Mathematics - 2018-10-06 (page 1 of 2)

Ultradark

00:05

@TedShifrin let's say you have an integral and another integral. The integrals are equivalent

Ted Shifrin

What does that mean?

Ultradark

the functions in the integrand are different

I mean obviously you can construct examples of this I guess

Ted Shifrin

As usual, I have no idea what you're talking about.

Ultradark

@TedShifrin I'll give you an example. I'm sure this is probably trivial

Ted Shifrin

I mean, $\int_0^1 x\,dx = \int_0^\pi \frac14\sin x\,dx$.

Ultradark

00:08

yeah^

but I'm looking at integrals with the same bounds

Ted Shifrin

So this has no content, honestly.

Ultradark

I guess it's not very interesting from a research level

Ted Shifrin

Look, you can take any function $f$ whatsoever and any interval $[a,b]$ and get $1=\int_a^b \frac1c f(x)\,dx$, where $c = \int_a^b f(x)\,dx$.

(Assuming $c\ne 0$.) Why is this interesting to you?

Ultradark

It's not really interesting to me lol

I don't really have an intuition for what's interesting or not in mathematics yet

Ted Shifrin

That's a lot a matter of taste. Most of the things you find interesting, I don't. And things I find interesting, you probably don't. Certainly mathein spends his time thinking about things that are not so interesting to me, etc.

Ultradark

00:18

how do you write a fraction in latex i always forget

Ted Shifrin

\frac

no, \frac x y or \frac 1 {12} if there's more than one digit.

Ultradark

$\int_0^1 e^{\frac 1 {\ln(x)}}*\frac 1 {\ln^2(x)}dx=\int_0^1 e^{\frac 1 {\ln(x)}}dx$

I checked this on wolfram alpha

wait a second

Ted Shifrin

This is definitely wrong. I can compute the first one explicitly. Not the second one.

Faust

02:44

@TedShifrin you still here?

Albas

03:44

Hi @WillHunting I am fine. How's it going with you

Secret

06:06

$$\int_a^b f(x) dx = \int_a^b g(x) dx$$

$$F(b)-F(a)=G(b)-G(a)$$

$$F(b)-G(b) = F(a)-G(a)$$

Well the obvious problem is that there are countably many such $F,G$ that can satisfy this relationship (because it is basically saying that the difference between the two primitives at the boundary points agree)

Holo

Secret, can't you start talking by stating the goal? Instead of throwing facts

Secret

Ok the above is really in response to this:

6 hours ago, by Ultradark

@TedShifrin let's say you have an integral and another integral. The integrals are equivalent

But I don't want to ping

It also does not help that because of stupid timezone issues, all that conversation before that integral took place while I am asleep, so usually most of the interesting things already took place in the past

and if I have to ping everyone whenever I tried to respond to something in the past more than 10 hours ago, the number of pings can pile up massively, which I don't know how much people will be annoyed seeing that mail box full of pings

Holo

Lol

Secret

Don't forget, I live in Sydney, while most of the active users live in the US

we are literally have non overlapping timezones

Holo

@Secret Also, there are uncountably many such functions. There are at least the amount of null sets in the interval

Secret

06:18

Right, but for continuous functions, there are only countably many

Holo

@Secret well, my solution is to not sleep :)

@Secret If continuity is a condition then yes

Secret

I really have to find a way to state all my assumptions, too often I am so used to thinking a certain condition of things that I often end up omitting the description of that condition

Holo

I noticed... It happens a lot

Secret

e.g. my mind works kinda like Given "integral operator" + "functions", output "continuous function"

and then "continuous" get omitted because the mind is so used to it it becomes subconscious

which frequently lead to talking past one another when your recipient don't have the same implicit rule of thinking

but it is often very hard to caught those trailing thoughts in their act because well... subsconscious bias

Holo

In summary, subsconscious screw you

Secret

06:22

Actually, its
Given:
Integral operator + Functions + (something about Ultradark's domain of interest)
Output:
Continuous function

Because if it is only Integral operator + Functions
then the output will be the set of all integrable functions, which is indeed $2^{\mathfrak{c}}$ in size

Holo

But what did Ultradark said about the domain?

