Mathematics - 2016-02-07 (page 1 of 3)

Forever Mozart

00:27

@MikeMiller

are you there?

I have a question

Mike Miller

sure

Forever Mozart

ok lets say $0$ is a point on $S^1$.

Mike Miller

sure

Forever Mozart

and $x,y\in S^1\setminus \{0\}$

is there a homeomorphism of $S^1$ which fixes $0$ and maps $x$ to $y$?

Mike Miller

Yes

Forever Mozart

00:29

ok because you can sort of flip the $S^1\setminus \{0\}$, right?

Mike Miller

Well I would say more because $S^1 \setminus \{0\}$ is homeomorphic to $\Bbb R$

Forever Mozart

this property you call $2$-transitive?

Mike Miller

Sure

Forever Mozart

if you can map an ordered pair to an ordered pair

I wonder if this true for all homogeneous compact connected metric spaces

oh forget compact

I want it to be true just for homogeneous connected metric spaces

Mike Miller

I would not expect that. I think another word people use is n-homogeneous.

All connected manifolds of dimension at least 2 are n-homogeneous for all n. In dimension 1, R is 2-homogeneous, but not 3-homogeneous.

(And the circle, 3-homogeneous but not 4.)

Forever Mozart

00:36

oh

Ted Shifrin

Goodnight @MikeM @Forever

Forever Mozart

later Ted

Ted Shifrin

LOL, no, that was my greeting

Forever Mozart

lol

you meant like Good Evening

in your alfred hitchcock voice

Ted Shifrin

This is the ritual that Mike and others of us go through on a daily basis.

Mike Miller

00:37

Ritual? I'm just saying hello.

guest

Hi Professor @TedShifrin

Ted Shifrin

hi skull

Forever Mozart

Alfred Hitchcock Intro Theme (Funeral March Of A Marionette)

►

Ted Shifrin

Yes, @Forever, I'm well acquainted therewith :)

guest

Have you seen The Crows?

Forever Mozart

00:40

Birds?

Ted Shifrin

I know of Birds and Raven.

guest

The movie.

Ted Shifrin

The Birds is the movie.

Forever Mozart

I think the birds were crows, but the movie is called The Birds

guest

That's it.

Ted Shifrin

00:41

Yup, the birds were crows.

Filmed in Bodega Bay, only a few miles from where my family had a cabin when I was a kid.

Forever Mozart

that is the wine country

guest

I never looked at a crow the same after seeing it.

Ted Shifrin

not really wine country ... wine country is inland a bit

Forever Mozart

my middle school teacher made us watch that movie

I forget why

Ted Shifrin

Hitchcock classic.

guest

00:44

Made me phobic.

Ted Shifrin

skull, don't worry — the crows never looked at you the same, either.

guest

:D

Semiclassical

oof, the birds

Ted Shifrin

hi @Semiclassic

Yes, this is our film studies class.

Semiclassical

not the creepiest hitchcock movie (hard to beat psycho in that department) but still

hi @ted

how goes your weekend?

Ted Shifrin

00:47

Don't forget Vertigo with the body hurtling down from the top of the tower at Mission San Juan Bautista. (I actually stopped there in September. There's no tower.)

Forever Mozart

Vertigo is my favorite of his, Rear Window close second

Semiclassical

Rear Window is good

i've only seen Vertigo once, and I probably didn't appreciate it at the time

Ted Shifrin

I just saw a new re-release in SF in September.

Semiclassical

North by Northwest isn't exactly creepy (though the fight scene on Mt Rushmore makes me gulp a bit)

but a good movie nonetheless

Ted Shifrin

oh, the plane flying only feet above ground almost decapitating him is pretty creepy.

Semiclassical

00:52

that's the cornfield scene?

Ted Shifrin

yup

Semiclassical

i remember it, but only just

i should see more movies

it's funny. my parents had two books of roger ebert reviews at home, and i devoured those. so i've waaaay read more reviews of movies than i've actually seen

guest

Tomorrow is the big show on turf.

