Mathematics - 2017-10-14 (page 1 of 4)

Leaky Nun

00:00

that means A normalizes B

Kasmir Khaan

hmm let me think one second

does this mean that AB = BA?

right?

N_G(B) = {g in G such that gBg' =B }

so in our case aBa' = B for all a in A

Leaky Nun

sure

Kasmir Khaan

so in other words

we define a map from A to AB/B

whose kernel is A intersect B

phi : A --> AB/B , ker (phi) is A intersect B

Need to check that all of the pieces are wel defined

Leaky Nun

ok

Kasmir Khaan

@LeakyNun i have a question =p

would it be correct if we had this

AB/A isomorphic to B / (A int Bb)

assuming that A, B are normal

Leaky Nun

00:15

sure

Kasmir Khaan

hmm nice =p

Can you give me a question so i try to solve?

either with use of first iso or second

Leaky Nun

I don't have any in mind

oh wait I do have

Kasmir Khaan

:D

Leaky Nun

Let $G = \Bbb R \times \Bbb R$ under addition.
(a) Show that $H = \{(x,y) \mid y=2x\}$ is a subgroup of $G$.
(b) Show that $G/H \cong \Bbb R$.

Kasmir Khaan

okay working on it now :D

Kasmir Khaan

00:30

@LeakyNun is the operation on R addition as well ?

Leaky Nun

@KasmirKhaan yes

Kasmir Khaan

okay so far i proved H is a normal subgroup

G/H makes sense with the group operation

I want to find a map that sends ( x,2x) to 0

let phi : RxR --> R

defined by ( x,y) |--> 2x-y

Leaky Nun

@KasmirKhaan good

Kasmir Khaan

that has as a kernel the elements of the form (x,2x)

:D

@LeakyNun so this is how we use those theorems :D

Leaky Nun

yes

Kasmir Khaan

00:33

then i need to show that phi is an isomorphism

hom + bijective

to complete the proof =p

Leaky Nun

think again

don't get carried away

Kasmir Khaan

okay hmm

well a found a map phi and its kernel is H

to G/H is isomorphic to R

by first iso theorem

I just found a map did not check yet it is a hom

and bijective

Hmm what is it am missing?

Leaky Nun

you said phi is bijective

Kasmir Khaan

hmm right that is not correct

let me think =P

Leaky Nun

(c) Explain what you have proved geometrically.

Kasmir Khaan

00:43

so the arguemnt should be

since phi is a surjective hom from G--> R

then there exists and map phi bar, from G/ker (phi) to im (f) such that f bar is an isomorphism

Steve

hi, need a quick hint there. consider the identity (z^5-1)/(z-1) = 1+z+...+z^4, how can we obtain a trig identity of cos(pi/5)sin(pi/10)=1/4?

Leaky Nun

@KasmirKhaan yes

Leaky Nun

00:57

@Steve Let a=sin(pi/10). We seek to find (1-2a^2)a = a-2a^3.
16a^5 - 20a^3 + 5a = sin(pi/2) = 1
16a^5 - 16 = 20a^3 - 5a - 15 = (a-1) (20a^2 + 20a + 15)
16(a^4 + a^3 + a^2 + a + 1) = 20a^2 + 20a + 15
16a^4 + 16a^3 - 4a^2 - 4a + 1 = 0
(4a^2 + 2a - 1)^2 = 0
4a^2 + 2a - 1 = 0
-2a^3 + a = (4a^2 + 2a - 1) (-0.5a + 0.25) + 1/4

sorry for solving your whole question :P

Faust

TACO

Leaky Nun

@Faust ?

Steve

@LeakyNun　thank you for that

Faust

@LeakyNun

Leaky Nun

@Steve the fact that sin(pi/10) is the root of a quadratic equation definitely amazed me

Steve

01:01

@LeakyNun　is there a way we can find this by complex numbers?

Leaky Nun

good question

anon

01:36

@Steve @LeakyNun Let $\xi=\exp(\pi i /5)$, a $10$th root of unity.

