Mathematics - 2020-03-16 (page 1 of 2)

Simple

00:21

Let $\lambda$ denote Lebesgue measure on $R$. Suppose $f : R \to R$ is a Borel measurable function such that $\int|f|d\lambda<\infty$. Then $\lim_{k\to\infty}\int_{[-k,k]}f=\int\,fd\lambda$

$E=\cup[-k,k]$, if $f$ is a simple function, by additiviy, we are done

Leaky Nun

00:35

DCT

Thorgott

DCT is the more economic option. An approximation argument works too: You first need to prove it for characteristic functions (Hint: continuity from below), then it follows for simple functions by additivity indeed and the general case follows by approximation.

Simple

I see. but not quite sure how to apply DCT, I don't see any pointwise convergent sequence

Thorgott

What's the definition of $\int_{[-k,k]}f$?

Simple

$\int\,\chi_{[-k,k]}f$

Thorgott

Indeed, do you see a sequence of functions here?

Simple

00:52

$\chi_{[-k,k]}f$?

Thorgott

Yup, does it convergence to anything pointwise?

Simple

it converges to $f$ since we can split $[-k,k]$ to many subsets

which is a approximation

Thorgott

the conclusion is true, but I don't understand that argument

Simple

$\chi_{[-k,k]}f=\sum\alpha_i\chi_{A\cap[-k,k]}$

Thorgott

01:20

I don't follow

Simple

then, I am not sure.

Adam L

04:40

Well. This is becoming a non event

Guru Vishnu

06:52

Hi. I found two questions on the main site which seem to ask about the same thing. Is it ok to flag the new question as a duplicate of the older one which is well-explained? I'm having this doubt because the "new" one was asked 5 years ago and wasn't closed.

Old one (original): https://math.stackexchange.com/q/11307/693070
New one (maybe a duplicate): https://math.stackexchange.com/q/1054086/693070

Ante

07:04

Is someone skilled in elementary number theory?

Balarka Sen

08:12

Someone must be, yeah

Leaky Nun

I guess Ramanujan would have been

@BalarkaSen wanna play?

Alessandro Codenotti

@BalarkaSen this is not the philosophy room

Ante

How to partition $\mathbb R$ into uncountably many uncountable sets?

Alessandro Codenotti

Pick a bijection $f:\Bbb R^2\to\Bbb R$ and look at $f(\{x\}\times\Bbb R)$

Balarka Sen

@LeakyNun Fuck I dunno I have forgotten chess

Ante

08:25

@AlessandroCodenotti Must that bijection be everywhere discontinuous?

Balarka Sen

@Alessandro Big if true

Leaky Nun

@BalarkaSen let's pick it up

Balarka Sen

i'm going home for two weeks because of corona, lets do it after i arrive there

Leaky Nun

sure

Alessandro Codenotti

@Ante hmm not sure

I can easily show that it cannot be continuous on an open set, but at a few isolated points? Dunno

Ante

08:33

yes, we are trying to extend theorem about mappings now with that question of mine

from $\mathbb R^n$ onto $\mathbb R^m$

bijections

Leaky Nun

space filling reacc only

Ante

it was something Cantor and Brouwer worked on, right?

@AlessandroCodenotti

about impossibility of continuous bijections

but can they be continuous at at least few points

let us solve that here and write a paper! :D

it it´s not known

@LeakyNun what would fill such bijection if continuous at one point?

Alessandro Codenotti

@LeakyNun those are not injective

Leaky Nun

oh right

Baire reacc only

open mapping reacc only

Ante

Osgood curve

In mathematics, an Osgood curve is a non-self-intersecting curve (either a Jordan curve or a Jordan arc) of positive area. More formally, these are curves in the Euclidean plane with positive two-dimensional Lebesgue measure. == History == The first examples of Osgood curves were found by William Fogg Osgood (1903) and Henri Lebesgue (1903). Both examples have positive area in parts of the curve, but zero area in other parts; this flaw was corrected by Knopp (1917), who found a curve that has positive area in every neighborhood of each of its points, based on an earlier construction of Wa...