Secret

not "domain" in mathematical usage, but "scope" of interest

English language ambiguity issues

By summarising what Ultradark/geocalc33/George Thompson's have been saying ever since I first aware of his existence, he is definitely onto some really nice functions, so I predict continuity is almost a must in his requirements

Holo

I don't understand a word that is not mathematically, I only speak logic symbols $\forall\!\!\!\exists$

I see, if so there are indeed countably many

Secret

In other (old) news

Jul 12 '17 at 4:59, by Secret

[To be investigated after all function spaces] $\Bbb{R}^{\Bbb{R}^{\Bbb{R}}}$

I seriously doubt I will get there before I die

Mar 12 '15 at 2:16, by David Wheeler

The space $\Bbb R^{\Bbb R}$ is ridiculously big. There are functions out there NO ONE HAS EVER SEEN

be reminded that:

Anyone who can visualise this space in its entirely, will ace quantum field theory

Holo

06:38

Well... This is not wrong... Yet...

There are function in the space $2^{\Bbb N}$ that no one even saw

Secret

That's true, since almost every real number is indefinable, not just uncomputable

(the cardinality of definable reals is necessary countable, since our language only has finite letters and hence only can form countably many possible descriptions)

Holo

So before visualise the space of the real valued function try visualise the space of the binary sequences and before that try to visualise quantum mechanics

@Secret I know, in the main site one of the most up voted answered I have is the proof that the number of proofs is aleph 0

Secret

$2^{\Bbb{N}}$ is actually pretty easy to visualise in its entirely. It is basically the binary tree of countable height so it is self similar in a sense

It's subsets?, not so really...

Holo

Yes, but you can't define most of those trees

Secret

yeah, for one such tree where the height labels are countable ordinals, we will run into problems at around height $\omega_{CK}^1$ I think. Otherwise, I don't think it can be defined

(Unrelated)

5

Q: Are there more Lebesgue measurable or more non Lebesgue measurable functions?

Are there more Lebesgue measurable or more non Lebesgue measurable functions? Does anybody see how to answer this. Please do tell.

Actually, I don't think I know the precise meaning of indefinable

user 2646

06:59

not definable :P

like division by 0 strictly in the set of real numbers

Holo

It means you can't reach "get" to that number using your language. For example, (R,+) we can't define any fractions because using the zero element and 1 and the addition function we can't get passed the positive integers

So in (R,+) the number 0.5 is indefinable

Secret

Ah I see

Holo

This also implies that the number is incomputable within the language

Secret

now that makes me wonder how transfiniteky long our string of description (and thus how much we need to expand our language) in order for all real numbers to be definable. That the reals has a certain cardinality seemed to suggest there has to be such maximum length unless incompleteness foil us to prove its existence

Holo

I think that (P(R), sup) is strong enough

But it doesn't really help us

Hmm

arxiv.org/abs/1105.4597

Apparently that even L=V is not enough for all the reals to be definable

But countable let models gives an option (not sure what exactly pointwise definable model of the reals is, so I sent a link)

11

Q: Is there an example for an undefinable number?

This question is motivated by a comment of Robert on the question Can any Real number be typed in a computer? : Can you "think of" an undefinable number? – Robert Israel I would like to reformulate this (ironic) question to: Can you give an (concrete) example for an undefinable number? ...

logic definition examples-counterexamples real-numbers philosophy

Look at the comments and at the first answer

Secret

07:17

O yeah. Chaitins constants are pretty much the upper limit of "describable": You even cannot reliably compute its digits

The simpler (R,+) example does make me wonder about something though:

is it even possible to expand our language using only the letters 0,1 and + and form a description (however long that is) that will basically pin down the operation /, thus adding / into this language?

because if we cannot, even transfinitely do that, then that will serve as a distinct obstacle to the proposal that one can simply make things longer to define more things so to say

Holo

Description? How do you define description?

