Mathematics - 2012-07-19 (page 1 of 6)

Peter Tamaroff

12:07 AM

@BrianMScott

Brian M. Scott

@PeterTamaroff What’s up?

Peter Tamaroff

@BrianM.Scott I'm doing some excercises on open and closed sets.

Chuck Fernández

@BrianM.Scott what was wrong with the centroid answer?

Peter Tamaroff

How are you doing?

Old John

@PeterTamaroff Did you get that topology problem nailed?

Brian M. Scott

12:14 AM

@PeterTamaroff Life’s been a bit hectic, so I’ve not been in here much.

Peter Tamaroff

@OldJohn Almost. I'm missing the last part.

@BrianM.Scott Oh. Well, I hope it wasn't bad-hectic.

Brian M. Scott

@ChuckFernández The centroid is where the net first moment is $0$, not where the areas on opposite sides of every line are equal.

Chuck Fernández

oh, right

Brian M. Scott

@PeterTamaroff So what’s the question?

Peter Tamaroff

@BrianM.Scott I have to prove that this

Let me find the question up in the lof

3 hours ago, by Peter Tamaroff

I need to prove that if $F$ is a closed set such that $A\subset F$ then $\overline A \subset F$, that is, $\overline A$ is the smallest closed set w.r.t. to inclusion that contains $A$ (I suppose?)

Darn, no. I already proved that.

3 hours ago, by Peter Tamaroff

And finally I have to show that $\overline A$ is the intersection of all such sets $F$ so that $\overline A$ is closed.

3 hours ago, by Peter Tamaroff

Let $F_{\alpha}$ be closed and $A\subset F_{\alpha}$ for each $\alpha$ in an indexing set $I$. We must show $$\overline A = \bigcap_{\alpha\in I} F_{\alpha}$$

Brian M. Scott

12:17 AM

What’s your definition of $\operatorname{cl}A$?

Peter Tamaroff

@BrianM.Scott Let $A'$ be the set of all limit points of $A$, and $A^i$ the set of isolated points of $A$. Then $\operatorname{cl}A=A'\cup A^i$

Brian M. Scott

Okay. I think that it’s a fairly silly way to define it, but we can work with that.

Peter Tamaroff

But I denote it by $\overline A$.

@BrianM.Scott Why?

Chuck Fernández

are you really 64 years old?

Brian M. Scott

The natural definition of that general type is that $\operatorname{cl}A$ is the set of points $x$ such that every nbhd of $x$ meets $A$. The other natural definition is that the theorem that you want to prove.

@ChuckFernández ’Fraid so.

Chuck Fernández

12:21 AM

do you get better or worse at math?

Peter Tamaroff

@BrianM.Scott You're almost as old as my dad.

Brian M. Scott

@ChuckFernández Well, I’ve more experience and more background, but I may not be quite so quick as when I was in grad school, for instance.

Peter Tamaroff

@BrianM.Scott I've seen it defined as $\overline A = A\cup A'$

That is, $A$ plus its limit points.

Chuck Fernández

do you think you would are able to more problems regardless of speed?

the would is a mistake in the sentence

Peter Tamaroff

@BrianM.Scott I've proved inclusion on one way.

Let $x\in \overline A$. Then $x$ is either a limit point or an isolated point of $A$. Assume the first, then $x$ is a limit point of each $F_{\alpha }$ so $x\in F_{\alpha}$ for each index since each $F$ is closed. If $x$ is an isolated point of $A$ then $A\subset F_{\alpha}$ so finally $x\in F_{\alpha}$ for each index so $\overline A \subset \bigcup_{\alpha \in I}F_{\alpha}$-

Brian M. Scott

12:24 AM

@PeterTamaroff Yes, that’s another; I just don’t normally see any great need to distinguish the points that are already in $A$ from those that are not. Eventually one wants to prove that all of these are equivalent anyway, but I prefer to pick a nice one as my definition.

Peter Tamaroff

@BrianM.Scott Your definition is what I'm trying to prove?

But if every nbhd of $x$ meets $A$ then $x\in A$ or $x$ is a limit point of $A$, right?

Brian M. Scott

@PeterTamaroff One of them is: one standard definition is that $\operatorname{cl}A=\bigcap\{F\subseteq A:F\text{ is closed}\}$.

Peter Tamaroff

@BrianM.Scott OK. Now I'm used to that notation because of Halmos' book. I'm trying to do some further reading on it

Brian M. Scott

@PeterTamaroff You said that you’d already shown that if $F$ is a closed set containing $A$, then $\operatorname{cl}A\subseteq F$; from this it’s immediate that $\operatorname{cl}A\subseteq\bigcap\{F\supseteq A:F\text{ is closed}\}$. (Note the typo in my earlier comment: that should have been $F\supseteq A$.)

Peter Tamaroff

@BrianM.Scott Yes, superset.

Chuck Fernández

12:30 AM

whats the best book in topology someone who has finished topology without tears can read?