Ted Shifrin

Sort of like we've read more textbooks than original papers ... but for movies it seems really wrong :D

Semiclassical

well, it should be said that Roger Ebert's reviews are pretty great to read. but yes

Mike Miller

00:56

@Ted: Nice Vertigo spoilers!

Ted Shifrin

Well, for all the people who come in here to read about movies, yes.

I didn't spill the most important beans, @MikeM.

Mike Miller

But who's to say I wasn't just about to go see it...?

Ted Shifrin

I haven't spoiled much for you.

guest

Who's your favorite actor @TedShifrin?

Ted Shifrin

I have no clue, skull. Long list including French actors.

Mike Miller

01:01

I'm kidding, of course.

Semiclassical

i really don't have enough range of experience to judge, but the name that comes to mind when it comes to 'favorite actor' is Jack Klugman.

Ted Shifrin

He would not be near the top of my list.

Semiclassical

admittedly, that's biased by him being in a certain Twilight Zone episode which I really really like

Mike Miller

I tend to like directors more than actors

Forever Mozart

muaaahhh the french... champagne has always been celebrated for its excellence. There is a California champagne by Paul Mason, inspired by that same french excellence. It's fermented in the bottle and like the best french spirits it's vintage dated.

thats all I remember

@MikeMiller

these two things are not equivalent

(i): for each $x,y\in X\setminus \{0\}$ there is a homeomorphism of $X\setminus \{0\}$ mapping x to y

(ii) for each $x,y\in X\setminus \{0\}$ there is a homeomorphism of $X$ fixing $0$, mapping x to y.

Mike Miller

01:09

Darn.

Forever Mozart

right?

because take $\mathbb R$ with $x=1$ and $y=2$

even for homogeneous spaces they are not equiv

Mike Miller

That is not a counterexample? Both are true.

Ted Shifrin

Any homeomorphism of $X$ fixing $0$ restricts to a homeomorphism of $X-\{0\}$?

Mike Miller

Yes, I had claimed in my flu-stricken stupor last night that they were equivalent, as opposed to (I)-> (II).

I don't know a counterezample in the other dirextion but I also don't know a proof

It is true if X is compact

(Proper maps induce maps on the 1-point compactification, and homeomorphisms are proper)

Ted Shifrin

Isn't it (II) -> (I)?

Mike Miller

01:33

Yes, sorry.

Ted Shifrin

My brain is pretty much gone, but not 100%.

David

01:51

I have a math question

Forever Mozart

02:02

yes?@David

PVAL

@MikeMiller There was a question on main about this recently math.stackexchange.com/questions/1631363/…

Forever Mozart

i am one lemma away from a big theorem

hope I can prove it

Secret

03:39

I had a feeling that the general rules of games of this type is that to change color to the color patches that have the least number of disjointed regions and not being surrounded by another color in all directions

So for example, there's only one disjointed region for orange, and it encircles yellow, while yellow has two disjointed regions and at least one encircles green

Thus the filling order is to turn orange to yellow, then yellow to green then green to red

I wonder how will this be written mathematically...?

usukidoll

03:52

anyone here?

AreaMan

04:12

I am here temporarily.

usukidoll

do you know accumulation points?

AreaMan

Something.

usukidoll

like hold on

AreaMan

Ok

usukidoll

Let S be a set of real numbers. A real number A is an accumulation
point of S iff every neighborhood of A contains infinitely many points
of S. That's the definition . I was given an example but I don't get it

AreaMan

04:14

Do you have a good mental picture of a "neighborhood"?

usukidoll

oh I forgot the neighborhood def
A set Q of real numbers is a neighborhood of a real number x
iff Q contains an interval of positive length centered at x-that is, iff there is
e > 0 such that (x-e,x+e) C Q.

AreaMan

That's not what I am asking... I mean, if someone says: A neighbourhood of 0 in the real numbers ... what do you imagine?