Secret

A string that uniquely specify the properties of the new letter we want to add into our language

e.g. Express all the properties of / (or equivalently (a,b)+(c,d)=(ad+bc,bd) and (a,b)(c,d)=(ac,bd)) usingly only 0,1,+ and the string can be transfinitely long if needed

Holo

Hmm, we can define equivalent relation to define when a/b=c/d, but for this we need to define first *

And for this we will need the infinite long string, as the definition of "repeating +" is unbounded

Like, we define a*2, a*3 and so on, which to finish this process we need the infinite string. Then we define a/b=c/d iff ad=cb

Secret

07:38

right and despite being unbounded, each description for each particular case of "repeating +" should be finite as if one does a+a+a+... then they fall outside the reals

so it seems quite doable to me

Holo

The problem is, we can't define a*n. We can only define it for specific n, so we need to go through all the n s

You can think of the definitions like the following: (0,a,a+a,a+a+a,...)

Secret

Ah at this point it might help to clarify a bit on what I am thinking:

(Assuming I recall correctly) Normally when we think about notions of "computability" and "definability" we often focus on strings of finite length, or only a manageable number of unbounded strings as descriptions

But because I am a kind of person who basically devoted all my free time in trying to comprehend infinities because they are so cool, the above will often be too restrictive to me. Thus when I was thinking about these problems, I am considering a more generalised case as follows:

1. Computability will be exactly the same as in the way it is normally defined, meaning that a program has to terminate in finite number of steps to output the true value of a proposition

2. There is also $\alpha$-computability, which is one of the notions of hyper computability where the program can run transfinitely long and terminates after $\alpha$ steps.

3. For definability I am looking for something more general (weaker?) as follows:

Given a language L and a new letter a we want to add into our language. Can we always express a in terms of strings $\kappa$ consists of letters in L where $\kappa$ is a known cardinal

So the only case where this can fail is when the value of $\kappa$ itself is unbounded, meaning the string is so long it failed to even be a set

Holo

What do you mean town cardinal?

Secret

Thus going back to the (R,+) case above, (0,a,a+a,a+a+a,...) is uncomputable, but it is a perfectly fine description that satisfy the requirement of 3, because the longest it can get is any finite ordinal (as any longer the sum will no longer be a real number)

@Holo Mac autocorrect "known" to "town"

Holo

Oh, so there is the mistake, when talking about definable we are in first order logic, so only finite strings allowed

Of we allowed infinite strings we can define * and / in (R,+)

Secret

07:52

Right I see

So definability is the wrong term I am thinking about

Holo

I don't know if there is a name to this, but let's call it $\kappa$-definable in L

For strings of length kappa

Secret

ok

So that means I am wondering whether there exists an lower bound to $\kappa$ such that all reals are $\kappa$- definable

Holo

You mean lower bound?

Secret

right sorry...

(cause upper bound can shoot all the way into a proper class lol, silly me)

Why that particular question interested me is because it will place fundamental limits on how much I can rely on the following philosophical thinking method to comprehend basically incomprehensible concepts

Operationalization

In research design, especially in psychology, social sciences, life sciences, and physics, operationalization is a process of defining the measurement of a phenomenon that is not directly measurable, though its existence is indicated by other phenomena. Operationalization is thus the process of defining a fuzzy concept so as to make it clearly distinguishable, measurable, and understandable in terms of empirical observations. In a wider sense, it refers to the process of specifying the extension of a concept—describing what is and is not an instance of that concept. For example, in medicine, the...

Holo

The hell is this? This is not math

Secret

07:59

If the answer to that question is negative, it means there exists maps $\Bbb{N} \to \Bbb{N}$ that cannot be pin down using a $\kappa$ long string and thus for those real numbers, I cannot try to understand them by just relying on how they interact with other known real numbers

@Holo That frame of mind is what I have been using so far to comprehend many crazy concepts in set theory, or more generally, anything I have not came across before during my learning

It has so far been very successful which caused me to wonder whether there are things such as mathematical things that it can fail

Holo

But $\mathfrak c$- definable in (R,+) we can define all of the rationals and the then define a sequence of rationals for each irrational and thus define all of R

Secret

The most important take home message about that is: It is what I used in my visual thinking and how I "visualise" infinite sets so far