Peter Tamaroff

@ChuckFernández That might be a good question for [reference-request] on main.

I think Mendelson's book is good, though.

Brian M. Scott

To get the opposite inclusion, merely note that $\operatorname{cl}A$ is one of the closed sets containing $A$, so $\bigcap\{F\supseteq A:F\text{ is closed}\}\supseteq\operatorname{cl}A$.

Chuck Fernández

i allready have that one remember?

Peter Tamaroff

@BrianM.Scott =D OH DEAR!

Brian M. Scott

@PeterTamaroff Some things really are easy. :-)

Peter Tamaroff

12:31 AM

@BrianM.Scott I need to get my set theory striaght!

@BrianM.Scott Wait wait.

But the question is asking me to show that the closure is indeed closed by showing it is the intersection of closed sets.

Brian M. Scott

@ChuckFernández Perhaps the best introductory graduate level text in general topology is Willard’s General Topology, which I believe is available in an inexpensive Dover edition.

Peter Tamaroff

=/

Chuck Fernández

does it have a lot of set theory notation and stuff?

Brian M. Scott

@ChuckFernández I’m not really sure what you mean by that.

Peter Tamaroff

@ChuckFernández If you're weak in Set THeory, I reccommend you strengthen it up.

Chuck Fernández

12:34 AM

Is it easier than John L Kelleys?

Brian M. Scott

@ChuckFernández I’ve mostly forgotten that one, since I never liked it, but I rather think that Willard is a little more comprehensive and a little more advanced; it’s certainly more modern in notation, terminology, and overall coverage.

Chuck Fernández

great

Brian M. Scott

@ChuckFernández Have you seen the book by Munkres?

Chuck Fernández

no

is it good?

Peter Tamaroff

@BrianM.Scott I forgot to tell you I don't have the assumption that $\operatorname{cl}A$ is closed. This exercise is aimed to prove it is closed by showing it is the intersection of closed sets.

Brian M. Scott

12:39 AM

@ChuckFernández I’ve a few objections to it, but overall it’s very good. This isn’t just my opinion, either: the book is generally considered to be the standard against which other undergraduate topology texts should be compared.

@PeterTamaroff Ah, okay. That makes things a little harder. The contrapositive is easiet. Suppose that $x\notin\operatorname{cl}A$. Then you should be able to show without much trouble that there is an open set $U$ such that $x\in U$ and $U\cap A=\varnothing$. Now let $F=X\setminus U$: $F$ is a closed set containing $A$, and $x\notin F$.

robjohn

@KannappanSampath: you're here again! How are things?

Peter Tamaroff

@BrianM.Scott Let me read that. THis is something we were discussing earlier: If $x=\lim \, x_n$ with each $x_n \in A$, then $x$ is either a limit point or an isolated point of $A$.

robjohn

@PeterTamaroff and it could only be isolated if all but finitely many of the $x_n=x$

Chuck Fernández

youre on page 57?

im reading the book right now

Peter Tamaroff

@ChuckFernández Yes.

Brian M. Scott

12:49 AM

@PeterTamaroff Don’t think about sequences: you’re trying to prove something that’s true in all spaces, even those in which sequences don’t determine the topology.

Chuck Fernández

what excercize

excercise

Brian M. Scott

exercise

Chuck Fernández

right

im on page 75

Peter Tamaroff

@BrianM.Scott But that just arose in the discussion of $x\in \overline A \iff x=\lim\;x_n $ with $_n \in A$. That is, $x$ is in the closure of $A$ if and only if $x$ is the limit of a sequence of points in $A$.

Chuck Fernández

theyre palindromes

Peter Tamaroff

12:52 AM

@BrianM.Scott I'm dealing with metric spaces now, but you're probably right.

Brian M. Scott

@PeterTamaroff That’s true in metric spaces (among others), but it’s not true in general, and you don’t need it.

robjohn

@PeterTamaroff then this is not a general discussion of topology, but must be limited to some special spaces. (what Brian said)

Peter Tamaroff

@BrianM.Scott Yes, I'm working with metric spaces now.

I am $2$ sections away of the chapter 3 which is Topological Spaces

Brian M. Scott

This result is true for spaces in general.

Peter Tamaroff

@BrianM.Scott I take your word, but I haven't the slightest idea of general topological settings. I'm sorry.

Brian M. Scott

12:55 AM

What’s your definition of ‘$x$ is a limit point of $A$’?

Peter Tamaroff

@BrianM.Scott Let $(X,d)$ be a metric space. $x \in X$ is a limit point of a set $A\subset X$ if every neighborhood of $x$ contains a point of $A$ other than $x$.

Brian M. Scott

Fine: that definition works for spaces in general, and that’s all you need to carry out the contrapositive argument that I sketched above.

Peter Tamaroff

@BrianM.Scott $X\setminus U$ is the complement of $U$ relative to $X$?

Brian M. Scott

Yes.

Peter Tamaroff

@BrianM.Scott I forgot to change the last $F$.