On the one hand, $$ \begin{array}{ll} \cos(\pi/5)\sin(\pi/10)& =\cos(\pi/5)\cos(\pi/2-\pi/10) \\ & =\cos(\pi/5)\cos(2\pi/5) \\ & = (\xi+\xi^{-1})/2\cdot(\xi^2+\xi^{-2})/2 \\ & = (\xi^3+\xi+\xi^{-1}+\xi^{-3})/4. \end{array} $$

On the other hand, $$ \begin{array}{ll} 0 & = \displaystyle \frac{\xi^{10}-1}{\xi^2-1} \\ & =1+\xi^2+\xi^4+\xi^6+\xi^8 \\ & =1+\xi^{-5}(\xi^3+\xi+\xi^{-1}+\xi^3) \\ & = 1-4\cos(\pi/5)\cos(2\pi/5). \end{array}$$

Leaky Nun

@anon damn I explored so long lol

oh that was clever as hell

Leaky Nun

02:09

sin(3pi/10) = 3a - 4a^3
We aim to find:
(sin(3pi/10) - sin(pi/10))/2
= Im(e^(3ipi/10) - e^(ipi/10))/2
= Re(e^(18ipi/10) - e^(16ipi/10))/2
= Re(e^(9ipi/5) + e^(3ipi/5))/2
= (e^(9ipi/5) + e^(7ipi/5) + e^(3ipi/5) + e^(ipi/5))/4
Let z = e^(ipi/5), the primitive 10th root of unity.
We are finding (z+z^3+z^7+z^9)/4.

Note that z, z^3, z^7, z^9 are the 4 roots of the 10th cyclotomic polynomial x^4 - x^3 + x^2 - x + 1.
By Vieta, their sum is 1.
Therefore, the required number is 1/4.

@anon after an hour of exploration I came up with what is essentially the same as your solution...

Leaky Nun

03:09

@Faust hi

Faust

hows leaky?

Leaky Nun

reading Galois at 4AM

Faust

anything ground breaking

Leaky Nun

not really

Faust

do u know the lucas lehmer test?

Leaky Nun

03:16

I don't

Faust

blarg

Leaky Nun

I just searched it now

0celo7

more algebra

Leaky Nun

guten Tag

0celo7

it's abend here

Leaky Nun

03:18

guten abend

0celo7

trying to decipher a paper

Leaky Nun

welche paper?

0celo7

arxiv.org/abs/1703.02619

Leaky Nun

runs away

0celo7

it's easier than that algebra crap above

Leaky Nun

03:19

definitely not

0celo7

definitely yes

Leaky Nun

nope

0celo7

I invoke "I am correct because I am older"

Leaky Nun

I invoke "I understand that algebra crap but not your paper"

0celo7

geometric flow theory is half handwaving anyway

Leaky Nun

03:21

explain it to me

0celo7

protip: place spheres appropriately and use the comparison principle

Leaky Nun

explain like I knew nothing

[I really know nothing]

0celo7

Alright

So we start with the empty set

Leaky Nun

If I can understand your explanation now at 4AM then you win

0celo7

Then we make 0 by taking a set containing the empty set

That might be wrong

Leaky Nun

03:23

0 = {{}}

0celo7

But then we make 1 by taking 0 and the empty set and putting that in a set

etc

Semiclassical

starting really from he basics, eh

0celo7

then you take quotients

Leaky Nun

so ZF ordinals but shifted by 1

0celo7

then cauchy sequences

Leaky Nun

03:23

oh come on

not from the basics

0celo7

ok so we just constructed R (suck it Wildberger)

Semiclassical

lol

Leaky Nun

lmao

Semiclassical

think you got trolled bro

Leaky Nun

@0celo7 come on

Semiclassical

03:25

moar liek 0cetroll

0celo7

@LeakyNun you take a compact hypersurface and flow it by its mean curvature in the inward normal direction

Leaky Nun

what is flow

0celo7

$df/dt=stuff$

some evolution equation

Leaky Nun

...

Yashas

How do I approach solving the following problem:

n! < (n + 1)2^n

I need to find where the inequality fails

0celo7

03:29

You have an immersion $X(\cdot,t):M^n\to\Bbb R^{n+1}$ and you define the flow $$\frac{\partial}{\partial t}X=Hn$$where $H$ is the mean curvature and $n$ is the unit inward normal vector

Yashas

The answer is 6 but I need to prove it

0celo7

@Yashas compute both sides for 6?

Leaky Nun

@Yashas 6! = 720 and (6+1)2^6 = 7*2^6 = 448

720 < 448 is false

Yashas

I am not allowed to do substitutions

Leaky Nun

(you only need one counter-example to disprove the claim)

@Yashas that's ridiculous

0celo7

03:30

huh?

