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12:18 AM
@Jonas: Oh, well. I think my answer is now as complete as I am going to make it, but the inner simplicity is lost in all the added detail.
@JM greetings fern-man
 
Hey rob! I'm actually due for a switch back to tori today... :)
 
@JM We will miss the fern.
@JM Okay, I've snagged the fern. :-)
 
Now that I've thought about it, rob... "WHIM" is a pretty nice acronym... :D
 
I thought so :-)
I thought of it during a fire drill at Apple.
 
You were thinking of math during a fire drill? Oy... :D
 
12:25 AM
Well, the acronym (or domain name). But what better to think about during down time than math?
 
Good point...
 
I think that Didier's objections, while they do point out details that were left out, were mostly misinterpreting my argument. I wasn't making the errors that he was indicating; I was leaving out details.
 
From the history of your answer... somehow I think there's a point where you really cannot and should not pack everything into an answer. It should probably already be a note/paper/something...
 
I have to go out tonight, and then do a lot of backed up work. I may be inactive for a while. Not that I have been very active recently. The end of the holidays is depressing.
@JM Yes, I agree. I was trying to point out the method of derivation of the formula, but as Didier points out, a proof was requested.
 
@robjohn "The end of the holidays is depressing." - Yep. Oh well...
 
12:33 AM
The method, which is an outline of a proof, is pretty easy to follow. When all the details are added to make a proof, it is a bit unwieldy and hard to follow.
 
FWIW: I can still read the current version of your answer, but it now takes some stamina on my part... and then the reason it doesn't take much stamina is that I've seen the first version.
 
I don't know if my rearrangement of the answer helps, but I hope it does.
I didn't rearrange too much, but some to make it flow better with the added information.
Hopefully those that upvoted previously, won't come back and look aghast at what I've done and rescind their votes :-)
 
Well, it's too bad I can't give another one...
 
If whoever (Didier?) downvoted rescinds their downvote, it will be 10 votes :-)
only 2 points, but it will make it a nice answer.
 
Yeah.
 
12:48 AM
Byron's answer is slick and answers the question asked. However, I don't think that it offers much motivation. Before my mods, I think my answer showed how one uses the ideas of stationary phase (when brought to the real line).
 
He does say that much in the preamble...
 
@JM yes, I know, and I know that the simplicity is what has gotten so many votes.
@Srivatsan: good evening!
 
@robjohn The difference is not that big, I suppose.
Hi robjohn
 
We have one more holiday get-together tonight at my uncle's house. I need to get a few things wrapped before we leave.
 
@robjohn Is there really a gap as Didier says?
 
12:52 AM
We are supposed to be there in 2-1/2 hours.
@Srivatsan a gap in my current answer? let me look.
 
@robjohn No, don't bother. I am just wondering why Didier is so bothered with your answer.
 
@Srivatsan I really don't know. It has puzzled me, too.
 
@robjohn =)
 
My current iteration is complete I think.
As I was telling JM, I think Didier misinterpreted some of my answer. I was not making the mistakes he was indicating, I was leaving out details.
I have filled in the details and now the answer is not as easy to follow.
 
@robjohn Hm, ok.
 
12:56 AM
@Srivatsan However, JM says that he can still follow it (at least because he has read the previous versions).
 
I only zipped through the previous version. I haven't checked the edits =)
 
In short: the only problem now with rob's answer is that it now requires some stamina to read properly...
 
@Srivatsan I hope that Didier does not turn out to be like Bill Dubuque was to me on sci.math. I got no breaks and he seemed to push hard on my answers.
 
Ignore him? Is that not an option?
 
In a negative way, not a positive, nurturing way.
 
1:00 AM
@robjohn I didn't like the tone either.
 
@Srivatsan If it moves in that direction, perhaps. I'd rather not, though, since he does have good things to say.
 
But the thing on sci.math, do you have any sample links?
@robjohn Yes, that's right too.
 
I will have to search, but it was usually of the form "your way hides the inner truth. See my post at http://..."
Bill never said "inner truth", but you get the idea.
 