Holo

I'll read about operationalization later today

Secret

but you are correct, it is not maths, but it is so central to my way of learning that some of my maths question can be inspired from that

Holo

I'm no good Outside of math :/

Secret

08:03

8 mins ago, by Secret

So that means I am wondering whether there exists an lower bound to $\kappa$ such that all reals are $\kappa$- definable

this is a maths question so we should be fine

Holo

Yep

Secret

I might try to dig around to see what I can get. My suspicion is the answer might be negative because you previously commented about V=L being not enough

Holo

3 mins ago, by Holo

But $\mathfrak c$- definable in (R,+) we can define all of the rationals and the then define a sequence of rationals for each irrational and thus define all of R

Secret

Ah I see

Holo

@Secret V=L while still using first order logic, now we are extending this

Now, this gives a bound, but it is not the lowest bound. And without assuming choice I don't think it is possible to get lowest bound

Secret

08:08

right

Holo

But I don't know how to show it either

For example, how would d-finite- definable will behave?

Secret

I think it will depends on which d finite cardinals you are working with, cause these can go from linearly ordered to pairwise incomparable

There are at least two kinds of d finite cardinals, even (a d finite set can be partitioned into two) and odd (partition into two plus a singleton)

and I recall in one of asaf answers it can get more wild than that

I don't know much about d finite cardinals otherwise, which is why I am reading about the d finite Borel set paper

Holo

Yes, there is strongly even/odd and weakly even/odd but I don't know much about this

But I never understood odd d-finite sets. Why adding the singleton to one of the sets won't make it even?

Secret

08:26

hmm... I never get into that deep level. Let's see if we can reason about it using what we knew about d finite sets:

Secret

08:42

We knew that all subsets of d finite sets are d finite (other d finite sets or finite sets)

Suppose A is d finite and odd. Then it can be partitioned into sets B,C,{d} such that $A = B \cup C \cup \{d\}$ of which at least B or C must be d finite

Now we are trying to determine the nature of $A \cup \{e\}$ where $e \not\in A$

Thus we have:

$$B \cup C \cup \{d\} \cup \{e\}$$

Assuming WLOG $B$ is d finite. Then the following can happen:

oops I got the wrong definitions

5

A: Construction of amorphous subset of $\Bbb R$

There is no amorphous set of reals - in fact, no amorphous set can be linearly ordered. (For further discussion of what kind of structure amorphous sets, and more generally Dedekind-finite sets can have, see this paper of Truss or Agatha Walczak-Typke's Ph.D. thesis.) To see this, suppose $A$ i...

Ok sorry:

Evinda

Hello!!!

I want to find the supremum, infimum of the set $\{1+\frac{(-1)^n}{n}: n=1,2, \dots \}$.

I have thought the following:

$1+\frac{(-1)^n}{n}=\left\{\begin{matrix}
1+\frac{1}{n}, &\text{ if n is even} \\ \\
1-\frac{1}{n}, & \text{ if n is odd.}
\end{matrix}\right.$

We have that for all $n \in \mathbb{N}$, $1\pm\frac{1}{n} \geq 1-\frac{1}{n} \geq 0$.

Also, we have that $1 \pm \frac{1}{n} \leq 1+\frac{1}{n}$.

And $1+\frac{1}{n} \to 0$, while $n \to +\infty$.

So is $0$ the infimum and $1$ the supremum? Or have I done something wrong?

Secret

Let $D$ be d finite and $A,B,\{c\},\{d,e\} \subset D$ such that $|A|=|B|$. Then:
Strongly even: $D = A \cup B$
Strongly odd: $D = A \cup B \cup \{c\}$
Weakly even: $D = \bigcup_{d,e} \{d,e\}$
Weakly odd: $D = \bigcup_{d,e} \{d,e\} \cup \{c\}$

Suppose $D$ is strongly odd. Then for $D \cup \{d\}$ where $\{d\} \not\subset D$, we have:

Evinda

Hey @LeakyNun
Did you see my question?

Secret

$$D = A \cup B \cup \{c\} \cup \{d\}$$

(interlude) Even Marystar moved away from that boring question model, so...

