Mathematics - 2012-07-02 (page 2 of 3)

user19161

18:00

@unNaturhal This is really too vague to be meaningfully answerable.

unNaturhal

@JasperLoy What do you mean?

Want to see the function?

user19161

@unNaturhal I meant your question is not really a question because we don't have the details. What the function is, what you are to do etc...

user19161

@Gigili Why? Who?

unNaturhal

@JasperLoy I avoided to write the function just because I should know a method to study discontinuity applicable to all functions, because I'm really confused about this :/

Gigili

@JasperLoy There are steps which apply to all kinds of functions.

Jordan

18:05

Hello

Gigili

@JasperLoy Called Ragib Zaman or something like that,

user19161

@unNaturhal It's not clear what you mean by "study discontinuity" here. You know the definition of continuity and using that you do whatever you need to do.

Jordan

How do I compute $(e^{x/4})^2$?

unNaturhal

@JasperLoy Mmmmh.. You are saying that continuity and discontinuity aren't the two faces of the same medal?

Gigili

1 min ago, by Gigili

@JasperLoy There are steps which apply to all kinds of functions.

user19161

18:09

Erm, never mind.

unNaturhal

@Gigili Exactly, this is what I'm asking for :P

user19161

I think we are on different wavelengths here. :-)

Gigili

@JasperLoy Don't make me post it again!

user19161

@Gigili See above.

unNaturhal

@JasperLoy Not necessarily :)

N3buchadnezzar

18:11

user19161

@N3buchadnezzar OMG. I handled those grenades before.

N3buchadnezzar

@JasperLoy Really?

user19161

@N3buchadnezzar Military service.

Gigili

@unNaturhal Didn't I answer your question? But you still need to ask someone else to make sure!

N3buchadnezzar

Cool. Well I guess he deserved it, I mean the microwave really?

unNaturhal

18:15

@Gigili Ehm.. you don't :/ But it's because I haven't understood all things you said...

Gigili

@unNaturhal Maybe we need an example!

unNaturhal

@Gigili Ok, but keep in mind that I'm looking a general method to apply to all functions :p

Wanna see the scan of my function?

user19161

@unNaturhal Erm, how would seeing the scan help? It's better to have the function itself. A picture is just an imperfect representation of the funtion.

Jonas Teuwen

@N3buchadnezzar Man, somebody trolled 4chan succesfully? Respect!

unNaturhal

@JasperLoy Ok :p But the scan contains all the things I do to 'till now :P

N3buchadnezzar

18:18

@JonasTeuwen Someone found a defect grenade? Respect!

unNaturhal

However, thi function is this:
$$\displaystyle{y = \frac{x^2}{1+\log|x|}}$$

Gigili

@unNaturhal Umm, OK. So long as it's not dangerous!

@unNaturhal At which point? $0$?

Jonas Teuwen

@N3buchadnezzar He actually did both! Troll 4chan and found a defect grenade.

Or he has The Balls.

N3buchadnezzar

@JonasTeuwen Not anymore!

Jonas Teuwen

@N3buchadnezzar Or maybe... that is all there is left?

unNaturhal

18:20

I studied the domain, and it is
$\forall x \in \mathbb{R}-{-\frac{1}{e}, 0, \frac{1}{e}}$ I think...

N3buchadnezzar

@JonasTeuwen Would they then fit that in a coffin for the funeral? Like his relatives comes to the open coffin to see just a pair of balls.

Gigili

Hum?

Jonas Teuwen

@N3buchadnezzar That's basically all he had as he apparently lacked a brain.

N3buchadnezzar

Some girls have breasts instead of brains. Some guys have balls instead of brains.

unNaturhal

Ok, I dunno how to write it, however it's "for all x except $-\frac{1}{e}$, $0$ and $\frac{1}{e}$"

user19161

18:23

@N3buchadnezzar Erm, I think they have both.

Gigili

@unNaturhal Sounds right so far. wolframalpha.com/input/?i=1%2Blog+%7Cx%7C+%3D0

Jonas Teuwen

@JasperLoy Non-functional brains are equivalent to no brain at all!

unNaturhal

@Gigili Ehm... why you put $1 + \log|x| = 0$?

Gigili

@unNaturhal Point Discontinuities arise when the function has a denominator that can be equal to zero.

Didn't you the same thing? How you calculated the domain?