@robjohn Drop it, don't have to inconvenience yourself now.
"The essence of the matter" =)
 
I also don't want to start up bad relations here. Bill seems to be accepting my methods better here. He even used a proof similar to one that I had used on sci.math and he had ragged on me about.
I was amused.
Combinatorics and number theory were the places we clashed mostly.
 
1:06 AM
I see. Makes sense.
 
I don't know how you were able to stomach sci.math.
 
@robjohn Bill doesn't answer combinatorics questions here. Interesting.
@DylanMoreland What do you mean exactly?
 
I liked to prove things using Bezout, because it ruled out the circularity that sometimes creeps in when using unique factorization (Fundamental Theorem of Arithmentic)
Bill hated that.
 
What!?! How could Bill hate Bézout?
 
He hated the way I used it. However, he used Bezout in much the same way that I used to in an answer here.
 
1:08 AM
We have some characters on the site but I'm pretty sure that "Archimedes Plutonium" would be banned here.
 
I have no clue what Dylan is talking about. Do you get the reference, rob? =)
@DylanMoreland Were/Are you at sci.math? Is it still functional?
 
Gah, he is still at it. Look at the first page, even.
@Srivatsan Just look at it.
The problems with it aren't subtle!
 
@DylanMoreland Yes, found Arch. Plu.
 
And the other questions on that first page aren't much better.
 
I don't understand. Why is there no effort to TeXify sci.math? Why put up with text? Or is there some retro charm in text that I am missing?
 
1:12 AM
@Srivatsan Oh, yes. AP is a thorn in sci.math's backside.
 
They're mostly pseudo-physics crank manifestos and spam.
 
@robjohn Hm, ok.
 
@Srivatsan sci.math is a usenet group, and they are pretty much limited to text
 
@DylanMoreland So I take it that you are into sci.math?
 
Not really.
 
1:14 AM
@Srivatsan some groups allow data uploads, but they are not carried on many servers.
 
I mean, when I first became interested in math I looked around for relevant communities and hence lurked on sci.math for a while. And I became convinced that it was a mess.
 
@robjohn OK. I'm surprised things haven't changed yet
 
Hello, I have a question about tangents if anyone is willing to help
 
I have a feeling this guy = this one = this other one = this yet another one. What do you think, @robjohn and @Dylan?
 
@Srivatsan He almost says as much.
 
1:23 AM
@DylanMoreland Where does he say anything?
 
@Srivatsan Some of the titles are related. The last user says in his one question: "As I have in my previous questions (sum of square numbers (difference way) - algebra precalculus) "
Etc etc. Incredible detective work by me.
 
Is this forum for math help or am I in the wrong place?
 
@DylanMoreland And he asks a lot about 0.000...1 stuff.
 
(errr 0.000...1 is nonsensical)
 
@user1123950 Sorry, you got ignored in our chat. How can we help you?
 
1:25 AM
@user1123950 You're free to ask here, of course. The main site is just about as fast, as you can see.
 
I asked in the main site but did not want to bother people by correcting my initial question
I have a function z = crazy math involving x and y
 
You should definitely go and correct the question in case there is a mistake or something.
 
and arbitrary point outside the function, r1 and r2
 
I was actually looking at your profile profile and then reading this. Which seems to have been dealt with.
 
trying to find a tangential approximation
The answer given, I realized, doesn't elucidate the problem for me since my function is technically of two variables
 
1:29 AM
Hm. What do you want to do then? Look at planes through a point and find out when they touch a surface?
 
Imagine that I am at a fixed altitude z and I am staring at a mountain (the function) and I swing a huge straight bar from where I am until it crashes against the mountain
 
You're saying pseudo-tangent and worrying that you can't calculate a derivative of this thing (or that it might not even be differentiable?). I'm not sure that I, personally, can say anything interesting in that case.
 
I want to find the impact point against the mountain
I just discovered that I can find the derivative, btw
I say pseudotangent if only because I don't know the proper jargon for such an intersection, as it may not be a true tangent but rather a point of first intersection or something
 
I see. So if you had 1 - x^2 - y^2 and were floating somewhere above the slopes of this, you'd want to get some parabola.
 
the parabola was an example; it's not really a parabola
The real thing is more like a circle if we're looking at a cross-section here for a particular z
 
1:41 AM
Well, I'm trying to think about another example. That's all. I'm not claiming that all you care about is parabolas.
 