Balarka Sen

08:42

Time to learn the proof of Chern-Gauss-Bonnet maybe

Balarka Sen

08:53

I forgot the forms formalism for differential geometry. I think it goes something like: Choose $(e_1, \cdots, e_n)$ local orthonormal frame on $M$ at $p$. Then $\nabla_{e_i} e_j = \sum_k \Gamma^k_{ij} e_k$, so the Christoffel symbols are $\Gamma^k_{ij} = \langle \nabla_{e_i} e_j, e_k \rangle$.

We can "dualize" the Christoffel symbols by letting $\omega_{ij}$ to be the $1$-form such that $\omega_{ij}(X) = \langle \nabla_{e_i} X, e_j \rangle$.

This is essentially rate of change of $X$ in the $e_i$-direction, felt in the $e_j$-component.

$\omega = (\omega_{ij})$ is a matrix of $1$-forms near $p$. This is the connection $1$-form if I recall right

OK, $d\omega + \omega \wedge \omega$ is the curvature 2-form I think

Secret

Sometimes I wonder if there is a relation between measure and topology

a set with positive measure vs a set with zero measure or is immeasurable seemed to have a lot to do with the topologies between borel sets and the target set

in particular, that the outer and inner measures mismatched in a nonmeasurable sets reminds me the non jordan curves on a torus

they got "stuck"

Ante

09:10

I almost do not understand ordinals in set theory at all, does someone knows how to easily explain that concept?

Secret

9

A: Definition of Ordinals in Set Theory in Layman Terms

Counting has two purposes, namely for specifying sizes and indices. These are directly related for finite quantities, because the number of natural numbers (including $0$) less than $n$ (before the position $n$) is $n$. But in set theory, when generalizing to infinite sets these two notions becom...

Ante

@Secret Thanks, still complicated to me. :) I am relatively new to strictly formal set theory and ordinals.

How much of general set theory I miss if I skip ordinal numbers? :(

Secret

How many kinds of infinity are there?

►

for an extremely gentle introduction without set theory stuff

Alessandro Codenotti

09:25

@Ante pretty much everywhere in set theory ordinals are used

Ante

:( why?

Alessandro Codenotti

Because they're useful

Ante

for what?

Alessandro Codenotti

All sorts of (transfinite) recursive constructions

Secret

hmm... ok, so nonmeasurability does not really have to do with homology

because eh... a line is going to stay a line regardless of what measure you use on it

@AlessandroCodenotti do you know of any mathematical terminology or related concepts that studies how much a given outer and inner measure of a set differs?

Ante

09:38

@Secret Can one exist and other not?

Secret

no...?

you always going to have a infimum and a supremum for a measure

and so there is always an inner and outer measure for any measure

Alessandro Codenotti

@Secret not really

Secret

hmm ok

Knight

14:12

@AlessandroCodenotti I was talking to someone (in real life) and in between our conversation he said that Mathematics education almost ends with Bachelors course, and when we enter Masters or PhD it’s more like a research rather than a study

As you are a masters student, can you please tell me if he was right ?

Alessandro Codenotti

14:24

Masters in the US and in Europe are two very different things, I can only talk about the European ones. They're pretty much the same as a bachelor, just with more advanced courses

Some programs are more research oriented and expect masters thesis to have something new research wise, but not even that is always a requirement

But you still have to take courses and pass exams

In good research oriented universities master courses might be about very recent developments in their field, but they're still standard courses with a lecturer and an exam rather than independent research

Knight

As far as I know very few people (few in comparison to other fields) take interest in the axiomatic set theory research, so if we take your example (f you don’t mind) then what is happening with you in Masters? Are you focusing much on recent developments or studying the things?