Personally I just imagine a little circle around x, with the mental indication that I am thinking of the points inside that circle. Does the connection to the definition of neighborhood make sense to you?

usukidoll

a number line

we have a lower bound................ and an upper bound
then we have a_n
and the little neighborhood is $a_{n-1}$ and $a_{n+1}$

so there's a circle around x and we have dots in the circle

AreaMan

Yes...

And a point x is an accumulation point of S if S has infinitely many points in every little circle that can be drawn around x.

BTW I don't understand why you have a sequence of a_n ...

just two numbers suffices to define an interval.

usukidoll

but what about on this example
Note: $a \in S$ doesn't imply that a is an accumulation point of S
so if we have
$S = (0,1) \lor \{ 33 \}$
33 is the only thing in common but it's not an accumulation point of the (32,34) interval. What does that mean

AreaMan

04:20

The annoying thing about this definition is that you have to assume that there infinitely many points of S near x other than x itself.

I mean, not assume, that is the definition.

usukidoll

the (32,34) is a subset of a real number (not an element of S)
only one point -> 33 in a set, but not an accumulation point. ummm? huh
this is from my notes on the last lecture btw. the book is so xsofdj34

AreaMan

So in the case of 33, there are not infinitely many points of S near 33 that are not 33.

usukidoll

could you draw that?

AreaMan

S..............................33.........................

usukidoll

hmmm so ... lost.

AreaMan

04:22

Let me give you this vague definition: an accumulation point of a set S is a point x so that there are infinitely many points of S near x.

Where near really means, arbitrarily near.

usukidoll

which is there are a bunch of points in the set S that can be near x

like super near

AreaMan

arbitrarily near

usukidoll

like very close

AreaMan

as near as you want. The point x= 1.000000000000000000000002 is not an accumultation point of [0,1]. This is because there is a little circle you can draw around x so that no other point of [0,1] is in it.

not just very close, arbitrarily close. This is the meaning of for all $\epsilon > 0$...

usukidoll

so the point x is a little over 1, and because of that it's not an accumulation point of [0,1] so I can draw a tiny circle around the x to make sure that any point that's in [0,1] is in it

AreaMan

04:25

Well, yes to the first part, no to the second - it doesn't have to be the case that any point that is in [0,1] is in the little circle...

the only point that is in all of the little circles is x...

The simplest interesting example : the set {1,1/2,1/3,1/4, ...} has zero as an accumulation point. Try to think on this example.

usukidoll

hmm I know the set decreases
1,0.5,0.333,0.25

AreaMan

Another definition of accumulation point may be helpful: x is a limit point of S if there is a sequence of points in S that converges to x. (i.e. x is a limit of points in S). x is an accumultation point if it is a limit point and you can find some sequence that isn't eventually (...,x,x,x,x,x,x,x....) (constantly x).

usukidoll

1,1/2,1/3,1/4,1/5,1/6,1/7,1/8

x is an accumulation point if the sequence doesn't repeat itself like having (1,1,1,1,1,1,1,) constant 1's ??

AreaMan

The point is that you can find a sequence converging to x that isn't just eventually x,x,x,x,x... I do not know where your 1 came from. In this example it is zero.

usukidoll

oh I see the sequence converging to x isn't going to be repetitive like x x x x x

AreaMan

04:31

Yes... because we want infinitely many elements of S near x, and being able to write the sequence x,x,x,x,x,x converging to x doesn't give us that.

Writing a non evnetually constant sequence convering to x does give us infinitely many points of S in any little disc around x....

usukidoll

so how does
the set
$\{ \frac{1}{n}: n =1,2,... \}$ have the accumulation point at 0. unless our n will have to be big to produce a fraction small enough to be close to 0 ??