(cont.)

Now there are many possibilities here:

$D = (A \cup \{c\}) \cup (B \cup \{d\})$. This is even since $|A|=|B|$ thus they have the same Dedekind cardinality, it follows that in this case $D$ can be partitioned into two equinumerous subset. Hence it is strongly even

$D = (A \cup \{c\}) \cup B \cup \{d\}$ is neither even nor odd because all 3 sets have different cardinalities

sorry typo: weakly even for the previous case

Evinda

09:01

Hey @Secret
Do you see a mistake at what I did?

Secret

$D = A \cup B \cup (\{c\} \cup \{d\})$ is even if $A$ and $B$ are both weakly even, or neither if $A,B$ both weakly odd

ok I tie myself up again. Start over

Let $D$ be d finite and $A,B,\{c\},\{d,e\} \subset D$ such that $|A|=|B|$. Then:
Strongly even: $D = A \cup B$
Strongly odd: $D = A \cup B \cup \{c\}$
Weakly even: $D = \bigcup_{d,e} \{d,e\}$
Weakly odd: $D = \bigcup_{d,e} \{d,e\} \cup \{c\}$

Suppose $D$ is strongly odd. Then for $D \cup \{d\}$ where $\{d\} \not\subset D$, we have:

$$D \cup \{d\}= A \cup B \cup \{c\} \cup \{d\}$$

Now since $|A|=|B|$, $A,B$ must be both weakly even, weakly odd, strongly even, strongly odd or neither

Case 1. $A,B$ weakly even. Then we have:

$D \cup \{d\} = A_1 \cup A_2 \cup B_1 \cup B_2 \cup \{c\} \cup \{d\}$

actually what am I doing. I only need there exists, not for all

Because we can group the union as follows:

$D \cup \{d\}= (A \cup \{c\}) \cup (B \cup \{d\})$

and because $|A|=|B|=\Delta$ for some Dedekind cardinality $\Delta$ and the union is disjoint. It follows that $|A \cup \{c\}| = |B \cup \{d\}|$. Hence we have a partition which has two equinumerous sets. Hence $D$ is strongly even QED

@Holo Let me knew if my proof miss something

By generalising the above proof, we should get:

weakly even + weakly even = weakly even

weakly even + weakly odd = weakly odd

Weakly odd + weakly odd = weakly even

strongly even but not weakly even + weakly odd = neither

strongly odd but not weakly odd + weakly odd = neither

strongly even but not weakly even + weakly even = neither
strongly odd but not weakly odd + weakly even = neither

strongly even + strongly even = strongly even

strongly odd + strongly odd = strongly even

strongly odd + strongly even = strongly odd

So in conclusion, it is like having two sets of natural numbers plus something that is not natural numbers

@Evinda don't see anything wrong. In fact, you can reason it much easier by considering the two subsequence that consists of even terms and odd terms respectively. Then you see the odd subsequence will converge to 0 from above and the even subsequence will converge to 1 from below

and therefore, this sequence is divergent as a whole because its limit at infinity does not agree at at least two subsequence

Secret

09:34

4

Q: Do Russell's socks form a Dedekind-finite infinite set?

A countable collection of pairwise disjoint two-element sets $(A_n)_{n<\omega}$ is called Russell-sequence if $\prod_{n<\omega} A_n$ is empty. (That is, there is no way to choose one of two socks from $A_n$ simultaneously.) The cardinality of $\bigcup_{n<\omega} A_n$ is called Russell cardinal. R...

set-theory

The Russel cardinals are an example of a weakly even d finite set

8

Q: Dedekind-finite arithmetic vs natural numbers arithmetic

It is known that the Dedekind-finite cardinals are closed under addition and multiplication, so one may do arithmetic in them, as opposed to only natural numbers. How much can those two arithmetics be different? For example, can there be a Diophantine equation which is not solvable in the natura...

nt.number-theory set-theory

for more information about dedekind cardinal arithmetic

Evinda

09:48

How do we get that the odd subsequence converges to 0? @Secret

Leaky Nun

10:00

@Secret no it's wrong

Secret

10:14

o sorry, somehow I misread 1-1/n as 1-1 as n tends to infinity for some reason

uh, that sequence actually converges to 1 (because both 1-1/n and 1+1/n converges to 1)

so the supremum and infirmum is determined by the other elements in the sequence

ugh I think I need coffee...