Alessandro Codenotti

I took two masters level set theory courses, the first one was an introduction to forcing and inner models, the second one was about more advanced applications of forcing (class forcing, iterated forcing, forcing axioms, in particular MA and PFA and how to build models of them, applications of forcing to the tree property, Prirky forcing, Namba forcing etc.). I also took a seminar on large cardinals and a seminar on independence results outside of set theory

None of this courses is cutting edge research level, even though the second set theory course is much more advanced than courses offered on average in masters programs

Apart from those I also took courses in topology and analysis to fulfill the credits requirements for the program

Knight

14:41

Wow! Thank you. We had a nice conversation

vesii

14:53

It's a very basic theorem but I can't seem to find a formal proof (from some book). The theorem: if there are $m$ equations and $n$ variable for $m<n$ then there are infinite many solutions or no solution. Can someone please point me where I can find a formal proof?

Michael Albanese

@vesii: Are you talking about linear equations?

vesii

yes

Michael Albanese

Do you know how to convert a system of linear equations into an augmented matrix?

vesii

I do

Michael Albanese

Finally, do you know about reduced row echelon form?

vesii

15:06

I have the general idea of how to solve it. My problem is how to write it "mathematically"

Michael Albanese

Consider the reduced row echelon form. It is either consistent or inconsistent.

If it is inconsistent, there are no solutions.

If it is consistent, we have to rule out a unique solution.

What is the maximum number of pivots (also called leading 1's) in the reduced row echelon form of an m x n matrix?

vesii

15:19

Why $n-r=n-m$?

Thorgott

Are you only working over $\mathbb{R}$? The theorem is false otherwise

vesii

why is it false?

Thorgott

Consider $0$ equations and $1$ variable in any finite field

vesii

is it possible to have zero equations and non zero variables?

also, why $n-r=n-m$ in that proof?

Michael Albanese

It isn't, but rather $r \leq m$ so $n - r \geq n - m > 0$.

Thorgott

15:23

it is, the zero equations are fulfilled by all variables (of which there are finitely many over a finite field)

this just corresponds to looking at the kernel of the trivial map that sends everything to zero

What's true over any field is the following: The inhomogenous system either has no solution, or the solution space is a translate of the solution space of the corresponding homogenous system. This is the kernel of a linear map $K^n\rightarrow K^m$, where $K$ is the underlying field, and rank-nullity together with $m<n$ implies that the kernel has positive dimension. If $K$ is infinite, then it necessarily has infinitely many elements.

Adam L

17:09

@Ante by elementary number theory do you mean number theory that doesn't involve plural quantification? Which is pretty much another way of saying not number theory but ideally Number theory ought to be elementary and algebraic

But no I don't know if im skilled I don't have a certificate saying so

geocalc33

19:16

Can anyone help with this question? math.stackexchange.com/questions/3582994/…

must I calculate the all angels in the space?

Adam L

19:50

What I find remarkable with people in mathematics is they can process layer after layer of abstraction within it's boundaries, but it's unthinkable for such a thing to be true in an applied sense, ridiculous to not take anything at face value

Astyx

Hey chat

geocalc33

hi^

@AdamL interesting

What is the name for a surface formed by taking the cross product of two vectors in $\Bbb R^2$ at each point and defining a new point directly above?

for example, if $S=(0,1)^2$ with a chart, and 90 degree angles between gridlines, one gets a plane, 1 unit above the surface

but if one deforms the mesh, $\psi: S \to S$ with a homeomorphism $\psi$, and then computes the cross-products at each point the plane is now deformed into a curved surface with boundary hugging the boundary $\partial S$

if $\psi$ is a made a continuous mapping, then you could run a program that lets you see how the surface deforms in real time. so you could associate a geometry to a set of warped curves on a plane

and the class of all homeomorphisms $\{\psi\}$ up to isotopy, could be given a group structure by quotienting by certain compatible objects. You could take the class of homotopies and do some cool stuff