Except when it can be cancelled out with a like term in numerator.

unNaturhal

@Gigili When I calculate the domain I set the condition of existence of all part of the functions that can give problem

In this case i set: Denominator different than zero and |x| > 0

Gigili

18:34

@unNaturhal That's what I said, but differently. You set it equal to zero, then the domain would be $\mathbb R$ except those values. Just like what you said beforehand.

Except $- \frac 1e$, etc.

unNaturhal

Mmmmh.. ok.. And now?

Gigili

And now, umm, you have the set of points of discontinuity.

I don't know what you're supposed to do.

unNaturhal

Ok, I know that the discontinuity point needs to be studied in the points that not belong to the Domain

So I tried to to this

Gigili

OK.

unNaturhal

I got $f(x_0)$, for the three points

And for $-\frac{1}{e}$ and $\frac{1}{e}$ it gives back $\infty$

But for $x_0 = 0$ it gives 0, so I got the limith both from right and left, but it gives $0$ too

Gigili

18:42

So if $f(0)= \lim_{x \to 0} f(x)$, it must be excluded.

unNaturhal

@Gigili Excluded from what?

You mean that it's not a discontinuity point?

Gigili

From the set of points of discontinuity of the function.

@unNaturhal Yes.

unNaturhal

@Gigili Ok, and what about the other points?

Gigili

@unNaturhal You do what you said above, if $f(x_0)=\lim_{x \to x_0} f(x)$ then it's not a discontinuity point.

1 hour ago, by unNaturhal

I'm sorry guys, I've a trouble... When I study discontinuity in a function, after doing $f(x_0)$ and the right and left limit for $x\to x_0$, how I have to continue?

Well, @Peter knows better!

Where is @Ben? I miss him!

unNaturhal

@Gigili So I finisced? I haven't to do nothing more?

Peter Tamaroff

18:51

@Gigili What's going on¿

Gigili

@unNaturhal I don't think so.

Do you have another example?

Peter Tamaroff

What's going on?

unNaturhal

@Gigili I dont' now... I'm not familiar with discontinuity, for this reason I do not have practice..

Eugene

literally after over 100 math lectures i still can't figure out when two trains travelling towards one another will collide

Peter Tamaroff

@Eugene Hahaha c'mon, think relatively!!!

Say train $A$ is moving at $X \rm km/h$ and train $B$ at $Y \rm km/h$. Then take train $A$ as a reference: $B$ is moving at $X+Y \rm km/h$ towards the "fixed point" of $A$. Profit.

Eugene

18:56

i remember once during my modular forms class we had to compute a large fraction addition our heads. three professors in the class and we all had to wait for the kid with the calculator for the answer.

Peter Tamaroff

@Eugene lol

Eugene

we aren't trained to do arithmetic is my theory

Gigili

@unNaturhal It's the same, you need to calculate the domain first and ...

unNaturhal

@Gigili Ok, thank you bery much :)

user19161

19:47

@unNaturhal Just to add to what gigili said, in general make sure the left limit exists, the right limit exist, the function is defined at the point, and all three are equal.

Gigili

@unNaturhal You're bery welcome.

Peter Tamaroff

@JasperLoy That is being redundant, but in general that is how it is taught.

Continuity is stated better as $$\lim\limits_{x \to c }f(x)=f(c)$$

this in turns implies $f(c)$ is defined, and that $\lim\limits_{x \to c }f(x)$ is defined, which in turn means the left and right limits exist.

user19161

@PeterTamaroff I know. I am just telling him the things to check for since he is looking at discontinuity.

Peter Tamaroff

@JasperLoy Oh. OK.

@JasperLoy Then I just add $f$ is continuous at $c$ $$\iff \lim\limits_{x \to c }f(x)=f(c)$$

@JasperLoy Question:

Gigili

@PeterTamaroff That's what I said last year.

Peter Tamaroff

19:57

Let $f$ be a relation that maps the each elements of $A$ to a unique element in $B$. That is, $f$ is a function from $A$ to $B$, which we denote $f: A \to B$. What would you call $B$?

@Gigili Last year?

user19161

@PeterTamaroff Joke.

Peter Tamaroff

@JasperLoy You answer the other question! =)

user19161

@PeterTamaroff There are various conventions for this. Some people call B the codomain and f(A) the range.

user19161

Other people call B the range and f(A) the image.

Gigili

@PeterTamaroff Codomain.

Peter Tamaroff

20:01

Right. That's why!