@BrianM.Scott You think I could consrtuct $U$ explicitly or just show it exists?

Brian M. Scott

1:01 AM

@PeterTamaroff You can’t construct it explicitly, but it’s easy to show that it exists. That’s practically the definition of $x$ not being in the closure of $A$.

Chuck Fernández

Is theophiles answer correct?

@BrianM.Scott

Brian M. Scott

@ChuckFernández Answer to what?

Chuck Fernández

it seems like copper hats answer to me

the triangle centroid thing

Peter Tamaroff

@BrianM.Scott If $x$ is not in the closure of $A$ then $x$ is not a limit point of $A$ neither it is an isolated point of $A$. Doesn't that mean $x\notin A$?

$x$ is isolated if there is a nbhd $N$ of $x$ such that $N\cap A=\{x\}$

Brian M. Scott

@ChuckFernández No, it has the same problem.

@PeterTamaroff The fact that $x$ is not a limit point of $A$ means that $x$ has a nbhd $N$ such that $N\cap A\subseteq\{x\}$. The fact that $x$ is not an isolated point of $A$ then implies that $N\cap A$ must be empty, since it can’t be $\{x\}$. So $N$ is the $U$ that I wanted.

Peter Tamaroff

1:06 AM

@BrianM.Scott OK. So if $x$ is not in the closure of $A$ then it can't be in $A$ either.

Brian M. Scott

@PeterTamaroff Indeed.

Peter Tamaroff

@BrianM.Scott Though thinking it over like you did is more enlightening.

@BrianM.Scott Thank you, Brian.

Brian M. Scott

Sure thing.

Peter Tamaroff

@BrianM.Scott I'm intrigued by something.

Brian M. Scott

What’s that?

Peter Tamaroff

1:09 AM

When you say $N\cap A\subseteq \{x\}$ you're implying that maybe the intersection is empty right?

Brian M. Scott

Yes: either it’s empty, or it’s $\{x\}$.

Peter Tamaroff

$\subseteq$ means improper subset right?

Brian M. Scott

@PeterTamaroff It means any subset; it allows the subset to be the whole thing but certainly doesn’t require it to be.

Peter Tamaroff

@BrianM.Scott OK. Sorry if I miss some basic set theoretic stuff.

Brian M. Scott

@PeterTamaroff $A\subseteq B$ simply means that every element of $A$ (if any) is an element of $B$. $A\subsetneqq B$ means that every element of $A$ is an element of $B$, and there is at least one element of $B$ that isn’t in $A$.

Peter Tamaroff

1:13 AM

@BrianM.Scott Adn $A\setminus B=A-B$, correct?

Brian M. Scott

The second notation is a bit old-fashioned, but yes.

Peter Tamaroff

@BrianM.Scott I know how you feel. Today I taught a 14 year old the arithmetic of fractions.

@BrianM.Scott Hehehe, that's because I read it in Halmos' book.

Brian M. Scott

@PeterTamaroff $14$ seems a little late.

Peter Tamaroff

@BrianM.Scott She failed the subject last year, that's why.

@BrianM.Scott This book I'm using had it first edition in 1962 so that should explain a lot.

Brian M. Scott

I figured that it was probably something like that.

Peter Tamaroff

1:15 AM

@BrianM.Scott Though it is a little sad someone failed to teach someone else the arithmetic of fractions.

@BrianM.Scott If I was a teacher I would see that as a personal failure.

Brian M. Scott

No one can be sure of reaching every student, especially when working with many students at once.

Peter Tamaroff

@BrianM.Scott Well, that is partly true. If a student is failing I'd reach to him/her.

Brian M. Scott

And if the student doesn’t reach back? Or if your teaching style just doesn’t match the student’s learning style? And what if there simply isn’t time?

Peter Tamaroff

@BrianM.Scott Well, if the student doesn't reach back then that's another issue. But well, I think you're the experienced teacher. I shouldn't be telling you what is or isn't true about teaching. =X

@BrianM.Scott How does it follow that $X-U$ is closed? Maybe some argument about $C(X-U)$ being open?

Chuck Fernández

I used to be the worst masth studant in my class. My teacher made me stay during recess to do substractions.

Brian M. Scott

1:23 AM

@PeterTamaroff I always found it distressing when a student just didn’t seem to grasp something no matter what I tried, and I think that many $-$ perhaps even most $-$ teachers feel the same way, but at some point most of us realize that it’s inevitable, and all we can hope to do is minimize it.

@PeterTamaroff The complement of an open set is closed.

Peter Tamaroff

@BrianM.Scott Oh, yes. But in this case, can't be write $X-U=C(U)$?

Chuck Fernández

i have a question

Brian M. Scott

@PeterTamaroff What do you mean by $C(U)$? If you simply mean the complement of $U$ in $X$, $X\setminus U$ is the normal notation.

Peter Tamaroff

$X-U$ is the complement of $U$ relative to $X$, but in this case $X$ is the "universe" of the space, so $C(X)=\varnothing$.