EnjoysMath

21:22

@geocalc33 math.stackexchange.com/questions/3583442/…

Mathphile

2

Q: Interesting patterns related to the sums of the remainders of integers

Let us define, $$r(b)=\sum_{k=1}^{\lfloor \frac{b-1}{2} \rfloor} (b \bmod{k})$$ After reading the posts Surprising fact about a certain number-theoretic function and Do primes have special sums... I decided to play around with the $r(b)$ function and after performing some computations using PARI...

geocalc33

@EnjoysMath nice

:L)

EnjoysMath

:>

@geocalc33 what cohomology does is just take $C^m$ a module of general functions into a ring (since you can pointwise add and multiply by scalars, you have a module) and defines a differential map betwene $C^m$ and $C^{m+1}$ such that $d^2 = 0$ the zero map. Then magical things happen

geocalc33

@EnjoysMath you might post that to mathoverflow?

looks complicated

nice!!

EnjoysMath

You define $H_n = \ker d_n / \text{im} d_{n-1}$ as the $n$th (co-)homology module

geocalc33

21:28

that sounds interesting

EnjoysMath

It measures things like how many remainder maps there are left after quotienting by a submodule of a certain type.

geocalc33

I need to decompose a manifold into rings

but I don't know how to do it

EnjoysMath

@geocalc33 Do you know essentially what a module is?

geocalc33

no I have no idea what a module is

EnjoysMath

It's like a vector space except the field of scalars is replaced by the ring of scalars.

geocalc33

21:29

oh

EnjoysMath

So you can't divide by a nozero scalar, but you can multiply by one

geocalc33

oh okay

EnjoysMath

So For instance $\Bbb{Z}$ is a ring, so that $\Bbb{Z}$ itself is a $R$ module where $R = \Bbb{Z}$ itself.

geocalc33

so I could easily decompose a manifold into a module

EnjoysMath

$\Bbb{Z}^2$ under componentwise addition and multiplication is a ring but also a $\Bbb{Z}$-module in disguise

See how it can be two things?

Also prove to me that every $\Bbb{Z}$-module is an abelian group and the converse holds. Therefore they are the same things.

It's like letting $\Bbb{Z}$ be an additive group, but when you say ring you're looking at more structure (including distributive mult.)

So a module is in between a group and a ring

Since every ring $\Bbb{R}$ is naturally an $R$-module.

geocalc33

21:32

and a field

EnjoysMath

No

A field is a ring except every nonzero is invertible multiplicatively

geocalc33

no I'm saying, a module is similar to a group ring and field

EnjoysMath

Thus you have $(F \setminus 0, \cdot)$ is a group and $(F, +)$ is a group (two groups interacting). Then you have a field

geocalc33

two groups interacting

EnjoysMath

But if $(F\setminus 0, \cdot)$ is just a semigroup then you have a ring

interacting distributively $a (b + c) = ab +ac$

Mathphile

21:34

11 mins ago, by Mathphile

2

Q: Interesting patterns related to the sums of the remainders of integers

Let us define, $$r(b)=\sum_{k=1}^{\lfloor \frac{b-1}{2} \rfloor} (b \bmod{k})$$ After reading the posts Surprising fact about a certain number-theoretic function and Do primes have special sums... I decided to play around with the $r(b)$ function and after performing some computations using PARI...

elementary-number-theory summation prime-numbers conjectures semiprimes

Can someone have a look at my question?