Gigili

Codomain

In mathematics, the codomain or target set of a function is the set into which all of the output of the function is constrained to fall. It is the set in the notation . The codomain is also sometimes referred to as the range but that term is ambiguous as it may also refer to the image. The codomain is part of the modern definition of a function as a triple , with a subset of the Cartesian product . The set of all elements of the form , where ranges over the elements of the domain , is called the image of . In general, the image of a function is a subset of its codomain. Thus, it may...

Oh, jasper said it.

Peter Tamaroff

@Gigili Oh, cool.

Gigili

I'm awesome like that.

user19161

Note that it is very common for one word to mean several things and several words to mean one thing in math.

user19161

You will know this after reading 9000 books.

Peter Tamaroff

20:01

I have this introductory book on topology (sets, metric spaces, topological spaces, connectedness, compactness) that calls $B$ the range.

@JasperLoy Why not OVER 9000 books? =D

user19161

@PeterTamaroff I was lazy to type over.

user19161

@PeterTamaroff Exactly, I don't talk rubbish here you know.

user19161

@PeterTamaroff Also note that that definition is for limit points. Depending on how you define continuity, continuity may hold vacuously at isolated points.

Eugene

@JasperLoy dude it's 4am man you should be asleep

user19161

@Eugene I don't sleep at regular hours. But I'll try to get back to it.

Gigili

20:11

@JasperLoy Is dude your first name?

And man your family name?

And what's your middle name?

Hero?

user19161

@Gigili hunk

Eugene

@JasperLoy well good luck trying to get back to bed. i'm going to study somemore.

bye all

user19161

@Eugene You should sleep at regular hours too. Only lunatics like me are excused.

Peter Tamaroff

@JasperLoy I work better at nighttimes too!

I'm going to watch a movie now. BBL

Gigili

@JasperLoy Dude Hunk Man. Delighted to make your acquaintance.

unNaturhal

20:35

@JasperLoy Perfect, thanks :)

@Gigili LOL xD I'm sorry xD

See you tomorrow guys ;) Good night!

David Wallace

@JasperLoy Dude hunk man who thinks he's Justin Bieber? Please!

Jordan

It is too hot to do math

user19161

@DavidWallace Sometimes I am Taylor Lautner too.

user19161

Hmm, time for new avatar then...

N3buchadnezzar

@JasperLoy Still gay then

Peter Tamaroff

21:08

@JasperLoy Are you around?

I want to prove this is true:

If ${A_1} \subseteq {A_2} \subseteq \cdots \subseteq {A_n} \subseteq \cdots $ are all finite non empty sets (of real numbers) then $$S=\bigcap_{n \in \Bbb N} A_n$$ is finite and non-empty.

Finiteness is evident since $$S \subset A_1$$

Non emptyness by contradiction.

Assume that $S = \emptyset$.

leo

hey

Peter Tamaroff

Then it must be the case that $\emptyset=A_i \cap A_j$ for some $i,j$ which is impossible.

@leo Hey

@anon

anon

hey

Peter Tamaroff

How's it going guys?

leo

is there a service in internet where you search a book or paper and then it returns the bibtex entrie of that book or paper?

like citeseer

Peter Tamaroff

21:15

@leo Prolly your uni's library...

anon

I'm thrashed from mulching my grandma's garden all day in monkey-high temperature

leo

but citeseer don't work with books

Peter Tamaroff

@anon Just wait for some non instant Karma.

leo

@PeterTamaroff oh that's true

:-)

gracias por recordarmelo

Peter Tamaroff

@leo De nada!

anon

21:17

@PeterTamaroff what makes you think my time today wasn't already karma's punishment for a prior transgression? :)

Peter Tamaroff

@anon Hahahahaha, well, that's another option!

@anon Seen what I wrote up there?

anon

about the intersection?

Peter Tamaroff

@leo Any thoughts on the above?

@anon Yup

leo

@anon enjoy it

let me see

anon

why would you use $\subseteq$ in the chain but only $\subset$ when writing $S\subset A_1$?

Peter Tamaroff

21:19

@anon Don't know.

Sorry.

I should have paid attention to that.