@BrianM.Scott Yes, I mean the complement of $U$ in XX$

Chuck Fernández

can you prove that the area of a triangle is the length of any side multiplied by half the length of its height ?

fixed at the same side

Peter Tamaroff

1:27 AM

@ChuckFernández Extend the triangle to a paralelogram,

Brian M. Scott

@ChuckFernández Sure: do a dissection of a rectangle with that side as base and that height as height.

Chuck Fernández

and then what?

wait, i see its impossible

without using calculus

because without calculus are doesnt exist

Peter Tamaroff

@ChuckFernández No.

You can do it.

Chuck Fernández

but without calculus area doesnt exist

Brian M. Scott

@ChuckFernández Nonsense. The calculus definition of area depends on the fact that the area of a rectangle is the product of its base and height.

Peter Tamaroff

1:30 AM

@ChuckFernández What do you mean?

@BrianM.Scott (+1)

Chuck Fernández

ok

Peter Tamaroff

@ChuckFernández Who told you/where did you read that?

Chuck Fernández

but the triangle is just an intersection of rectangles

Peter Tamaroff

Is it safe to buy used books in Amazon?

Brian M. Scott

@PeterTamaroff If you pay attention to the seller ratings. I’ve done it a few times with excellent results.

Chuck Fernández

1:32 AM

but isnt a triangle just an intersection of rectangles??

Peter Tamaroff

@BrianM.Scott OK. Brian, I proved that the graph of a continuous function $f:X\to Y$ where $X$ and $Y$ and metric spaces is closed by a sequence argument.

Brian M. Scott

Okay.

Peter Tamaroff

Is it true in general?

I ask because of
" **Don’t** think about sequences: you’re trying to prove something that’s true in all spaces, even those in which sequences don’t determine the topology."

Chuck Fernández

so the area of a triangle can in fact be described without calculus then?

Peter Tamaroff

@ChuckFernández What is your point?

Chuck Fernández

1:34 AM

can the sam e be made with the moment of a circle?

Peter Tamaroff

@ChuckFernández Moment? That is no longer an area, is it?

Chuck Fernández

no

can it?

Peter Tamaroff

@ChuckFernández If I recall correctly, the moment of a circle is defined as a quotient of integrals.

Chuck Fernández

yup

Peter Tamaroff

@ChuckFernández Well, try and figure out based on that.

Chuck Fernández

1:36 AM

but area is also defined as an idefinite integral

definite*

Peter Tamaroff

@ChuckFernández Not really.

Chuck Fernández

why not

Peter Tamaroff

@ChuckFernández You say "area" is defined as {blah}

But

Brian M. Scott

@PeterTamaroff It’s true in general provided that $Y$ is Hausdorff, meaning that distinct points have disjoint nbhds.

Peter Tamaroff

the notion of area precedes that of the integral.

Chuck Fernández

1:37 AM

then what is area?

Peter Tamaroff

@BrianM.Scott Oh, OK. Long way till Hausdorff spaces here.

@BrianM.Scott Do you have any publications I could find useful? (Apart from your amazing repertoire in MSE)

@ChuckFernández Well, we can start by thinking what length is.,

And don't tell me length is $\int_a^b \sqrt{1+f'^2}dx$, please!

Brian M. Scott

@PeterTamaroff But you know that in a metric space distinct points have disjoint nbhds; it’s possible to use no more than that fact about metric spaces in your proof $-$ no sequences at all $-$ and then the proof will carry over pretty much verbatim to the general setting.

Chuck Fernández

hey brian have you seen this? ratemyprofessors.com/ShowRatings.jsp?tid=370076

Peter Tamaroff

@BrianM.Scott OK. Let me think about how that would go.

Brian M. Scott

@PeterTamaroff Probably not, though there’s one that you might find interesting after you’ve had a bit more general topology. It’s A ‘More Topological’ Proof of the Tietze Extension Theorem, The American Mathematical Monthly, 85 (1978), 192-3.

Chuck Fernández

1:45 AM

wikipedia seems to agree with you

Peter Tamaroff

@ChuckFernández ????

Brian M. Scott

@ChuckFernández I’ve made it a point not to look at that site. I know what my students thought of me over the years, both good and bad. I expected quite a bit; some students liked this, or at least liked the fact that they learned a lot, and many (predictably!) disliked it.

Chuck Fernández

Area

Area is a quantity that expresses the extent of a two-dimensional surface or shape, or planar lamina, in the plane. Area can be understood as the amount of material with a given thickness that would be necessary to fashion a model of the shape, or the amount of paint necessary to cover the surface with a single coat. It is the two-dimensional analog of the length of a curve (a one-dimensional concept) or the volume of a solid (a three-dimensional concept). The area of a shape can be measured by comparing the shape to squares of a fixed size. A shape with an area of three square metres...

Peter Tamaroff

@BrianM.Scott What is "the curve" the guys talk about.

