

Fargle

Maybe the contrapositive would be easier? Although I'm not sure, I don't really see how to tackle this one

Lucas Henrique

Yeah, probably.

Vrouvrou

11:52

hello, please how to show that $Z=\{f\in\mathcal{C}([-1,1],\mathbb{R}), f(0)=0\}$ is arc connected ? where the space is given with the. norm sup

Lucas Henrique

I’m also trying to prove that the cardinality of disjoint sets is additive (supposing that any finite set has a bijection with a natural number)

Akiva Weinberger

@Vrouvrou You can connect $f$ and $g$ with $t\mapsto (1-t)f+tg$ I think

Vrouvrou

yes I write $h(t)=(1-t)f +t g$ but how to prove that it stay in Z?

how to write ? h(t)(0) ??

Secret

Suppose there is an infinite set $A$ such that there exists a subset $B$ that does not biject with the natural numbers. Then $B$ does not inject or does not surject to any $n \in \Bbb{N}$. Since all natural numbers are well ordered by construction (By considering the von neumann ordinal construction (not sure how to handle nonstandard models)), by induction if $B$ inject but does not surject to $n$, then $|B| < |n|$ but this is impossible as the smallest possible $|B|$ is zero, which bijects with $0$, so there will always be a $B$ that bijects with a natural number in this case

Akiva Weinberger

@Vrouvrou That looks weird

but I guess it's correct

A better way would be $h_t(0)$ maybe

$h_t=(1-t)f+tg$

Vrouvrou

11:56

we consider $h_t(x)=(1-t)f(x)+t g(x)$ for $x\in [-1,1]$ ?

Akiva Weinberger

Yeah

Vrouvrou

as it is arc connected it is connected right ?

Lucas Henrique

@Secret I’m sorry but I don’t understand why you made this hypothesis. Is it equivalent (contrapositive) to the negation is the theorem? (Since it’s a proof by contradiction)

Akiva Weinberger

Yeah @Vrouvrou

Secret

I actually never done that particular proof before, and for some reason I am trying to do a direct proof by contradiction of the statement instead of its contrapositive

ok I stuffed up the first step of the proof, forget it

because I have not ruled out $|B|=|n|$

Lucas Henrique

12:24

So $\not \forall A(A \, \text{is finite} /implies \exists n \in \Bbb N(|A| = |n|) \iff \exists A(\forall n(|A| /neq |n|))$ right?

Secret

yeah that will be the contrapositive

Lucas Henrique

I’m typing at mobile and, omg, this is totally wrong lmao

LaTeX without real-time parsing is pure pain.

Leaky Nun

13:01

@LucasHenrique trust the force, luke

Secret

Attempt at direct proof (trial 2):
Suppose by contradiction there exists finite set $B$ such that it has no bijection with any $n \in \Bbb{N}$.
Case 1: $B \hookrightarrow n$ but $B \not\twoheadrightarrow n$
Then since all $n$ are ordinals by construction, and thus its elements are all $k < n$, there must exists at least one $k$ such that $B \twoheadrightarrow k$. Any such injection is also a bijection since its range is all of $k$. Thus we found a $B \leftrightarrow k$ where $k$ is a natural number. Contradiction.

Therefore all finite sets $B$ must biject to some natural number QED

typo $\aleph_0$, not $\aleph_{\alpha}$

...

Seriously it will be really cool if set theory can allow a set such that finite = infinite + finite and these two sets are disjoint

But then it is easy to see how that will break logic because if you have a $m = n + \kappa$ and the sets corresponds to $\kappa$ and $n$ are disjoint, then it basically means that it takes infinite number of elements to add $m-n$ elements into the set

so peano arithmetic will break

In fact it will also suggest that $m-n$ bijects with $\kappa$ which is pretty absurd

unless you can have a set that is infinite and finite at the same time

Algebraically it is not that disasterous, because it basically means:

$$(((n+1)+1)+1) \neq n+(1+1+1)$$

PS In ZF in the above proof, case 2's failure is where the dedekind finite sets, amorphous sets etc. can pop up

3

Q: Dedekind Finite set contains Dedekind Finite subsets

I would be grateful for some help in proving: If a set is Dedekind Finite then every subset of it must be Dedekind finite too. I tried a reductio ad absurdum way of thinking but I can't seem to find anything absurd. Thanks in advance

set-theory cardinals

Dedekind finite sets are like onions because their subsets can only ever be dedekind finite (this also include finite subsets)

It's literally an onion that you can peel it forever and shrink indefinitely

user 2646

13:17

@Loong hi

Loong

hi

user 2646

how's life?

Loong

a lot of work

user 2646

and no play?

Lucas Henrique

Thank you so much, @Secret!

Secret

13:34

The link contains an overkilled version of proofing the finite set proposition above

(other unverified and unrelated stuff)

Presumably we can well order the reals the following way:

Pick a countable set of binary sequences e.g.:

10000...
01000...
00100...
00010...
00001...
...

Call this set $a_0$

Next, define the map $f$ to be the procedure "take the unique diagonal of whatever it is in its arguments"

Then $a_1 = f(a_0)$ has order type $\omega +1$

$f(a_1)$ has order type $\omega +2$ and so on

$f(a_{\alpha})$ thus will have order type $\omega + \alpha + 1 = \alpha + 1$ if $\alpha > \omega$

Holo

Alpha+1* no?

Secret

Then it is easy to see that if we pick any $\beta \in [\omega_1,\omega_2)$ then there will be some $f(\beta)$ that will exhaust the reals since the reals under ZFC+CH hasd cardinality $\aleph_1$

Thus the cantor diagonal operation plus choosing a $\beta$ will well order the reals

This means, in order for this to be consistent, the above has to be equivalent to the axiom of choice somehow...

(since it is consistent in ZF that there is no well ordering of the reals)

Holo

@Secret but if we are choosing beta=\=omega[1] how can you be sure that it does not becomes a constant before beta?

Secret

I mean, there will be some beta in that ordinal interval which will cause $f$ to no longer produce new elements

Holo

So, it is not any beta, it is exists beta?

Secret

13:49

The main trouble here is really $f$ becomes uncomputable for $a_{\omega_{CK}^1}$, so I don't know if we can ever pin down $\beta$

As for whether in principle we can do that, my set theory level is not high enough to investigate uncomputable functions

Holo

I'm pretty sure we can't, because won't if give us formula for well ordering or the reals? And we know that although the ordering exists the formula doesn't in ZFC(If I remember correctly)

Secret

Yeah, so somehow the above had to be equivalent to the axiom of choice, but I am not sure which step I have used choice here, since I did start with a well ordered countable set of sequences and each diagonal operation should only produce one new sequence

Holo

No, pinning down $\beta$ is stronger then choice

CaptainAmerica16

@WillHunting Prove that for a function $f:X \rightarrow Y$, we always have $A \subseteq B \Rightarrow f(A) \subseteq f(B)$. Let $y \in f(A)$. Assume $A \subseteq B$. There exists $x \in A$ such that $f(x)=y$. Since $A \subseteq B$, we can create the statement $ x \in B$. From this we have the statement $f(x) \in f(B)$, so $f(A) \subseteq f(B)$.

Secret

@Holo what if we only can say that $\omega_1 \leq \beta < \omega_2$ will it still be stronger than choice?

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