No eating, no breathing, no sleeping; parabolas reign supreme above all else
 
This would be a little messy, but maybe you could still use what JM does in his answer. You look at the various planes passing through your point P (containing the vector that goes straight up from P?) and do the 2D analysis there.
I'm obviously no help with this. The main site might be, although you would want to be able to articulate this well.
A drawing helps loads.
 
"A drawing helps loads." - indeed.
 
Yeah; I'll have to do that next time
 
@Srivatsan I am surprised there are no Polo's
 
1:48 AM
@user1123950 But, please post the question about your actual function as a new question.
 
2:18 AM
Don't mean to create a new thread, but how do you study if there are no answers at the back and no solution manual. How do you know if you are on the right track? :)
 
2:35 AM
Well, I think there's an important distinction between "I don't know how to do this" and "I thought I solved this, but there's a gap or missing justification that I don't see."
Which one are you worried about?
 
I also think that if you're doing exercises out of a book and you can't do the majority of them, then probably something has gone wrong. If there are just a few that seem very hard, then you think about it off and on and then maybe you ask someone. This site is good for that.
 
But you do have to get used to it. Most books don't have solutions and it seems like a minor miracle to find a book that has a solution manual and is any good.
There's Spivak, for example. I can't think of many others.
 
Which Spivak?
 
2:42 AM
I should have been more precise: Calculus.
 
I still can't do this tangent thing
I am shocked there isn't much literature on this
 
No no, that was probably implied. I didn't realize there was a solutions manual.
 
@CamMcLeman There is! Pricier than I remember.
 
JM said the line tangent to y=f(x) at x=h is y=f(h) + f'(h)*(x-h), but what about for a line tangent to z=f(x,y) at z=x1,y1?
 
@DylanMoreland Got it...
 
2:50 AM
@user1123950 There's a whole tangent plane. You could read something like this: tutorial.math.lamar.edu/Classes/CalcIII/TangentPlanes.aspx
 
i just need the tangent line for my level of z; no need for a plane
again it's like a man swinging a bar into a mountain and asking what the intersection point is
 
A tangent line will be in that plane. If you put in another constraint...
 
it basically says z = f(x0,y0) + fx(x0,y0)*(x-x0) + fy(x0,y0)*(y-y0) describes the tangent plane but i have no idea what this means
I am guessing my arbitrary point is x0, y0 and f() is my height function?
but then no idea what fx and fy mean
 
Those are partial derivatives with respect to x and y.
 
oh bugger this is tangent at a point on the function
I am after a tangent *through a point connecting to the function
is there a way i can just approximate it with iteration or something
 
3:28 AM
There's a whole family of tangent lines, and they comprise a plane. Unless you put in further constraints, we can't have a unique line.
@user1123950 Right, the man can be "anywhere" at the foot of the mountain, yes? That's exactly why your problem is underdetermined; you haven't said where the guy's swinging his bar from.
 
Most tangent tutorials ask for something tangent to a function at a point on the function
I know where he's swinging from
I know his x, y, and z
And I am swinging a bar into the mountain at the same z
and z is a function of x and y for the mountain
 
So, you can now obtain a parametric equation for the beam...
 
How?
 
You know how to set up the equation for the tangent plane, yes?
Unfortunately I have to leave now; hopefully someone else can step in to help you. Good day.
 
3:47 AM
fffffuuu---
 
@user1123950 Do you mean that the bar can't move up and down?
Like, I hold it out parallel to the xy "ground" and then just swing it left and right until I hit something?
Otherwise I'm not sure what you mean by "same z".
If this is the case, then you're back to the sort of 2D problem that JM explained.
 
Right, the bar is held out parallel
I know where I am in space, and I know where the mountain is, and I can describe the mountain in terms of z = some function of x and y
 
There will be some confusion with implicit differentiation, and you'll have to be more careful because you're not working with an explicit example.
 
I swing the bar from my spot and record where the bar connects to the mountain
 
Oh. Well then there's nothing 3D about this really.
Let me fix some notation. You have a function giving z-values from (x, y) pairs. f(x, y).
 