I would love to see some ideas

EnjoysMath

S would we, I would break that down into one question and then show an attempt at answering it. That way it would get more attention for sure

geocalc33

agreed

Mathphile

@EnjoysMath what if I don't know where to start at all

EnjoysMath

@Mathphile where you having trouble, identify the textual location where your brain goes wtf

Mathphile

okay so here is my thought process

EnjoysMath

21:38

Okay, first break into cases so that the arithmetic works out nicely

even / odd

Mathphile

looks at question for 1 hour

Brain: Is this even provable?

maybe it's because I don't know enough math

EnjoysMath

K

here

Look at their proof here: math.stackexchange.com/a/3582676/26327

Mathphile

the highest math I have done so far is calculus II

EnjoysMath

Nvm

I see what they're doing now

they're employing a trick called "working backwards from the desired equation"

Mathphile

okay

EnjoysMath

21:41

So write down here in LaTeX the formula you want to show

then I can help you

Say for $b$ even, ...

Essentially given an equation in the ring $\Bbb{Z}/(k)$ (the integer modulo $k$ form a ring - look that up first)

Mathphile

$\sum_{k=1}^{\frac{b}{2}-1} (b \bmod{k})$?

EnjoysMath

Yes, equals?

In the setting of an equation in a ring, you can always do a few things.

Mathphile

What is a ring?

EnjoysMath

$A = B \iff aA = aB, a + A = a + B, A - a = B - a$ and so on, understanding this?

where $a \neq 0$

Mathphile

Sorry for being so dumb

EnjoysMath

21:45

A ring is a set of elements $R$, such that there are two binary operations on $R$

Mathphile

I really appreciate you trying to help me

EnjoysMath

And they interact distributively and are typically written as mult. and addition.

$a (b + c) = ab + ac$ is one of the defininig conditions (axioms) of the definition of a ring

that's implicitly for all $a, b, c \in R$.

So prove first that $\Bbb{Z}/(k)$ is a ring

$\Bbb{Z}/(k) = \{0, 1, \dots, k-1\}$ where $k \in \Bbb{Z}$.

$\Bbb{Z}/(2) = \{0, 1\}$ for instance has $\text{XOR} = +, \text{AND} = \cdot$

Mathphile

so would $\Bbb{Z}^n$ also be a ring?

EnjoysMath

Since $1 + 1 = 2 = 0 \pmod 2$, right?

$1 \ \text{XOR} \ 1 = 0$ in circuit logic

Yes, you've identified

The compontwise product of rings

and in this case the $n$-fold product of the ring $\Bbb{Z}$

Mathphile

okay

EnjoysMath

21:48

$\Bbb{Z}/(k)$ is a ring quotient, so you'll have to know about quotienting a ring by an ideal

and why it still forms a ring

An ideal is a subring $I \subset R$ such that $rI \subset I$ for all $r \in R$ and that's called absorbing the whole ring

Mathphile

hmm

EnjoysMath

That makes $(k)$ an ideal, but a principal ideal since it's generated by $1$ element.

By generation I mean $(k) = \{ r k : r \in R\}$ literally

Mathphile

@EnjoysMath I don't think I am understanding much

EnjoysMath

so $(3) = \{3x : x \in \Bbb{Z}\}$

and so on

Look at my example

$3\Bbb{Z}$

Mathphile

I think I won't ask such questions anymore until I start some advanced math classes

EnjoysMath

21:51

is an ideal. You'll want to end up proving that $n\Bbb{Z} \equiv (n)$ where $\equiv$ means they're definitionally equal; are the only ideals of $\Bbb{Z}$

Mathphile

@EnjoysMath I really appreciate your help tho

EnjoysMath

Okay, do you know what a group is?

Mathphile

nop

EnjoysMath

All the structures I mentioned so far are categorized as abstract algebraic structures

Mathphile

I am a CS major btw

EnjoysMath

21:52

the most basic (and most interesting) is the group.

Yes, they use group theory in CS see PRIMES is in P

15 page paper by AKS team in India

Why do groups occur everywhere? It could be because our world is in 4D space time or how we naturally view it and in that space rotations about a point form a kind of group

Mathphile

@EnjoysMath I have heard of these structures but never studied them

EnjoysMath

There's a rotational symmetry of space

geocalc33

Spacetime is 4d?