The min is 1-1=0, the max is 1+1/2=3/2. The limit of this sequence is 1. Thus all the points are in [0,3/2]. Hence our supremum and infirmum has to be the max and min

Leaky Nun

@Secret correct

Secret

10:35

Btw. It's pretty:

That's $1+\frac{(-1)^n}{n}$

user2236

12:36

@Secret you call that "pretty"?

LegionMammal978

Just to make sure I'm understanding this correctly: There does exist some ordinal $\alpha$ such that $2^{\aleph_0}=\aleph_\alpha$, and the CH is equivalent to saying that $\alpha=1$. Is this correct?

Secret

yes

Leaky Nun

13:02

@LegionMammal978 the former requires choice

Holo

13:16

@Secret sorry, I had an Emergency, I'll read everything now

Secret

it's ok no rush

Holo

13:36

@Secret How can strongly even set can not be weakly even? let $f:A\to B$ be bijective and $D=A\cup B$ d-finite, then $D=\bigcup_{a\in A}\{a,f(a)\}$

Secret

ok I never thought I can pair them that way. I really need to spice up my ability to find bijections

right so even if $A,B$ is strongly or weakly odd, this will still work, thus all strongly even are also weakly even

Holo

And we can do the same with strongly odd

$D=\{b\}\cup\bigcup_{a\in A}\{a,f(a)\}$

Secret

Meanwhile I am still trying to prove there are no d finite sets that is both weakly even and weakly odd. Somehow I have to show there is no bijection between $\{b\} \cup \bigcup_{a \in A} \{a,f(a)\}$ and $\bigcup_{a \in A} \{a,f(a)\}$. I mean, writing it this way it is obvious there cannot be a bijection, but what exactly prevent us to split up uncountably many pairs to allow us to pair up that singleton?

Holo

13:57

Hmm, I think this is a bad notation @Secret

You should use $\bigcup\{a,b\}$ and $\{b\} \cup \bigcup\{a,b\}$, because the existence of $f$ is a choice function to the family $\{a,b\}$ indexed by $a$(which we can't say whether or not it exists)

@Secret : assume that $D$ is weakly even and $C$ is weakly odd, then there exists bijective $f$. Let $C=\{c\}\cup\bigcup\{a,b\}$, then what can we say about $\bigcup\{f(a),f(b)\}$?

Secret

just to check, $f$ is given as a bijection between $D$ and $C$?

Mats Granvik

How many equally sized tetrahedrons do you need to construct a cube?

Holo

@Secret Yes

I should use new names instead of using $f$ always, this is the curse that analysis course cast on me

Secret

$\bigcup\{f(a),f(b)\}$ is a union of pairs, thus it is weakly even, and each of these pairs uses elements in $C$ which is weakly odd, hmm...

Holo

And because $|\bigcup\{f(a),f(b)\}|=|\bigcup\{a,b\}|$ then $|\bigcup\{f(a),f(b)\}|<C$

As C is d-finite

Now I want to show that $\bigcup\{f(a),f(b)\}=D$ but I don't know if we can or how

Secret

14:13

how do we conclude $|\bigcup\{f(a),f(b)\}|=|\bigcup\{a,b\}|$, isn't one of these $f(a), f(b)$ have to be equal to $c$ by our assumption of $f$?

Holo

No, $f$ is bijective from $\{c\}\cup\bigcup\{a,b\}$, so $g\equiv f\restriction_{\bigcup\{a,b\}}$ is also bijective, now I am looking at $g[\bigcup\{a,b\}]=\bigcup\{f(a),f(b)\}$($f$ is bijective from $C$ to $D$, not from $D$ to $C$)

Secret

ah I see

Holo

The problem is that I don't know what to do now... all I can think of is to create descending sequence to show $1=0$ and thus contradiction, but because we are working with d-finite we can't define such sequence

Secret

yeah because it can be infinitely decreasing

Holo

We also don't have choice so we can't use Zorn's lemma

Secret

14:19

I am also wondering what $f(c)$ equals to since by assumption $f$ bijects between $C$ and $D$

besides the step in proving $\bigcup\{f(a),f(b)\}=D$

hmm...