Yashas

It isn't a claim

The question is to find in what range the inequality holds

0celo7

what is a good bewildered emoji/emoticon

Leaky Nun

you substitute to check that it fails

@0celo7 @_@

0celo7

@LeakyNun no, that's "helplessly confused"

Leaky Nun

@Yashas so you should have phrased it better

you need to find where the inequality holds

Yashas

03:30

I need to find the range where it holds. Not check if it is always true or not

Yes sorry

Faust

lol

Yashas

n! increases much faster than 2^n

So the two graphs will intersect at most once

Leaky Nun

nope, it still fails for 7

Yashas

Yes it fails for every number larger than 5

Leaky Nun

@Yashas that isn't what you said

Yashas

03:32

The answer is 1 to 5 inclusive

Leaky Nun

The fifth panel also applies to postmodernists.

0celo7

@LeakyNun Kevin wants to show that for initial data "close", the time of first singularities must be "close"

Yashas

I need to prove it mathematically

Leaky Nun

@Yashas ^^^

Semiclassical

@LeakyNun ahh, that old standard

Leaky Nun

03:33

> I need to prove that for every n>5, n! > (n+1) 2^n

would have been a much better question

for which the answer would be "induction"

Semiclassical

now I find myself wondering how a proof sans induction would work, heh

Yashas

I guessed 6

Semiclassical

not very well, i imagine

Yashas

I am supposed to find that 6 without guessing

Leaky Nun

@Yashas no you aren't

0celo7

03:35

@Semiclassical the best proofs are proofs that you don't think use induction but really do because induction is so natural

Semiclassical

heh, like proofs without words?

0celo7

No, those are awful

Semiclassical

psh

0celo7

Any proof that contains something like "and so on"

Semiclassical

hmm

0celo7

03:36

Sometimes the induction is so natural that it doesn't have to be stated

Then there are other induction proofs, like Sard's theorem...

Leaky Nun

The number of subsets of a set of n elements is 2^n, because when you start with 0 elements there is only 1 subset; and when you add 1 element, you double the number of subsets.

There, implicit induction

Semiclassical

I think it depends what you're proving.

the inductive proof of, say, 1+2+3+...+n = n(n+1)/2 is pretty boring

Yashas

What if the inequality started failing at 58695369556

Leaky Nun

@Yashas it won't

Yashas

I wouldn't be able to find that number

Semiclassical

03:39

whereas Gauss's observation that (1+2+...+ n) +(n+(n-1)+...+1) = (n+1)*n is quite nice.

0celo7

@Semiclassical Or the positive mass theorem from GR. Lee and Parker reduced it to induction but the proof of the inductive step was folklore until earlier this year.

Yashas

I know how to prove it with induction after finding where the inequality fails but I need to find the number first

0celo7

It was an open problem for 30 years

Leaky Nun

@Yashas you know that the starting point would be small

since their growth rate differs by like a lot

Yashas

Yes but I am not allowed to do that

0celo7

03:40

You are supposed to show that an inequality is false without a computation?

Yashas

Yes

Semiclassical

you really are. you take a certain n as your base case for the inequality failing, and use induction to show that it indeed fails for all cases above this.

0celo7

fail the course then

@Semiclassical uh

it's true for the large cases

Semiclassical

durp

i though it was the other direction -_-

anyways. once you've checked that it's true for all n>5, then there's only 5 cases remaining

so check those and confirm that the inequality fails for all those cases.

0celo7

@Semiclassical he's not allowed to do that

Semiclassical

03:43

riiight

0celo7

@LeakyNun so this paper has several nontrivial arguments, and some new methods

Semiclassical

i suppose a slightly interesting question would be to consider the inequality n! > a*p(n)*c^n for a given polynomial p(n) and c>1

0celo7

one person proposed to use a more complicated but better understood surgery method

Semiclassical

and ask for the largest n for which it's false.

0celo7

I'm taking a look at the actual paper because the talk was...unconvincing

Semiclassical

03:47

but, eh, i'm not sure one can give an interesting estimate for that

oh. i guess the a is superfluous since it can be absorbed into p(n)

Secret

[Random]

2

Q: Is a Thomson's lamp physically realistic?