EnjoysMath

Well if you take a triangle and center it at $O$ on the $\Bbb{R}^2$ plane

It's set of reflective transformations, together with rotational transformations (about $O$) form a group

geocalc33

yeah

EnjoysMath

21:55

It's a finite group with $6$ elements called the Dihedral group of order $D_{2n} = 6$, where $n = 3$ the number of points in your regular polygon

So I'm just listing examples of groups, just to demonstrate they're everywhere. That is in no way a lesson on those groups specifically

geocalc33

oh

yeah

EnjoysMath

@Mathphile define a group for me. What is an abstract definition of a group?

Mathphile

@EnjoysMath a group is a set with a binary operation?

EnjoysMath

It's a set $G$ with a binary operation $\cdot$ sometimes written as "juxtaposition" $ab$ of two elements, such that $a^{-1}$ exists for each element which means you can left or right multiply by $a^{-1}$ and get as value the multiplicative identity $1$ of $G$. So in your problem equation $a = b$. You can do $1 = a^{-1} a = a^{-1} b$ and that goes in the other direction since you can multiply all thru by $a$, right?

So you say $a = b \iff ca = cb$ for all $a, b,c \in G$ and that's by definition of equality and "binary operation" (a function $G \times G \to G$).

@Mathphile almost. It's what you said, with the inclusion of all inverses with respect to the operation

Mathphile

okay

EnjoysMath

21:58

Also there's a third axiom

It's $a(bc) = (ab)c$ or the associative law

without it, you only have what's called a magma which means you have to keep track of parentheses

but when $a(b(cd)) = (ab)(cd) = a(bc)d = \dots$ they're all equal by associtivity rule applications, so you can just drop all parentheses

That does not mean $a b = ba$. That's a condition called "abelianism". If you $G$ is defined as an abstract abelian group, then you add in the axiom $ab = ba \forall a,b, \in G$?

Does that make sense?

An abelian group is simply a commutative group.

commutative $\iff ab = ba, \ \forall a,b \in G$.

Mathphile

@EnjoysMath kind of

EnjoysMath

Thus in addition to being a group, what is $(\Bbb{Z}, +)$ as a structure?

It's an abelian group. Clearly as we commute addition of integers, right? $m + n = n + m$ naturally.

Mathphile

I'll brb in 10

Thorgott

without associativity, you still have a loop

EnjoysMath

@Mathphile solve the abelian group equation $a + b = c$ for $a$.

We'll keep working equations since your problem required working a modulo equation

Mathphile

22:04

ah damn

EnjoysMath

WHat's up?

Mathphile

I think I gtg for now

EnjoysMath

Cool cool

Mathphile

I hope to see you later

EnjoysMath

Same

@JackOhara solved DLP yet?

Leaky Nun

22:10

@AlessandroCodenotti play?

geocalc33

@EnjoysMath what's your strategy to solve DLP?

EnjoysMath

None, I give up

:D

Alessandro Codenotti

@LeakyNun I can play a couple of games when I finish this one

geocalc33

What was your strategy before?

EnjoysMath

It failed

Alessandro Codenotti

22:12

Alright send me a challenge @Leaky

Leaky Nun

done and done

@AlessandroCodenotti can you see it?

geocalc33

0

Q: Surface formed from cross product of vectors

What is the name for a surface formed by taking the cross-products of pairs of vectors, tangent to a grid, in $(0,1)^2$ at each point and defining a new point above the base space? For example, if $S=(0,1)^2$ with a specific chart, and 90 degree angles between gridlines, one computes the cro...

calculus reference-request surfaces curves

@EnjoysMath Still what was the main idea?

EnjoysMath

Yes, it failed horribly. So I don't want to polute the waters with any vague idea that it might be solvable that way

geocalc33

ok

Jack Ohara

@EnjoysMath Hey mate

I am sorry I have been busy couple days

too much work

and corona is messing with the my world

How are you?