Holo

@Secret Well, we don't know the nature of the bijective, if we knew then we could find the nature of $C$, which it is impossible

@Secret Wait no, we should prove that $|\bigcup\{f(a),f(b)\}|=|D|$, because the bijective needs no to be unique

Secret

right, forgot the absolute signs

Leaky Nun

hi @WillHunting

Will Hunting

@LeakyNun Hi Kenny.

Secret

@KennyLau There are two different notions of "even" for arbitrary sets - partitionable into pairs, and partitionable into two equinumerous sets. With choice these are the same, but without choice the former is strictly weaker than the latter (this is just the idea of Russell's shoes and socks). The arithmetic structure on Dedekind-finite sets, assuming that they are linearly ordered, corresponds to the strong notion of evenness (etc.), so every Dedekind-finite set in such a model is either strongly even or strongly odd. — Noah Schweber Sep 16 '17 at 19:19

Manolis Lyviakis

14:25

im taking a graduate course on introduction to differentiall manifolds and the proff said we can do a presentation to a subject if we want for extra grade. Any topic that combines Algebra+Manifolds?(like lie groups or something)

Holo

@Secret do you think or know that weakly even can not be weakly odd?

Secret

Well, according to Noah's comments, he said you cannot have d finite sets that are both even and odd in any way

which is why I became curious on how to prove it

Holo

Where do you see it?

Secret

digging that up, give me a sec (I linked that before but I need to find it again)

Secret

14:45

34

A: What's between the finite and the infinite?

There's a few things I can think of which might fit the bill: We could work in a non-$\omega$ model of ZFC. In such a model, there are sets the model thinks are finite, but which are actually infinite; so there's a distinction between "internally infinite" and "externally infinite." (A similar ...

> for instance, you can show that a Dedekind-finite set can be even (= partitionable into pairs) or odd (= partitionable into pairs and one singleton) or neither but not both.

so Noah is saying we can prove there exists no d finite set that is both weakly even and weakly odd

3

A: Existence of sequence converging to infimum in a choiceless universe

Yes, some choice is needed. Suppose that $A\subseteq(0,1)$ is a dense Dedekind-finite set (this was shown consistent with the failure of the axiom of choice by Cohen). Now $\inf A$ is $0$, but since any decreasing sequence from $A$ is eventually constant, so there is no sequence in $A$ convergi...

Ok I made a wrong comment above. We cannot have a sequence because then we have a countable subset

so any attempt to build a sequence it can only have finite number of terms

not sure how that will help us yet though

Holo

I don't think it will, I am trying to think about back and forth kind of proof but nothing pops to my mind

mercio

@MatsGranvik 3 ?

Holo

The problem, I have no intuition about d-finite sets...

tarit goswami

15:39

If $G$ is not an abelian group, there is an orbit of order at least $2$. Is this statement true?

Leaky Nun

@taritgoswami yes, for a certain interpretation

tarit goswami

Ooh I forgot to mention the action

Here $G$ is acting on itself, and the action is conjugation, means $g\ast a=gag^{-1}$

Leaky Nun

then yes

tarit goswami

@LeakyNun can you explain it?

Leaky Nun

it's just unfolding definitions

tarit goswami

15:53

Ok got it. If $G$ acts on itself by action of left multiplication, can we always say that - There exists an abelian group G such that there are exactly 2 orbits ?

Li357

I have a quick question. If I'm proving an iff statement (say $A \Longleftrightarrow B$), can I use information I gained from $A \Longrightarrow B$ in proving $A \Longleftarrow B$?

Leaky Nun

@Li357 yes

Li357

Okay, thanks!

Leaky Nun

@taritgoswami I'm confused.

tarit goswami

@LeakyNun Me also.