I'm puzzled by some comments, I think your expositions here are great.

Brian M. Scott

@PeterTamaroff As in grading on the curve, or he curves the grades?

Peter Tamaroff

1:48 AM

@BrianM.Scott Possibly both. Someone says "the curve is large, but don't rely on it"

Chuck Fernández

my favorite teachers are allways hated by most of the students

Brian M. Scott

@PeterTamaroff Most instructors establish set percentage cutoffs for the various letter grades (A, B, C, D, F) at the beginning of the course. If the cutoff for an A is $92$%, a student who averages at least $92$% on graded work will receive an A. This requires the instructor to design exams to match his grading scale. I never cared to do this. Like my father, I preferred to design exams to test what I wanted to test and then interpret the results.

Chuck Fernández

besides my GPA is 3.3

Brian M. Scott

I’d guess that over my entire teaching career something around $80$% was a typical cutoff for an A, and a $50$% could be a C. Students didn’t like this, partly because they weren’t accustomed to it, and partly because they couldn’t tell exactly how well they were doing.

Chuck Fernández

lol

why do people compain then?

you know in mexico you need 90 to get an a no matter what

Peter Tamaroff

1:52 AM

@BrianM.Scott I see.

Brian M. Scott

@ChuckFernández Because my exams were hard, and because they’d internalized the idea that anything below $60$% just had to be a failing grade.

Peter Tamaroff

@BrianM.Scott Yes, true, I don't trust numbers.

@ChuckFernández What is GPA?

Chuck Fernández

what americans use as average

for grades

Peter Tamaroff

@BrianM.Scott I'd probably be very demanding from my students.

Brian M. Scott

Many teachers will tell you that preset cutoffs are more objective, but they’re fooling themselves: they just transfer the subjectivity from the grading to the composing of exams.

Peter Tamaroff

1:53 AM

@ChuckFernández What does GPA mean?

@BrianM.Scott Right.

Brian M. Scott

@PeterTamaroff It stands for Grade Point Average.

Chuck Fernández

yup

Peter Tamaroff

@BrianM.Scott And what does it measure?

Chuck Fernández

do they grade from 0-10 in Argentina?

Peter Tamaroff

@ChuckFernández Yes.

Chuck Fernández

1:55 AM

did you go to the imo with argentina?

Peter Tamaroff

And in general with $.5$s

I've rarely seen $.25$'s too.

@ChuckFernández No. I don't participate in $IMO$s and I am not really interested either.

Chuck Fernández

inquiry.princetonreview.com/leadgentemplate/GPA_popup.asp

Brian M. Scott

@PeterTamaroff Academic performance. The most common system is the four-point system: each A is worth $4$, each B is worth $3$, each C is worth $2$, each D is worth $1$, and an F (failure) is worth nothing. Add up the values of your grades in the courses that you’ve taken and divide by the number of courses, and that’s your GPA.

Actually, I’ve deliberately oversimplified a bit.

Peter Tamaroff

@BrianM.Scott I see.

I think that numbers are not really good mirrors of acadmic performance.

Brian M. Scott

Different courses may have different weights: one course might be a three-credit course, while another was a five-credit course. In practice, then, one takes a weighted average of the grades.

Peter Tamaroff

1:57 AM

I mean, in some sense it does, but not in the general case.

Brian M. Scott

@PeterTamaroff Used well, they can be a decent first approximation. One just shouldn’t expect much more than that.

Peter Tamaroff

@BrianM.Scott And SAT is like a very important test there?

Chuck Fernández

one of your tests is take home?

Brian M. Scott

@PeterTamaroff Pretty much anyone who wants to get into college has to take either the SAT or the ACT, yes.

Chuck Fernández

did you ever see your questions in math.se?

Peter Tamaroff

2:01 AM

@ChuckFernández Who are you asking?

Chuck Fernández

brian

Brian M. Scott

@ChuckFernández I didn’t find MSE until after I retired. I did once or twice see one of my questions in one of the Usenet math help groups.

Dylan Moreland

@BrianM.Scott I don't have a lot of data, but it seems like kids these days are taking just the one that will make them look better. When I was going through this process (Six years ago, I guess? Yikes.) I got the impression that you took both no matter what.

Peter Tamaroff

@DylanMoreland What does SAT evaluate?

Chuck Fernández

math and english

Brian M. Scott

2:03 AM

@DylanMoreland When I was going through it (1964) anyone who was serious took the SAT.

Peter Tamaroff

@ChuckFernández But what if you want to be a Lawyer? Why should you care about math?

Chuck Fernández

it consists of three parts: multiple choice math multiple choice english and an essay

Dylan Moreland

@PeterTamaroff Standardized testing isn't supposed to make sense :)

Brian M. Scott

@ChuckFernández Analytical, mathematical, and English skills.

Chuck Fernández

pierre de fermat was a lawyer

Dylan Moreland

2:04 AM

Right, it's different now with the essay and such. Even then I remember the ACT being a lot easier.