The Thomson's lamp is a philosophical puzzle that can be describe as follows: Consider a lamp with a toggle switch. Flicking the switch once turns the lamp on. Another flick will turn the lamp off. Now suppose that there is a being able to perform the following task: starting a timer, he turns t...

Now can someone came up with a quantum Thomson lamp

Semiclassical

sounds like it'd be a time-dependent quantum mechanics problem

and those problems are usually more annoying

Faust

quantum

funny word

Secret

My guess it should end up similar to the classical case, since the flipping of the state is effectively performing countably many measurements

so by the logic of the PSE, it should average out

A supertask is basically asking whether bounded countable sequence of events in time is physically possible

Faust

04:09

math.stackexchange.com/questions/2471511/… if anyone's bored

Silent

@LeakyNun I first thought that it is true for every metric space, but then I got to know that even for Hausdorff space, it is true. Is this true more generally?

Leaky Nun

53

A: Intersection of finite number of compact sets is compact?

For Hausdorff spaces your statement is true, since compact sets in a Hausdorff space must be closed and a closed subset of a compact set is compact. In fact, in this case, the intersection of any family of compact sets is compact (by the same argument). However, in general it is false. Take $\ma...

0celo7

Just never work with a non-Hausdorff space

is that bad german?

Leaky Nun

what

no it isn't

0celo7

I have no idea what it means

Leaky Nun

04:17

wait

it was just bad translation :P

0celo7

um, ok

Leaky Nun

@0celo7 hau ab

0celo7

@LeakyNun That's Not Nice.

Leaky Nun

@0celo7 :c

Secret

I need non Hausdorff spaces on a daily basis

Leaky Nun

04:23

lol

Secret

its the only way to model time travel

Leaky Nun

@Secret I thought you're a chemist

Secret

I am, but I like anything weird

thus I spent heaps on time on things people said weird

Foundations of mathematics will help me to understand infinity

and once that is out of the way, half of mathematics should become easier

(But I do agree, I think I spent way too much time on things not related to my work, oh well...)

Silent

@LeakyNun Wow thanks for that link!

Leaky Nun

no problem

Semiclassical

04:27

Memo to self: persuade/trick Secret into learning resurgence theory.

Secret

Luckily, Cantor did most of the hard work thus there is no danger of going insane from touching infinity

Daminark

Oi how's it going?

Leaky Nun

@Semiclassical what is resurgence theory?

0celo7

@Secret Learn Ricci flow with surgeries and teach me please

Semiclassical

it's basically asymptotic analysis on steroids.

0celo7

04:28

@Semiclassical same...ish?

Leaky Nun

@Semiclassical steroid?

Semiclassical

a performance enhancing drug, basically

Leaky Nun

asymptotic analysis?

0celo7

wot

you don't know what steroids are?

Leaky Nun

@0celo7 I know, but I thought it would have another meaning because "asymptotic analysis" looks like maths

Semiclassical

04:30

it's an idiom

0celo7

"X on steroids" is a common saying...

Semiclassical

"used to suggest a highly exaggerated, enhanced, or accelerated version of something."

Leaky Nun

@Semiclassical oh, lol

Secret

How much differential geometry background I need to do Ricci flow stuff...?

0celo7

hard to say

Secret

04:32

I still have hardly any given how much I suck at analysis

Leaky Nun

it's 5:32 AM and i'm doing galois

Semiclassical

whyyyy

Leaky Nun

because I'm trying to solve a problem I set for myself

Secret

There's a more practical reason why I want to understand infinity: Limits are my worst enemy

abstract algebra is slightly more tractable because there isn't much epsilon delta stuff

Semiclassical

nCatlab has a page on resurgence theory: ncatlab.org/nlab/show/resurgence+theory

Daminark

04:33

@Secret woo!

Semiclassical

though, given that it's got an nctablab page but not a Wikipedia page

that probably tells you how esoteric it is

0celo7

@Semiclassical Oh, so you're working on $(1,\infty)$-topoi of bundle functors.

Semiclassical

lol

Leaky Nun

... more than half an hour was spent dealing with apparent contradictions due to my misdiagnosis of the Galois group of $x^4-4x^2+2$ over $\Bbb Q$ as $\Bbb V_4$ rather than $\Bbb Z_4$

Semiclassical

i'm referencing ncatlab, so that must be right

Secret

04:35

epsilon delta stuff makes a lot of things happen at a very short time. Given how much I suck at driving (an example of a task that requires rapid multitasking) and drove to the opposite lane, you can imagine why I suck at analysis

0celo7

@Secret you're a stereotypical bad asian driver?