3:56 AM
it's 2D technically because I am only operating within one z-slice
 
You have a starting point (x_0, y_0, z_0).
 
but i don't know how it's all set up etc
yes
 
Technicalities can be very helpful. So now you're working in the plane z = z_0.
 
right
 
The intersection of this with your surface is some curve described by z_0 = f(x, y).
Say (x, y, z_0) is on that curve. You can write down the line between that point and (x_0, y_0, z_0). You can also write down the equation for the tangent line to the curve at the point (x, y, z_0).
You more or less want the slopes of those two lines to match up, right?
 
4:00 AM
two lines? shouldn't it be one line?
er I guess it's two lines, as there are potentially two tangent connections depending on which direction I swing
 
Maybe I can draw this and scan it. But I have to take the dog out for a bit.
 
but yes the slopes match
 
I'm not so much worried about direction as
two other points.
(1) I can draw two lines at (x, y, z_0). One is the tangent line. The other is the line going through (x, y, z_0) and (x_0, y_0, z_0). I want these to coincide.
 
sure
 
(2) I'm worried that we could pick up "interior points" this way. Like if my slice at z_0 looked like a washer
I would get four points from this process, it would seem. And it seems like you would only want the two outer ones?
It seems like you would be able to discard these. It wouldn't be pretty, but it's very doable.
 
4:04 AM
the slice does not look like a washer -- an odd disc, more like
i.e. like a mountain crosssection
no interior caves or anything like that
 
Even then I think I could think of something funny.
I think.
I just wanted an example that I didn't have to draw.
 
So I mean we've got a "true" tangent line on the mountain that may or may not intersect me, and a direct-connect line from me to that point on the mountain, yes?
 
Regarding my (2), what if you had a cross section that looked like the boundary of this thing: clker.com/cliparts/Z/M/3/7/J/7/purple-splotch-md.png
 
oh, no
nothing like that
it's basically spherical
 
If your fixed point were to the bottom right of this and you came in from the left, you'd probably want to hit that bottom left tendril
 
4:09 AM
not a circle, but just a curvey thing
 
But you'd probably pick up the valleys and some other tendrils as well.
I'm not sure.
 
in this case, the algorithm for hitting this circle should be the same algo that hits the lower left tendril
it's literally just swinging out and hitting the first thing it connects
 
That's what I mean. So there's more work involved there.
You could probably compute the angles and take the "leftmost" and "rightmost" ones.
At any rate, it would be good to work this naive approach out. I might be able to talk more later, but chat is so lively that maybe you'll have hammered this out by the time I get back.
Dogs are great but the upkeep is a little nuts sometimes.
 
cats = lower maintainance
albeit not as cool
 
4:23 AM
Hello
anyone notice that the question on diophantine equations has been deleted?
 
@BenjaminLim which one?
One by V.?
 
@tb yeah
what happened?
too many downvotes?
man I got to say some of the comments in there were funny
 
What happened?
 
@BenjaminLim I can't find a diophantine question that was deleted in the mod-tools.
 
sorry, not diophantine equations
what am I talking about
goldbach's conjecture
 
4:27 AM
what kind of comments were there
 
@BenjaminLim It was deleted by some high-rep users and a moderator a few hours after the meta thread was opened.
 
do you know why it was deleted?
i'll be back, I need to help my grandmother with the garden :D
 
@BenjaminLim No I don't. I guess it wasn't really a question and it was attracting too much attention while not being constructive at all. You don't have to give a reason for voting to delete a question.
Have fun. I always forget that it's summer over there (I wish it were summer here, too :))
 
Is that the most downvoted question here?
 
You can sort questions according to votes and go to the last page to get a more complete picture.
 
4:50 AM
Ah. I'm amazed they managed to get so many downvotes. I thought questions with < -3 become invisible on most pages.
 
Well, usually what goes below -3 is a community reaction on the attitude more than anything else.
On a more interesting note: how's your Sphere-thing doing?
 
Eh, I decided I wasn't likely to get the sort of answer I wanted, so I'll just be happy with what the machinery gives...
 