Dylan Moreland

@PeterTamaroff You mean to reverse these inclusions, right? Otherwise the intersection is just $A_1$.

anon

haha, woops

Peter Tamaroff

@DylanMoreland Darn, yes!!!

$${A_1} \supseteq {A_2} \supseteq \cdots \supseteq {A_n} \supseteq \cdots $$

Jordan

....if I am trying to find the area of a shaded region on a polar coordinate system with $r = \sqrt{\theta}$ and it crosses into all four regions I have to break it up into 4 integrals right?

anon

I would prove there is an $n$ such that $A_n=A_{n+1}=A_{n+2}=\cdots$.

Peter Tamaroff

21:21

@DylanMoreland Apart from that, is the reasoning OK?

Dylan Moreland

It is probably not important that these are real numbers. Although that suggests using compactness (I would not recommend this; it's completely overkill).

Peter Tamaroff

@DylanMoreland Oh, no no, no way. That is far from this chapter.

@DylanMoreland Yeah that's why it is in parenthesis

(I purposedly put them in parenthesis, thought it wasn't parenthesised in the text)

anon

@Jordan that depends kinda on the shaded region

Dylan Moreland

@PeterTamaroff I don't see enough reasoning for my taste, I guess.

leo

@Jordan possibly. We can't know with only that information

Jordan

21:22

@anon iti s in all 4 reghions

Peter Tamaroff

@DylanMoreland How would you complete it? (What's missing?)

I will complete it. Give me a sec.

anon

@Jordan that doesn't tell me much. is it $\theta$ from $0$ to $2\pi$, bounded by the interval $[0,\sqrt{2\pi}]$ on the $x$-axis?

Dylan Moreland

For contradiction: for each element $b_i \in A_1 = \{b_1, \ldots, b_k\}$ you could do the following. Since $S = \varnothing$ there exists an $n(i)$ such that $b_i \notin A_{n(i)}$.

You get finitely many indices. Take the maximum of those.

leo

something like pingeonhole principle

Dylan Moreland

ChatJax is still a little wonky. Am I going to have to relearn JavaScript :(

Peter Tamaroff

21:26

@DylanMoreland It's working fine for me.

Dylan Moreland

Does the above make sense? Gotta walk the dog for a bit, so hopefully it does.

Peter Tamaroff

@DylanMoreland But isn't it enough to state that if $S = \varnothing$ then at least one of the sets is disjoint¿

@anon Why should that be the case?

@Jordan Try writing it in cartesian coords, maybe.

anon

the first set is finite, and their cardinalities form a nonincreasing sequence of positive integers. so at some point the sequence will be constant...

Jordan

All the information I have is $r = \sqrt{\theta}$

Peter Tamaroff

@Jordan Have you tried graphing it?

anon

21:30

can't do anything without bounds on $\theta$

though $[0,2\pi]$ is a reasonable guess

Peter Tamaroff

@anon I get you point, but is my reasoning correct?

Dylan said it was insufficient

anon

and if it's the case, then the integral (done using polar coordinate formulae) does not need to be split

why must it be the case that $\emptyset = A_i\cap A_j$ for some $i,j$? (I mean, you have to build up the reasoning)

Peter Tamaroff

@anon From a scale from 1 to 10, I think that is a 5 trivial. Maybe a 7.

N3buchadnezzar

How is it going guys?

anon

2 days ago, by anon

if you say so bro

Peter Tamaroff

21:34

@anon Hay, but you're right!!!

Jordan

@PeterTamaroff I have a graph

Peter Tamaroff

Though it is so evident I can't prove it.

Jordan

This problem is really pissing me off actually

N3buchadnezzar

@Jordan I have icream =)

Peter Tamaroff

@Jordan Calm down.

Is it in a book?

Jordan

21:36

It is 100 degrees here and no AC, too hot to be doing math for a class I already failed

yes

Peter Tamaroff

I think I have Stewart's Calculus in my "digital library".

Jordan

section 10.4 #5 I think

N3buchadnezzar

@PeterTamaroff You mean "Yarr Yarr, shiver me timbers pirates ahead folder"?

Peter Tamaroff

@Jordan Edition?

Jordan

7e

Peter Tamaroff

21:40

@Jordan It says $0 \leq \theta \leq \pi /4$!!!!!!!!!!!!!!

@N3buchadnezzar LOL

Jordan

Mine doesn't say that

Peter Tamaroff

@Jordan Odd.

Well, there you go.

Jordan

What if it was just a standard circle on a cartesian graph

how would I find the circle if it wasn't centered on the origin?

or ellipse or whatever

some shape that isnt symetric

anon

in cartesian coordinates?