Secret

Abstract algebra does not require a lot of rapid time decisions, thus I can take time to figure how to solve it

Leaky Nun

no, that isn't even right

0celo7

I've read this lemma 5 times and still don't get it

Leaky Nun

gg I can't even determine the Galois group of $x^4-4x^2+2$ how am I going to solve my problem

0celo7

04:37

it might be time for me to retire

Secret

@0celo7 well it has to improve soon else I will be in trouble as driving is essential in the western world

Leaky Nun

@0celo7 which lemma?

Semiclassical

now i wish i knew galois theory, if only so that i could pull a "oh, of course it's Z4" without talking out of my arse

0celo7

2.1

I know what it's saying

but I also don't

I'm tired

Semiclassical

@0celo7 [insert schrodinger cat joke here]

Leaky Nun

04:38

I don't think Z4 is right either

now I'm thoroughly confused

Secret

Hmm, a brief look at the ncat article suggest I am not ready for resurgence theory because it involve resummation of an infinite series, and I am still not very good at infinite series yet despite the many practice with waiting and co.

while I know a bit more about special functions now, it is still not enough

Semiclassical

tbh the only resurgence theory stuff i understand is zero-dimensional in nature

Leaky Nun

15

A: Galois Group of $\sqrt{2+\sqrt{2}}$ over $\mathbb{Q}$

$\alpha^2-2=\sqrt 2\in F$ by closure of multiplication and addition. $$\frac{\sqrt 2}{\sqrt{2+\sqrt{2}}}=\frac{\sqrt 2 \cdot\sqrt{2-\sqrt 2}}{\sqrt{2+\sqrt{2}}\sqrt{2-\sqrt 2}}=\frac{\sqrt 2\cdot\sqrt{2-\sqrt 2}}{\sqrt{4-2}}=\sqrt{2-\sqrt 2}$$ Since $F$ is a field, it has multiplicative inverse...

So Z4 it is

Semiclassical

which is a problem given that it's apparently in QFT where people really want to make use of resurgence theory

and that's not zero-dimensional, therefore I don't understand it :)

Secret

They want to tame those infinities

Leaky Nun

04:43

gg it's already 05:43 AM now

Semiclassical

something like that

0celo7

@LeakyNun no use sleeping now

Leaky Nun

@0celo7 why?

Semiclassical

maybe a better way to put it is that they know how to tame the infinities when you're doing perturbative calculations. but if you want to capture nonperturbative phenomena ten that doesn't really help you

0celo7

it's morning

Leaky Nun

04:44

@0celo7 well it's saturday

I sleep whenever I want

@Semiclassical @Perturbative

Secret

my current maths focus in my self study is type theory. That will mean I might soon end up in that mclane book thus if the trend goes like that, I might have enough background to understand most of the ncat articles

This is because homotopy type theory and intuitionist type theory are closely tied to categories

But again given I need to balance my chemistry research I need to figure out how to do them both together without sacrificing the quality of either

While I had most of my jobs automated, Phase III involves 500+ calculations that are not very similar thus hard to write a template to automate them

inmight try that attempt out later in the python code to see if I can automate them...

It is possible I might need another interface to store my data, as excel is currently too laggy

The chemistry is interesting, but not that computer infrastructure

which is why I am kinda sluggish in the research.

There are already 6 calculations done in Phase III, but this has to be sped up in order to caught up lost time

Phase IV will involve transition states (saddle point) calculations, which will be a bit more difficult but manageable

Adter that, there should be enough data to establish a trend, and then the wet lab component will begin

Leaky Nun

In a group of order $2^m 3^n$ must the 3-subgroup be normal?

0celo7

Use Sylow

Faust

@LeakyNun wth is it that you want $s^9 \equiv 0 $ ?

Leaky Nun

04:59

@Faust I said $s_9$

Faust

why 9?

Leaky Nun

11-2

Faust

that seems super arbitrary

why 11-2

Leaky Nun

p-2

read wiki

Faust

@LeakyNun oh thats what it says

Silent

05:01

What stronger thing does Rudin show by not simply applying Archimedean property as claimed here?

Secret

05:18

> There exists a model of ZF¬C which has an infinite set of real numbers without a countably infinite subset.