AreaMan

Yes, the set is for all natural numbers n, so in particular very big n...

usukidoll

like 1/1000000000000000000000

AreaMan

Even bigger...

arbitrarily big

usukidoll

wow.... 1/100000000000000000000000000000000000000000000000000000000000000000000000000000‌0000000000000

AreaMan

04:35

even bigger!!!

usukidoll

1/100000000000000000000000000000000000000000000000000000000000000000000000000000‌000000000000000000000000000000000000000000000000000000000000000000000000000000000‌00000000000000000000000000000000000000000000000000000000000000000000

AreaMan

Arbitrarily big! Because we want the fraction 1/n to get arbitrarily close to 0.

usukidoll

THAT will be too huge to type

AreaMan

You can't type it ... there are infinitely many elements in the set, none of them are equal to zero, and yet for any circle around zero you draw, you can find some elements of your set inside of it.

usukidoll

ah .. this is starting to make sense a bit

AreaMan

04:38

That makes zero an accumulation / limit point of your set...

usukidoll

I'm also reading a pdf that made it easier than the book I have

AreaMan

Okay, great.

I really liked abbots understanding analysis when I was learning this stuff.

usukidoll

but... I'm still unsure on some things..
It's like one minute I get it and then the next I'm like what is this again I forgot :O

so for any circle around zero, there will be some element of the set inside that circle

AreaMan

yeah...

usukidoll

ok.. I just got to practice a bit before tackling those last proofs that involve accumulation points

AreaMan

04:40

Draw some pictures. You should try to have mental pictures for each of these concepts ... neighborhood, limit point, closure... and mental picture really means, knowing the characteristic example of that concept.

In the case of limit/accumulation point, the characteristic example is {1,1/2,1/3,1/4, ...} with limit point 0. Also, you should think of the word accumulation... it is pretty descriptive.

Okay, it seems that limit point actually means what we were calling accumulation point, and that accumulation point means that there is just some sequence converging to that point. This is a mildly annoying thing - I can never remember which is which.

But what matters is to put the idea in your mind.

usukidoll

thanks. I need to read more about it before I do anything... I'm just thinking about if what I did to prove the sum of a_n and b_n is Cauchy is right or I might have done something illegal

AreaMan

Here are some exercises to help: A finite subset of R has no limit points. Every real number is a limit point of the set of rational numbers.

Well, you can ask someone else. I am going to bed.

Good luck understanding this stuff. Draw lots of pictures...

it is all very visual.

usukidoll

ok :)

thanks for helping

AreaMan

np

Secret

07:14

What property of the polynomial space$\mathcal{P}_n$ is unique to it and does not have analogue in $\mathbb{R}^n$?

Overly Excessive

08:22

Hi, please help

Abhimanyu Aryan

Its a question related to Maths and computer science:

I am trying to understand the max size of 32-Bit Interger

I did `2^32` and answer came out to be `4294967296`. The actual max size of 32-bit integer is `2,147,483,647` but I think this only includes unsigned int(+ve values)............So i added `2,147,483,647 + 2,147,483,647` and the answer came out to be `4294967294`...Its 2 less than what I got with 2^32 any reasoning? @BenHoffstein

manetsus

09:27

How to find maximum value of x+y in the equation ax + by = c where a,b,c are given?

Albas

09:56

@manetsus the equation you have given is a linear equation for which given any value of x I can always find a value for y. That means x+y will not have a specific maximum

manetsus

@Albas ow, sorry, one other constraint x and y both should be non-negative as well as given constants.

Albas

Still @manetsus it remains as a linear equation. For ex 3x+4y=7, pick a value of x you can always find a value of y. Does not matter how high that value of x is there is a corresponding value of y. So there is no specific maximum

manetsus

@Albas you can not get the maximum of x+y more than 2 for the example you mentioned. Sorry, if you are thinking about fraction. Only integers would be accepted.

Albas

Oh you said nonnegative integers. Sorry for that

manetsus

@Albas Actually it is a real world problem and x and y are two counters of things. So, it must be integer. Sorry, I had missed the point earlier.