One way that might be a worthwhile answer would be to look at harmonic forms. Unfortunately I forgot about the details and I currently don't remember where I saw them (it might have been in a Séminaire Cartan). Basically you can compute the cohomology of any symmetric space using $G$-invariant forms without taking cohomology at all.
The point is that harmonic forms are automatically closed.
 
Ha...
I have no intuition for that at all.
 
Can anyone help me with this tangent problem?
 
5:06 AM
@ZhenLin Maybe this helps? Doesn't look too bad, but I haven't read it.
@user1123950: what tangent problem?
 
The best way I can describe it: A man sits at arbitrary point (x,y,z) and there is a mountain nearby, describable with z = f(x,y). The man takes out a bar, holds it parallel, and swings it counterclockwise until it connects with the mountain at the same level z. I'm trying to find the point of connection.
 
@tb There's some discussion if you scroll up just a little.
 
Eep, codifferentials.
 
@ZhenLin what's wrong with them?
 
I haven't learned about them either.
 
5:14 AM
:/
 
@user1123950 Let's call the point $(x_0, y_0, z_0)$. Now you need to determine the curve described by $f(x,y) = z_0$. If you choose a coordinate system in the horizontal plane your problem reduces to finding the tangent through the point to a function, which is pretty easy. The difficult thing is finding a good description of the curve implicitly described by $f(x,y) = z_0$.
 
Yes, I just don't know how to set this up properly
I don't know how to set up a tangent function to z=f(x,y), for starters
*tangent line
 
Do you have a concrete function $f$?
 
yes, and i have its derivative too
 
5:32 AM
That's not yet good enough: you need to parameterize the curve given by the intersection of your mountain with the plane at height z_0
 
5:44 AM
That's a very good point. Surely there's literature on explicitly getting parametrizations for implicitly defined curves.
 
5:55 AM
@user1123950 This doesn't circumvent the difficulty that t.b. brings up, but see this formula.
You can find dy/dx along g(x, y) = 0 (without solving for y in terms of x!) using implicit differentiation, which is more or less the chain rule.
 
@DylanMoreland Well, yes, but the difficulty is to find the point on the curve through which the tangent line is supposed to pass. I don't know how to do that without actually solving the implicit equation.
 
@tb I'm confused. Are we really disagreeing at all?
 
I overlooked the comment linking to the formula with implicit differentiation.
So I guess we aren't :)
 
Aha. I was responding to something he said about setting up the tangent line.
If you have a point in hand, I think what I said above can tell you whether it's good or not. I just saw Steven's answer and that seems to be a nicer way of doing it, especially if one wants to shove this into a computer in the end.
He addresses the implicit stuff in passing at the start. I wonder what he would have to say about it.
There are general results like the implicit function theorem. I use that from time to time. I don't really know how to turn it into a calculation, though.
 
What Steven says is basically what I had in mind.
 
6:06 AM
@user1123950 You could try leaving a comment on his answer, asking him to expand on the first parenthetical.
 
6:17 AM
His answer only takes into account the f(x) case though unfortunately
mine is z=f(x,y)
i think i am just giving up on this
doesn't seem realistically feasible
thanks for trying anyways though, guys
have a good night
 
Actually, now I don't know what to make of his comment at all. I thought he was discarding the case of a curve defined by f(x, y) = 0, but he instead talks about y = f(x). Which is pretty simple.
So I don't know if he has any ideas on that at all.
Ah well. Maybe you'll find something. The real difficulty here seems to be: if I have f(x, y) = z_0 (which may as well be f(x, y) = 0) then how can I solve for y or x or just get some way of writing (x, y) as a function of some single variable?
The implicit function theorem can guarantee you existence in many cases, but you want something effective.
Ah well. Good luck.
 
Dylan, I think it's hard to say anything more without knowing specifics about this crazy math function that z = f(x,y) is supposed to be...
 
6:43 AM
i'm back
 
How is the garden doing?
 
not bad, some of the flowers are dying from heat...
Man I got to say the comments in your "third prize" are incredible
 
Ben, I would really appreciate it if you didn't revive such threads: editing them bumps them to the front page. Let sleeping dogs lie.
 