Reals without singletons?

I think that's the easiest example of an infinite dedekind finite set

Leaky Nun

@Secret there is singletons of course

but they aren't countably infinite

@Secret yes, that

Secret

Wait, so there are finite number of singletons in such models?

Leaky Nun

@Secret infinitely many, I think

you just can't group them into one set, I think

or if you could, the set cannot be countable, I think

Secret

also how can such reals contain the naturals since the naturals are countable?

Leaky Nun

@Secret so they can't

maybe you misread the statement

it means infinite subset of the real numbers

Faust

05:36

anyone know of a sequence that has a convergence sub-sequence to every real number?

Leaky Nun

@Faust take an enumeration of $\Bbb Q$

Faust

i know that one

i wanted something i could write down

Leaky Nun

you can write down an enumeration of $\Bbb Q$ lol

Faust

well you can draw a funny arrow along Q

Leaky Nun

@Faust $(\tan n)$

should do the job

Faust

05:38

that was my next thought

Leaky Nun

@Faust good

Faust

lol

Leaky Nun

SLOPPY APPLICATION OF GALOIS THEORY - too ashamed to sign my name here

We know that $\sqrt{2+\sqrt{3}} = \sqrt{\frac12(4+2\sqrt3)} = \frac12(\sqrt2+\sqrt6)$. Then, it is only natural to ask whether $\sqrt{2+\sqrt2}$ can be expressed similarly.

DEFINITION OF PROBLEM:
We wish to express $\sqrt{2+\sqrt2}$ as sum of roots of rational numbers, i.e. find a (normal and separable) field $L=\Bbb Q(a_i,a_2,\cdots,a_n)$ such that for each $i$ there exists $j=b_i$ with $a_i^j = c_i \in \Bbb Q$, and that $\sqrt{2+\sqrt2} \in L$.

@MatheiBoulomenos

Secret

06:06

Actually, thinking about it, the set of irrationals can be infinite dedekind finite subset of the reals. This is because lacking axiom of choice, we cannot inject the reals to it

moreover, we can generate a family of infinite dedekind finite sets by union of the irrationals to a finite collection of rationals

and they will be strictly not smaller than the previous ones

MatheiBoulomenos

06:18

chat.stackexchange.com/transcript/message/40539168#40539168 this doesn't work if $n_7=15$, the normalizer of a 7-sylow has 28 elements

@LeakyNun

Alessandro Codenotti

@Secret why can't we do that?

MatheiBoulomenos

06:40

@LeakyNun there's no reason to expect that such an expression exists for $\sqrt{2+\sqrt{2}}$. It follows from Kummer theory that such an expression must exists over $\Bbb Q(i)$, however. (Well $\sqrt[4]{i}=\zeta_{16}$). Note that the case would be different if we had that Galois group $V_4$. For the roots of polynomial whose Galois group is $(\Bbb Z_2)^{n}$ for some $n$, there exists always an expression as a rational linear combination of square roots

Daminark

06:55

What is Kummer theory?

MatheiBoulomenos

It describes the possible form of certain Galois extensions

Daminark

I see, that sounds nifty

MatheiBoulomenos

The basic theorem of Kummer theory is that if $n \cdot 1 \neq 0$ in $K$ and $\zeta_n \in K$, then a Galois extension of $K$ is of the form $K(\sqrt[n]{a_1}, \dots, \sqrt[n]{a_k})$ iff the Galois group $G$ is Abelian and satisfies $g^n=1$ for all $g\in G$

This is used in the proof that there are polynomials of degree $\geq 5$ whose roots may not be expressed as radicals (at least that's the proof I learned)

Daminark

The proof I have heard of (though I dunno any algebra yet) is that your sorta look at the Galois group of the field extension associated with the polynomial and see if that's solvable

MatheiBoulomenos

Yes, what I meant was the proof that the roots may be expresed as radicals iff the Galois group is solvable

Daminark

07:18

I see

Anyway, good night!

MatheiBoulomenos

Good night!

Secret

@AlessandroCodenotti Uh if both sets lack a countable subset, then in the absence of a choice function, there should be no way to write down an injection from one to another since they are sets that don't inject into the naturals?

Without axiom of choice, it should not be possible to pick out a countable sequence in the set of irrationals?

Alessandro Codenotti

07:56

$\pi+n,n\in\Bbb N$ seems like a countable sequence

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