Albas

10:10

Hmm.. Then you can do this. Write the function in the form y=f(x) and set $y\geq0$. Solve the inequality and then take only integer values.@manetsus

manetsus

@Albas but the values of x,y could be very large, say $10^18$.

Albas

But for a very large $x$ to be compatible with your end result y will have to be negative.
Eg:3x+4y=7 for $x=17,y=-11$ and you do not want negative values of y or x@manetsus

user276387

A fair die is rolled twice. If the two results are the same, a coin is tossed. Why is the total number of different possible outcomes of this experiment $42$?

Albas

@user276387 draw the tree diagram for the above experiment

@manetsus is there any problem?

user276387

@Albas will do. Is there a way to get the answer through symbolic calculation, though?

manetsus

10:18

@Albas Actually a,b,c all are in range 1 to $10^18$

@Albas So, x and y could be very large. Say, $ x + 2y = 10^18$, How would be your approach?

Albas

I would do the same thing again. And will get a range for the values of x . Then choose the highest integer value

manetsus

@Albas It would not serve all cases.

Albas

Can you give me an example?

manetsus

@Albas Sorry, I wanted to say I would take a long time like billions of years to get the result in the fastest computer of the world as the range of x or you could be very large.

Albas

See if you wanted to find that maximum value of xy for the given linear equation we could use first and second order differentiation

But we have x+y

And I don't think so it should take much time. You will get an equation like x\leq a for some positive real number a. Then you can take the closest integer value to that number a if a is not an integer@manetsus

@user276387 thus us not actually a simplification but if you want you.can break it down into Cartesian products

manetsus

10:54

@Albas Say $3x + 11y = 4033564838744$ What would be the output?

Aldon

11:32

Hi all I have a question

If I have to normalize a piecewise function ($\int_{-\infty}^{\infty} f(x)^2\,dx$), do I have to separate the integral piecewise?

And also, I wouldn't have to integrate from $-\infty$ to $\infty$ right? Just from the bounds of the piecewise function?

Huy

13:13

@MikeMiller: what's the natural bijection between the cosets of $\langle \alpha \rangle$ in $\pi_1(S)$ and the set of lifts of $\alpha$ to its universal cover? the construction that came to my mind requires a lift to start with: fix any lift $\tilde{\alpha}$ of $\alpha$. let $\gamma \langle \alpha \rangle \in \pi_1(S) / \langle \alpha \rangle$. let $\phi \in \operatorname{Deck}(p)$ be the deck transformation corresponding to $\gamma \in \pi_1(S)$. then compose the fixed lift with $\phi$.

@MikeMiller: this doesn't seem all that natural to me though, but maybe that is indeed what is meant by natural bijection between those sets? can you confirm or tell me what I should be looking for?

user19405892

Can someone help me with this question: Given two congruent triangles A_1A_2A_3 and B_1B_2B_3 prove that there exists a plane such that the orthogonal projections of these triangles noto it are congruent and equally oriented.

that should say onto not noto

Jake1234

13:37

Hey guys, on wikipedia site about quadratic forms in my language, the definition is that when considering $L(a,b)$ a bilinear form, $q(h) = L(h,h)$ is it's corresponding quadratic form.

But on english wiki, it's defined this way for symmetric bilinear forms only.

Is the first definition uncommon/weird?

Anubhav.K

14:13

hii @MikeMiller can you tell me whether this is true or false... If a closed manifold can be covered using two chart then it is homeomorphic with $S^n$??? I thought about this qes but couldnot able to find anything beteer.

Huy

14:42

@Anubhav.K: I think you need connectedness, and then it's true

Anubhav.K

@Huy I mean to say closed and connected manifold

Huy

@Anubhav.K: check this out

Anubhav.K

@Huy thanks,

Mike Miller

15:04

@Anubhav.K If your charts are homeomorphic to $\Bbb R^n$, yes.