Jordan

yes

Peter Tamaroff

21:43

@Jordan A circle has the general eqn

Jordan

fuck circles

Peter Tamaroff

$(x-h)^2+(y-k)^2=r^2$

The center is at $(h,k)$ and the radius is $r$.

anon

More generally, if $F(x,y)=0$ describes some curve in the plane, then $F(x-a,y-b)=0$ describes the same curve but translated by the vector $(a,b)$.

Peter Tamaroff

@anon Let's say, moved right $a$ units in the $x$-axis and up $b$ units in $y$-axis for simplicity's sake.

anon

as you wish

Peter Tamaroff

21:47

@anon I like yours better, but just not to overcomplicate things.

Jordan

I have no idea how to do this problem

nor does the book

Peter Tamaroff

@Jordan Which one?

@Jordan How can the book know!?

Jordan

the same one

By displaying the steps

Peter Tamaroff

@Jordan You need $0 \leq \theta \leq \pi/4$. The area of the polar graph of $p(\theta)$ is given by what integral, in general?

Jordan

$r = \sqrt{\theta}$

and no bounds are given, it is a spiral thing that ends on 0 or 2pi

Peter Tamaroff

21:49

@Jordan Focus! You know that the are from $\theta = a$ to $\theta = b$ is $$\int_a^b \frac 1 2 \rho(\theta)^2 d \theta$$

in this case you have $0$ to $\frac \pi 4$

And your functin is

$\rho(\theta) = \sqrt{ \theta}$

So you get

Jordan

why pi/4?

Peter Tamaroff

$$\int_0^{\pi /4} \frac 1 2 \theta d \theta$$

We just said the bound are those $0$ and $\pi /4$

Jordan

but why

the book doesn't specify any bounds

Peter Tamaroff

@Jordan The book says that! I just told you!

@Jordan I feel like you're trolling me now, dude.

Jordan

Mine doesn't though

You have a different version than I do

And yeah anyone who isn't a genius at math just is a troll, why is that such a popular belief here?

Peter Tamaroff

21:53

@Jordan I'm not saying that.

Just some minutes ago we said the bound were $0 \leq \theta \leq \pi/4$

Jordan

I am sure there are some people here who are horrible with tools, who don't even know what a breaker bar or crescent wrench is

Peter Tamaroff

Seems you're not paying enought attention, not that you're not a genious at math.

Jordan

and that would be absurd for most people to hear

Peter Tamaroff

I'm not a genious at math either.

Jordan

But my book doesn't give that specification

You are like 18 and doing third year college math, close enough to genius

Peter Tamaroff

21:54

@Jordan I'm not even in first year, Jordan. I told you already.

Jordan

college algebra is first year math

Peter Tamaroff

@Jordan We have to choose some bounds. You're book seems to have an error, so let's choose one, to move on. Don't you think it is logical to do so?

There is no sense is asking for the area if no bounds are given.

(For this function)

Jordan

Sure but to me the bounds looks like that are 0 to 2pi

Peter Tamaroff

@Jordan Well, solve it with that. I'm off to study.

Jordan

wolframalpha.com/input/?i=cartesian+r+%3D+%5Csqrt%7Btheta%7D+

Thank you for trying to help

Peter Tamaroff

21:58

@Jordan Welcome.

ymar

I've just asked a really stupid question. Do you think I should ask the only answerer to delete his answer so I can delete my question too? That would save me some embarrassment, but also relieve the main page of one idiotic question.

Peter Tamaroff

@ymar Don't delete it, it might help others later. I don't think it is stupid.

ymar

@PeterTamaroff OK, I was afraid I might hear that... Thanks.

anon

I have a handful of stupid questions. Ya gotta ask them sometime or later.

In fact I cringe at a few of the questions I've asked...

ymar

@anon Thankfully, I've asked enough questions not to see this one when I open my user page. :)

Jordan

22:09

I am just giving up in my calculus class, there is no point

I am going to fail anyways and it is just exhausting and too frustrating to try and learn it this fast

N3buchadnezzar

Chat is so irritating now

Were are all the images of cute kittens?

anon

therestoftheinternet.stackexchange.com

N3buchadnezzar

?

Peter Tamaroff

@N3buchadnezzar There's actually a cluster point for those: icanhascheezburger.com

Dylan Moreland

@anon That sounds good.