EnjoysMath

22:25

I'm good, haven't had a close encounter with the virus yet

I'm working on a python code, state machine editor

So you can write state machine entry/exit functions in python and drag / place nodes and transitions

I want to release it in a month

It will have built-in support for PyQt5 signals as transition events

Edward Evans

yo

EnjoysMath

Hey, "oof"

:D

I like when you say "oof"

Edward Evans

lol how come?

EnjoysMath

it's expressive

Leaky Nun

@AlessandroCodenotti that wasn't resignable...

SF says -0.1

Astyx

22:35

Hi chat

Alessandro Codenotti

SF also says I was at +6, but it doesn't matter if I'm playing like a monkey tonight

EnjoysMath

high, how are you?

Leaky Nun

I mean, it was a drawn endgame

I can't win with one pawn

Alessandro Codenotti

Whatever, I was annoyed

Leaky Nun

ok fair enough

Alessandro Codenotti

22:36

I could have promoted to a queen

But only noticed that as I made a different move

Leaky Nun

where?

Alessandro Codenotti

Uhm I just closed it

Before you put your rook behind my passed pawn

I could have sacked my rook checking your king

Followed by promotion

That's a very common pattern, when you have a pawn on the 7th rank with your rook in front of it, if the promoting square is not defended and you can check with the rook somehow then you can also promote

Leaky Nun

ah!

Alessandro Codenotti

@LeakyNun Move 30 and 31

Leaky Nun

thanks

Alessandro Codenotti

22:39

I noticed it in the moment I moved to recapture the bishop :/

Leaky Nun

rip

topologicalorientablesurface

Consider $\mathbb{R}$ and $[a,b]$. Then $[a,b]$ is totally bounded. Let $\epsilon>0$. Choose $N$ sufficiently large such that $\frac{(b-a)}{N}<\epsilon$. For $i=0,1....,N$, define
$x_i=a+i\frac{(b-a)}{N}$. Observe, $x_i\in [a,b]$, since $x_0<x_1<...<x_N=b$.
Then How do I show that $[a,b] \subseteq \bigcup_{i=0}^N[x_{i-1},x_{i}]$ formally?

EnjoysMath

You could say obvious there just since $x_0 \lt \dots \lt x_N$

topologicalorientablesurface

pigeon hole, right?

EnjoysMath

And each interval clearly overlaps

topologicalorientablesurface

22:43

@EnjoysMath

EnjoysMath

I don't see pigeonholing required

topologicalorientablesurface

hmm. the lecturer bracketed pigeon hole

EnjoysMath

There is a transitivity going on

Leaky Nun

@AlessandroCodenotti SF says 6. Nc3 in our first game was a blunder (0 -> -2) but I have no idea why

I tried following the SF line but I still don't understand

EnjoysMath

$[x_1, x_2] \cup [x_2, x_3] = [x_1, x_3]$. So by transitivity usage and induction, you have your proof

Astyx

22:44

You could look, for each $x\in [a,b]$, at $\{i, x_i<x\}$

That set is finite and non empty so has a max, say $M$

then $x_M \lt x \le x_{M+1}$

There are extreme cases that you have to care about

EnjoysMath

Oh, I probably misunderstood

topologicalorientablesurface

Is $[a,b]\times [c,d]$ totally bounded?

Astyx

What does totally bounded mean ?

topologicalorientablesurface

@Astyx $X$ is totally bounded if for each $\epsilon>0$ there exists finite many balls of radius $\epsilon$ which cover $X$

equivalently, if for all epsilon, $X$ can be covered by finitely many sets, each of diameter less than epsilon

Astyx

It is

Alessandro Codenotti

22:56

@topologicalorientablesurface yes

topologicalorientablesurface

if $X$ and $Y$ are metric space and $X$ is totally bounded and $Y$ is totally bounded, then is $X\times Y$ totally bounded? (where $X\times Y$ is equipped with the standard metric)

?

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$Mathematics$ Mathematics