Li357

15:55

Wait, I had the symbols messed up...

Leaky Nun

@taritgoswami ...

I'm confused by your question because you worded it poorly.

tarit goswami

@LeakyNun OK, I mean consider that $G$ acting on itself by left multiplication, means $g\ast a=ga$ where $g,a\in G$. Now is this statement always correct - "There exists an abelian group G such that there are exactly 2 orbits" ?

Leaky Nun

I don't understand what you mean by "always"

also you introduced a group $G$ and then introduced the action, and now how can you change the group again

tarit goswami

means for every group $G$

Leaky Nun

you asked about whether there exists

tarit goswami

15:59

yes

Leaky Nun

then how can you talk about "for every"

your use of quantifiers is very confusing

the action of a group on itself by left multiplication is always transitive, so there is always only 1 orbit

tarit goswami

Sorry for that, omit the "always" , is it ok now?

Leaky Nun

yes

tarit goswami

@LeakyNun is there any theorem ? I haven't studied that until now..

Leaky Nun

again, you should unfold definitions and prove it yourself

Rudi_Birnbaum

16:05

@mercio the volume would confirm it at least.

LegionMammal978

@LeakyNun Ah, thanks, that makes sense

Secret

17:18

$\omega_1$ really deserve that dramatic music

In ZFC, it is the 2nd ordinal where you need a net with the domain of the same cardinality as itself to reach it

So, if $\omega_1$ is a tower of the same height, approaching it at any $\aleph_0$ steps the tower won't get into view, but should you move as fast as $\omega_1$ you will be greeted by a view on how mindboggingly tall it is at an instant

Mancala

Can anyone explain to me the intuitive difference of an Embedding for a Proper Immersion?

Ted Shifrin

@Mancala: An immersion might not be one-to-one.

Secret

boy surface is a classic example

it has a triple point

Leaky Nun

@Secret take the ordinal 7

Faust

@TedShifrin

Secret

17:24

::secret just got pwned by Leaky::

Ted Shifrin

hi @Faust

mercio

you could make that pic with any number of countable ordinals like ω^ω or ε0 in place of ω1

Faust

i'm probably gonna be gone for awhile, not sure ill ever come back so wanted to say goodbye.

Thanks for all your help ^^

Ted Shifrin

oh no, @Faust.

Secret

@LeakyNun The lesson: Do not underestimate the weirdness of finite sets

Ted Shifrin

17:28

Why won't you come back?

Please keep in touch on email, at least.

Please get healthy, man.

Faust

My life got kinda complicated, i dont think ill be able to do math anymore

Mancala

@Secret A boy's surface is an example of proper immersion that is not an embedding?

Secret

nope

Ted Shifrin

@Faust: Please communicate with me by email. I hope you'll be OK.

Secret

because it self intersects, meaning there are points that get mapped to the same point in the image

Ted Shifrin

17:29

@Mancala: Any time you have a compact manifold, properness is automatic.

Faust

Ok, i should be ok eventually i will keep in touch via email.

Ted Shifrin

Mapping a circle to a figure 8 is the easiest immersion. You can see it is not one-to-one.

@Faust: OK. I am worried for you now, so please keep in touch.

Secret

Ok I don't know the difference between immersion vs proper immersion

I knew an embedding has to be one to one

Ted Shifrin

When you're mapping on a compact space, properness is automatic.

But you do need to check that you have an immersion, Secret.

Secret

17:42

(Unrelated) You cannot divide by zero if you are using a field, a ring, a semiring, some near rings etc. You need to go really hardcore to either redefine distributive law (wheels), or blow up associativity (and I mean, BLOWING UP everything except power associativity) to even get not boring structure of them. Pretty much the most pathological algebraic structure that can exist just before one start to put heavy doses of set theory, category theory and other really really abstract nonsense stuff

and the surreals is a field that is a proper class

and finally, no love from Vihart about the ZF infinite sets and the ZFC nonmeasurables

Ok I ramble too much, better do some chemistry

Balarka Sen

17:59

Hi @Ted

Ted Shifrin

@Balarka!!!

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$Mathematics$ Mathematics