N3buchadnezzar

22:15

@PeterTamaroff They need to be more spread out

there stof the internet huh

Peter Tamaroff

@DylanMoreland I was telling anon that I claim that if $S = \varnothing$ then it must be the case $A_j \cap A_i = \varnothing$ for some $i,j$.

But then I have to prove that claim.

N3buchadnezzar

Is it more common to have the equal sign on the left or the right side, or does it deppend?

Peter Tamaroff

I think it is the analogous to $a \cdot b =0 \rightarrow a=0 \vee b=0$ in integral domains. Thought this is not true for an arbitrary collection of sets in general, in this particular case it is.

Since each set is contained in its predecessor.

Jordan

I am two weeks behind in my calc class, I think I am just going to stop trying and focus on something more productive

Dylan Moreland

22:32

@PeterTamaroff There's some justification to make. And it's important to realize how this is different from, say, the fact that $\bigcap_n (0, 1/n) = \varnothing$ and yet any two of these intersect.

So it isn't just some easy property of intersections.

It's even a decreasing sequence!

Peter Tamaroff

@DylanMoreland But that interval is uncountable!

I'm talking about finite sets here.

Dylan Moreland

I know you are.

Peter Tamaroff

Moreover, that interval is not closed. If it were we'd have $\cap \neq \varnothing$

Dylan Moreland

I'm just pointing out why I think you have to justify something.

Peter Tamaroff

@DylanMoreland I can't figure out what it is. I told anon before: the result seems so obvious I can't seem to figure out how to prove it.

I'm not disregarding your claims, I appreciate them.

Dylan Moreland

22:41

This claim about $A_i \cap A_j$ being empty, you mean?

leo

What about the anon's suggestion: prove that the sets are the same starting with some index

otherwise there is a infinite sequence in $A_1$ which is finite

Jordan

This is pointless trying to learn calc 2

I don't understand how they expect you to know all this stuff about shifted conic sections without even teaching it in this class

I took college algebra like 2 years ago, how am I suppose to remember all the miscelaneous formulas?

Dylan Moreland

@leo Ooh, I like that one.

Peter Tamaroff

@DylanMoreland What I'm saying is that if $A_1 \supseteq A_2 \supset \cdots \supset A_n \supset \cdots$ and $\cap A_n = \varnothing$ then it must be the case $A_i \cap A_j = \varnothing$ for some $i,j$, but that is impossible.

@leo Yeah, but I like to do it my way. I understood that one, now I want to understand mine =)

leo

oh see

@PeterTamaroff why is it impossible?

Peter Tamaroff

22:49

@leo I added the premise.

David Wallace

@JasperLoy Is that a man or a woman?

leo

what I don't understand is $\cap A_n = \varnothing$ implies it must be the case $A_i \cap A_j = \varnothing$ for some $i,j$

Peter Tamaroff

@leo Because ${A_1} \supseteq {A_2} \supseteq \cdots \supseteq {A_n} \supseteq \cdots $

Dylan Moreland

I mean, $\{A_n\}$ decreasing and each $A_n$ finite really implies that the $A_n$ are eventually $\varnothing$, which is even better. So that's why it seems like a weird conclusion.

Peter Tamaroff

@DylanMoreland Or eventually "constant" as anon says.

Dylan Moreland

22:54

Sorry, I meant to add the condition that the intersection was empty.

I was trying to find the most economical way of stating the setup and managed to forget one of the parts!

Peter Tamaroff

@DylanMoreland We're assuming each $A_n$ is non empty, sorry.

Dylan Moreland

I get that.

But you're trying to prove something by contradiction.

Peter Tamaroff

@DylanMoreland Yeah.

Dylan Moreland

And I find it easier to think about the contrapositive of a related statement. It's just less confusing.

Peter Tamaroff

Basically from the premise, no set is disjoint to the other.

So that the intersection can't be empty.

Dylan Moreland

22:57

I mean, we've already spent way more time on this than it deserves.

Peter Tamaroff

And since no set is disjoint, $\bigcap A_n=\varnothing$ actually means $A_n = \varnothing$ for some $A_n$ which is false by hypothesis.

Dylan Moreland

I think you could modify any of the three arguments given to prove your claim.

Peter Tamaroff

@DylanMoreland Yeah, you're right.

Dylan Moreland

:)

Peter Tamaroff

@DylanMoreland Have you seen the book "Understandying Analysis" by Steven Abbott?

Jordan

22:59

I can't study math for more than a minute, it feels so pointless

Transcript for

$Mathematics$ Mathematics