Chuck Fernández

each part costs 800 points

is worth*

Brian M. Scott

@DylanMoreland In 1964 there was an optional writing sample, which I think most serious candidates did.

Chuck Fernández

i got 1476 in my sat without the essay

Peter Tamaroff

@ChuckFernández 1476/1600 ?

Chuck Fernández

yes

Brian M. Scott

2:07 AM

787/774 (verbal/math), for a total of 1561. 800 on the math and physics subject exams, 732 on the German.

Chuck Fernández

what is the following in the sequence of penguin,carpet and banana

option a)dog b)lawyer c)salami

Peter Tamaroff

@ChuckFernández I'd go with potato.

Chuck Fernández

some of my questions where like that

i have a question

Peter Tamaroff

@BrianM.Scott Onte $\Gamma_f$ being closed

I'm thinking about the implications of the continuity of $f$.

Every nbhd in $\Gamma_f$ is the product of two neighborhood in $X$ and $Y$, correct?

Brian M. Scott

No, not at all. Every basic nbhd in $\Gamma_f$ is the intersection of such a product with $\Gamma_f$.

Peter Tamaroff

2:14 AM

@BrianM.Scott What do you mean by basic?

Brian M. Scott

Members of the usual base for the topology of $X\times Y$.

Peter Tamaroff

@BrianM.Scott I'm just trying to see how to show that every limit point of $\Gamma_f$ is in it.

Chuck Fernández

whats nbhd

Peter Tamaroff

@ChuckFernández Vecindario =P

Chuck Fernández

oh

Brian M. Scott

2:15 AM

@PeterTamaroff It’s easier to show that if $x\notin\Gamma_f$, then $x$ has an open nbhd disjoint from $\Gamma_f$; from that it follows immediately that $\Gamma_f$ is closed.

@ChuckFernández I get tired of writing out neighborhood all the time; both nbhd and nhd are common abbreviations.

Peter Tamaroff

@BrianM.Scott Know that feeling.

@BrianM.Scott I see.

If every $x \notin \Gamma_f$ has a disjoint nbhd from $\Gamma_f$ then no limit point of $\Gamma_f$ is outside it, right?

Brian M. Scott

Exactly.

Peter Tamaroff

@BrianM.Scott Could it happen $\Gamma_f$ has no limit points?

Brian M. Scott

Yes, if $X$ is discrete.

Peter Tamaroff

@BrianM.Scott Then $\Gamma_f$ would be what, vacuously closed?

Brian M. Scott

2:20 AM

Just closed.

But that’s true of any closed set.

In my view the notion of closed set is more fundamental than the notion of limit point.

Peter Tamaroff

@BrianM.Scott I see. How do you define closed sets?

Brian M. Scott

I would define closed sets to be complements of open sets.

Peter Tamaroff

@BrianM.Scott OK. Mendelson does that too.

Then he proves $F$ is closed iff it contains all its limit points, and iff for every convergent sequence of points of $F$, $\lim a_n \in F$.

@BrianM.Scott Can I choose the open ball about $x$ with radius $\delta < d(x,\Gamma_f)$ or am I being just silly there?

Brian M. Scott

@PeterTamaroff That last bit with the sequences doesn’t generalize to topological spaces in general, but the rest does.

Peter Tamaroff

@BrianM.Scott Good to know.

Brian M. Scott

2:26 AM

@PeterTamaroff Until you prove that $\Gamma_f$ is closed, you don’t know that $d(x,\Gamma_f)>0$.

Suppose that $p=\langle x,y\rangle\in(X\times Y)\setminus\Gamma_f$. Let $U$ and $V$ be disjoint open nbhds of $y$ and $f(x)$ in $Y$.

Since $f$ is continuous, there is an open nbhd $W$ of $x$ in $X$ such that $f[W]\subseteq V$. Now $W\times U$ is an open nbhd of $p$, and ...

Peter Tamaroff

@BrianM.Scott I usually use that if $f$ is continuous then for every nbhd $N$ of $f(a)$ there is a nbgd $M$ of $a$ such that $f(M)\subset N$.

You're using that right?

Yes yes.

Stupid question,

Maybe I'm just repeating it to fixate it.

@BrianM.Scott But $(W\times U)\cap \Gamma_f =\varnothing$?

Brian M. Scott

Exactly.

Peter Tamaroff

@BrianM.Scott I reasoned it out by some graphing =P

Brian M. Scott

Nothing wrong with using sketches to aid one’s thinking.

Peter Tamaroff

@BrianM.Scott Hehehe, true.

Former_Math_Addict

2:35 AM

@BrianM.Scott What do you think causes the "inevitability" of some students not to be able to grasp somethings?

Peter Tamaroff

@BrianM.Scott Why do you say open nbhd?

Is the hypothesis the nbhd is open vital?

Brian M. Scott

@PeterTamaroff No, but I find it simpler to work with open sets when possible rather than with any old ugly nbhd.