@Huy: I agree that bijection only works if you pick a lift first. Indeed if you have a bijextion between lifts and cosets you can easily pull out a canonical lift as the lift corresponding to $\langle \alpha\rangle$

Mike Miller

15:31

I guess the natural way to say it is "There is a natural action of $\pi_1(S)$ on the set of lifts that has point-stabilizer $\langle \alpha\rangle$

venkat

hi... this is venkat... I ve one doubt.. if f(x) is nth degree polynomial n>2, how to find roots of f(x) mod p =0, where p is prime number

hhh

Are all of them subrings?

venkat

coefficients of f(x) belongs to [0,10]

hhh

Actually, only $R_4:=\{f: [0,1]\rightarrow | \lim_{x\rightarrow 1^{-}} f(x)=0\}$ is a subring since $a-b$ must belong to the subring $S$

If you have a=0 and b=1 such that f(a)=f(b)=0, we can choose $0-1 \not\in [0,1]$

venkat

15:49

sorry i am not getting... could take any simple quadratic equation and explain me... or give some reference...

Albas

16:15

math.stackexchange.com/questions/1595807/… can I use lagrange multipliers for the above problem

Paradox 101

@MikeMiller can you please explain how do we use the fact that a function is injective to prove that $f(C \cap D) = f(C) \cap f(D)$ where $C$ and $D$ belong to $A$ which is the domain of the function?

Andrew Thompson

@Paradox101 What have you tried?

Paradox 101

@AndrewThompson I need to show that $f(C \cap D) $ is contained in $f(C) \cap f(D)$ and vice versa right? So if for the first implication I say $x$ belongs to $C \cap D$ can I say that $f(x)$ belongs to $C \cap D$? I mean I don't get how to use the injectivity here

Andrew Thompson

You can't. Why would $f(x)$ belong in $C \cap D$?

Paradox 101

I'm not sure about this I have to include the function somehow. I can't use the concept of inverse functions here can I? Doesn't the function need to be a bijection for that?

Andrew Thompson

16:28

Well, you can. Let $f \colon A \to B$ be a function which is injective. It might not hit everything in $B$, however you can consider the same function onto its image, i.e. $f \colon A \to f(A)$. If $f$ is injective, this will be a bijection.

The textbook example for this is the following: let $f(x) = x^2$ considered as a function $\Bbb R \to \Bbb R$. Is this surjective? Is it injective?

mr eyeglasses

16:59

hi @AndrewThompson and @Paradox101

Andrew Thompson

Hi @mreyeglasses.

Paradox 101

It's not injective but it's surjective? @AndrewThompson

Hi @mreyeglasses

Andrew Thompson

Why do you think its surjective, @Paradox101?

r9m

@Albas maybe .. but I doubt Michael's problems can be dealt with L-multipliers ;) ..

Paradox 101

wait no it's not surjective either @AndrewThompson

Andrew Thompson

17:03

Great. Now consider the same function, but as a function $[0, \infty) \to \Bbb R$. Is it surjective? Injective?

r9m

@robjohn hi :-)

robjohn

@r9m Hey there... haven't seen you for a while.

Mathematician

Hello

I'm stuck in a little step (i suppose) for a problem, if a set A contains more than half elements of a group G, then every element of G is a product of 2 elements of A, obviously the subgroup generated by A should be G so every element is a product of elements from A but how to show the product of 2?

Andrew Thompson

@robjohn You ruined my pedagogical build-up :'(

mr eyeglasses

RIP

robjohn

17:09

@AndrewThompson Sorry... I just saw the question next to a comment I was reading.

Andrew Thompson

@robjohn Just joking. @Paradox101 Now you have the easy direction. Note that injectivity was not needed. Try the other direction and note where injectivity is needed.

Mathematician

any help?

Paradox 101

@AndrewThompson it would be injective but not surjective?

r9m

@robjohn yup .. I was inactive for some time. :-) How are you?

robjohn

@r9m Work has been busy lately. I haven't had time to do as much as usual.