@Former_Math_Addict No instructor can be the right instructor for every student. And not every student wants to learn.

Former_Math_Addict

@BrianM.Scott Fair enough :-)

Thank you.

Brian M. Scott

You’re welcome.

Peter Tamaroff

@BrianM.Scott Then I should be saying "Thus, no point not in $\Gamma_f$ is a limit point of $\Gamma_f$ so that $\Gamma_f$ must contain all it's limit points, from where $\Gamma_f$ is closed.

Brian M. Scott

2:42 AM

Yes, that’ll do it.

Peter Tamaroff

I'm just wondering: What if $\Gamma_f$ has no limit points? (You mentioned this can be true if $X$ is discrete)

Brian M. Scott

@PeterTamaroff What of it? It would fail to be closed iff it had a limit point that wasn’t in it. If it has no limit points at all, then it certainly has none that aren’t in it!

Peter Tamaroff

@BrianM.Scott OK. I'm asking because I have that a set is closed $\iff$ it contains all its limit points.

Brian M. Scott

@PeterTamaroff Which it does vacuously if it has none.

Peter Tamaroff

@BrianM.Scott Right

28 mins ago, by Peter Tamaroff

@BrianM.Scott Then $\Gamma_f$ would be what, vacuously closed?

Maybe the wording was not good.

=P

Brian M. Scott

2:49 AM

One doesn’t speak of a set being vacuously closed. It’s only your definition of closed set that tempts you to do so.

Former_Math_Addict

Pardon my interruption... maybe the wording of "if and only if" is not good :-D

Peter Tamaroff

@BrianM.Scott Yes, that was my mistake.

@Former_Math_Addict ?¿

Former_Math_Addict

@PeterTamaroff That was meant as a joke :-D

Peter Tamaroff

@BrianM.Scott Do you think Rudin's Princples is a good text?

Brian M. Scott

@PeterTamaroff For some people. It’s exceedingly compactly organized, so that sometimes you don’t see the point of a result until many pages later, but anyone who knows everything in baby Rudin has a superb grounding.

Peter Tamaroff

2:54 AM

@BrianM.Scott I see. I'm reading it in paralell to Mendelson's but it doesn't seem to be very comprehensive, for example.

(Speciallly when defining like 11 topological concepts at once :P)

Brian M. Scott

I don’t know the Mendelson book. The Rudin is very specifically aimed at analysis.

Peter Tamaroff

@BrianM.Scott See that

Brian M. Scott

Bleagh. That’s not really very good pædagogy.

Peter Tamaroff

@BrianM.Scott I rest my case, your honor.

@BrianM.Scott For example, in the defnition of open set, I have it defined it has a set that is a nbhd of all of its points.

And I have defined nbhd of a point $a$ as a set containing an open ball about $a$.

Brian M. Scott

@PeterTamaroff Sorry: I had to step away for a moment and didn’t finish. It is, however, a good example of Rudin’s concision: everything that you need is there, but you have to think about it in order to use it effectively.

(By the way, that’s 2.20 in my edition of the book!)

Peter Tamaroff

3:01 AM

@BrianM.Scott I'm trying to buy it but I can't find it here in Argentina. I might have to get it abroad.

@BrianM.Scott Good to know.

Brian M. Scott

@PeterTamaroff That works for metric spaces. In a more general setting one starts with open sets, and an open nbhd of $x$ is simply an open set containing $x$. Then a set $N$ is a nbhd of $x$ (not necessarily open) if there is an open set $U$ such that $x\in\U\subseteq N$.

Peter Tamaroff

@BrianM.Scott What is puzzling is that right away he talks about connectedness and compactness, while that are the last chapters of Mendelson's book.

Brian M. Scott

@PeterTamaroff They’re pretty fundamental to analysis.

Think of the intermediate value and extreme value theorems.

Peter Tamaroff

@BrianM.Scott Right.

Well, maybe I can try and study from Mendelson first and then move swiftier through Rudin.

I randomly open a section of Chapter 4 - Connectedness. Section 7 is Homotopic paths and the Fundamental GRoup.

@BrianM.Scott Is that serious business?

Kannappan Sampath

@robjohn I am doing good. Pleased to hear from you. How are you doing? Will drop in more often.

Brian M. Scott

3:06 AM

@PeterTamaroff Well, it’s pretty important material.

Peter Tamaroff

@BrianM.Scott I'm talking about "Homotopic paths and the Fundamental GRoup." not IVT, XD

Brian M. Scott

@PeterTamaroff So was I.

Peter Tamaroff

@BrianM.Scott Oh, sorry.

@BrianM.Scott My real question is: how far into the rabbit hole is that?

Brian M. Scott

@PeterTamaroff In what sense? It really depends on the direction in which you want to go.

Peter Tamaroff

@BrianM.Scott I'm quite interested in Topology.

But so far I haven't even started first year college.

Brian M. Scott

3:11 AM

@PeterTamaroff Then you’ll definitely want to learn that material at some point. It’s useful in some parts of general topology and absolutely essential for algebraic topology. But you’ll want to get a good grounding in basic general topology first.

Peter Tamaroff

@BrianM.Scott OK. Guess I'll sleep now. Thanks for the talk.

Brian M. Scott

Sleep well!

Peter Tamaroff

@BrianM.Scott Oh, one last thing!

Brian M. Scott

Yes?

Peter Tamaroff

I proved this yesterday if I'm not wrong:

Let $(X_i,d_i)$ be metric spaces for $i=1,\dots,n$ and convert $X=\prod_{i=1}^n X_i$ into a metric space $ (X,d)$ by setting $d(x,y)=\max\limits_{1\leq i\leq n}\{d_i(x_i,y_i)\}$. Let $O_i\subset X_i$ be open for each $i$. Prove that $O=\prod_{i=1}^n O_i$ is an open subset of $X$ and that each open subset of $X$ is a union of sets of this form.

Basically by the following:

Brian M. Scott

3:14 AM

@PeterTamaroff In fact that’s the standard base for the product topology, and what you’re proving is that the metric that you’ve defined generates the product topology.

Kannappan Sampath

Yay! Installing Inkscape.

Peter Tamaroff

yesterday, by Peter Tamaroff

43 mins ago, by Peter Tamaroff

OK. This is my train of thought so far:
$(a)$ Every open ball in $X$ is the product of open balls of the $X_i$.
$(b)$ Every open subset $O_i$ is the union of open balls in $X_i$.
$(c)$ $O$ is the product of the union of open balls in each $X_i$, so $O$ is the union of the product of open balls in each $X_i$, which are open balls in $X$.
$(d)$ Thus, $O$ is the union of open balls of $X$, so it is an open subset of $X$.
$(e)$ Freaking put all of the above in a proof.

@BrianM.Scott Maybe that'll clear out when I study Topological Spaces.

Brian M. Scott

You might find it useful to ask yourself exactly what $B_d(x,\epsilon)$ looks like for $x\in X$: $y\in B_d(x,\epsilon)$ iff $d(x,y)<\epsilon$ iff $\max_i d_i(x_i,y_i)<\epsilon$ iff $d_i(x_i,y_i)<\epsilon$ for all $i$ iff $y\in\prod_i B_{d_i}(x_i,\epsilon)$. Thus, $B_d(x,\epsilon)=\prod_i B_{d_i}(x_i,\epsilon)$.

Peter Tamaroff

@BrianM.Scott That's how I proved $(a)$.

$(a)$ was an exercise in the section about open balls and nbhds.

Brian M. Scott

Once you have $(a)$, everything else follows easily. For instance, if $x\in O$, then $x_i\in O_i$ for each $i$, so for each $i$ there is an $\epsilon_i>0$ such that $B_{d_i}(x_i,\epsilon_i)\subseteq O_i$. Now let $\epsilon=\min_i \epsilon_i$. Then $B_d(x,\epsilon)\subseteq O$. Since $x$ was an arbitrary point of $O$, $O$ is open.

Peter Tamaroff

3:26 AM

@BrianM.Scott The succintness overwhelms me, Prof.

Thanks, and now, yes, night.

Brian M. Scott

G’night!

Peter Tamaroff

@BrianM.Scott Night. I may owe you someupvotes fromall this =P

robjohn

3:44 AM

@KannappanSampath Good to hear. We were out to dinner.

Kannappan Sampath

@robjohn to the nearby chinese restaurant?

robjohn

@KannappanSampath Japanese, but yes

Kannappan Sampath

@robjohn Hmm, I should not rely on my memory. :(

But, so, how did you like the food?

robjohn

@KannappanSampath very well, we always like the food there.

Kannappan Sampath

:)

@robjohn Interested in a question about Euler's totient?

robjohn

3:50 AM

@KannappanSampath sure

Former_Math_Addict

I'm still having trouble understanding this.

Kannappan Sampath

This is from John Coates' Example Sheet: If $n$ is not a perfect square and satisfies, \begin{align} n-n^{2/3} < \varphi(n) < n-1 \tag{1} \end{align} show that $n$ is a product of two primes.

@robjohn I don't know where to start. For one, we know $n$ is not a prime. But, that's just an observation, a priori irrelevant.

Oh: Since $n$ is not a prime, it has atleast two prime factors. Since $n$ is not a perfect square, enough to conclude that $n$ has atmost two factors. Is this fine @rob?

robjohn

@KannappanSampath If $n=\prod_k p_k^{e_k}$, then $\varphi(n)=\prod_k (p-1)p_k^{e_k-1}=n\prod_k\frac{p_k-1}{p_k}$

Kannappan Sampath

@robjohn Isn't that $p$ also $p_k$?

Yes. Agreed. The first expression comes from the multiplicativeness of $\varphi$. (of course, it requires us to know what $\varphi (p^k)$ is for a prime $p$.)

The next is fiddling the first.

Transcript for

$Mathematics$ Mathematics