Albas

17:18

@r9m though the equations that I get are pretty huge

Mathematician

@robjohn I'm stuck in a little step (i suppose) for a problem, if a set A contains more than half elements of a group G, then every element of G is a product of 2 elements of A, obviously the subgroup generated by A should be G so every element is a product of elements from A but how to show the product of 2?

r9m

@robjohn I guessed that .. :) Your rep tree forest is loosing it's green heights! :)

JeSuis

@robjohn Hi, I have a linear map $A:\Bbb{R}[x]\mapsto \Bbb{R}$ such that $A(P)=P(1)$. I would like to prove that $A$ is not continuous for the norm $p>1$. So I tried use sequences (as "usual"). But I am stuck finding the right sequence...

Albas

You have any other idea@r9m? It is cyclic inequality so can we use any facts about that?

robjohn

@JeSuis which norms are you looking at?

r9m

17:21

@Albas not at the moment .. last time I tried one of Michael's problems I remember I ended up giving up on it .. :( (not complaining though). I just ain't all that good with inequalities ..

JeSuis

$\Vert P\Vert=\bigl(\sum_{i=1}^n\vert a_i\vert^p\bigr)^{1/p}$ @robjohn with $p>1$ because it works for $p=1$.

Idle001

hey @r9m long time

r9m

scratching head .. @Idle001 ya long time indeed ..

Idle001

@r9m im agawa (my dynamic version)

r9m

@Idle001 aha!! Now I remember :D Hi pal ..

robjohn

17:37

@Mathematician For each element $g\in G$, consider the set $gA$ what is the size of this set? What is the size of $A$? What can you conclude?

r9m

@I'manartist .. that monster!! :D Hi ..

robjohn

@JeSuis what are the $a_i$ and how do they relate to $A$?

r9m

@I'manartist maybe $H_n$ representation-chans can help us here? :-)

JeSuis

@robjohn If I write a polynomial $P=\sum_{k=1}^n a_iX^i$ the $P(1)=\sum_{k=1}^n a_i$ then for the case $p=1$; $A$ is continuous.

(Triangle inequality) but for the case $p>1$ I don't see how can I prove that $A$ is not continuous

robjohn

@JeSuis Ah, so the $P$ are polynomials.

JeSuis

17:47

yes :)

Andrew Thompson

@Paradox101. Yup. Same question and function, now for $[0, \infty) \to [0, \infty)$.

Paradox 101

@AndrewThompson then it's a bijection?

Andrew Thompson

Correct. Now you should have some intuition. Prove first that if $f \colon A \to B$ is any function, and $C,D \subset A$, then $f(C \cap D) \subset f(C) \cap f(D)$.

JeSuis

@TedShifrin Hello

Paradox 101

How is injection going to be used here? If $f \colon A \to B$ is injective then so would the functions on $C$ and $D$ be as they are subsets of $A$?

Ted Shifrin

17:56

hi, @AndrewT, @Paradox, @robjohn; bonjour, @JeSuis

JeSuis

@TedShifrin what's up ?

Paradox 101

Hi @TedShifrin

Andrew Thompson

Hi @TedShifrin. @Paradox101 Did I ever say the function was injective?

mr eyeglasses

@TedShifrin don't forget about me

r9m

@TedShifrin Hi professor ... how are you? :)

mr eyeglasses

17:57

so many @TedShifrin LOL

Ted Shifrin

oh, hi, @mreyeglasses :)

mr eyeglasses

@TedShifrin you are so popular

Ted Shifrin

heya @r9m ... haven't seen you in a while

That's one word for it, @mreyeglasses

r9m

@TedShifrin yup .. I had been laying eggs as you can see (my avatar :P) ..

Ted Shifrin

glad to know you're being productive one way or the other, @r9m

r9m

17:59

@TedShifrin yup .. that's always the plan .. keep producing no matter what :D

Transcript for

$Mathematics$ Mathematics