Ok I forgot for a moment that when you edit something it bumps up to the front page
I laughed so hard at some of the comments :D
 
7:00 AM
this question is interesting in that its bit strange
math.stackexchange.com/questions/96563/error-when-using-mathematica-to-solve-differential-equation
 
7:31 AM
<rant> These constant minimalistic edit suggestions really go on my nerves. For heaven's sake: if there are three lines to edit it isn't to hard to do it in one go. If you can't do it, just leave it be </rant>
 
Hey theo,
what's wrong with the following reasoning?
 
@AsafKaragila and no induction step this time? very good.
 
@tb In the statement of the Baire Category Theorem for metric spaces
it says that the if we have a countable collection of non empty, dense open sets $\{G_a\}$ of a complete metric space $X$
 
Can I find a point on arc? where Arc is the part of circle
 
7:36 AM
Then $\cap_{i=1}^\infty G_i$ is not equal to the empty set
@coure2011 what are you referring to?
 
@BenjaminLim actually it's a dense $G_\delta$.
(that's much more)
 
Yes, for the moment I just need the intersection to be non-empty
but what I am thinking is this, I can show that in the collection $\{G_a\}$, the intersection of any two of sets in here is non-empty
 
@AsafKaragila The König should decree that this is required reading :/
 
can't I proceed to say then that the intersection of any three is non-empty, and so on?
 
Can I find a point Px,Py on arc. Known values are Start, End, angle and radius, where Px,Py is length L away from Start
 
7:40 AM
@coure2011 why not?
do you know the formulas $s = r \theta$
and $x = r\cos \theta$
 
can you please guide me
 
$y = r \sin \theta$
 
yes, but in the end you're taking a countable intersection. If you look at $G_n = (0,1/n)$ for example: any finite intersection is empty while the countable intersection is empty.
 
@tb I was thinking about writing a Chat Etiquette post on the meta. I was just lazy...
 
x code theta and y sin theta works only when the center is at origin
I dont know the center point and also it could be anywhere
 
7:42 AM
@AsafKaragila Rule 1: take at least one minute before jumping in and asking random nonsense. Rule 2: you're only allowed to ping two users within a span of 10 minutes, unless there's good reason to ping more. Rule 3: drawing attention to a question on the main site doesn't count as good reason.
 
@tb 4. ALL GLORY TO THE HYPNOTOAD.
 
@tb I see the point, and that's what I was thinking, maybe like this:
pick a point $ x \in G_1 \cap G_2$ say
 
Rule 4: Dont consider questions as home work assignment and for that reason just providing hints
 
the intersection of these two open sets is open
 
@BenjaminLim: any idea?
 
7:46 AM
@tb The hint given in Rudin is to find sequence of neighbourhoods $E_n$ such that $\bar{E_n} \subset G_n$
and then apply the famous thing on diameters of sets going to zero
@coure2011 well I don't really understand your question
 
@BenjaminLim yes that's it. What is the question?
 
Are you saying if you have a point $(a,b)$ on a circle
 
Basically the nested intervals principle.
 
@tb To prove the statement of the Baire Category Theorem that I typed above
@coure2011 you know the radius of the circle $r$
 
I want to find a point on arc, I know the start point, end point, angle, radius, the point I need to find is 's' distance away from start-point
 
7:48 AM
you know the angle theta when you write $(a,b) = (r \cos \theta, r \sin \theta)$
 
yes i know the r
yes
 
well
$(a,b)$ is your start point
then if you want to find the end point that is of arc length $L$ away from $(a,b)$
 
the arc's center is not at 0,0... r-cos-theta and r-sin-theta works only if the center is at 0,0
 
@coure2011 well you did not tell me that
anyway it is trivial
 
@BenjaminLim well, do you ask me to give the proof? Just start with an arbitrary $E_1$ as you specified and construct $E_n$ inductively.
 
7:50 AM
@tb no, I would not learn anything like that
@tb Well my $E_1$ is entirely contained within $G_1 \cap G_2$
 
ok so its not possible, right?
 
You didn't say how you chose $E_1$.
 
My idea is we know $G_1 \cap G_2$ is not empty
so say $x \in G_1 \cap G_2$
then since this intersection is open we can choose a neighbourhood $E_1$ that is entirely contained in the intersection
 
but this doesn't guarantee that its closure is contained in $G_1$.
 
@coure2011 the point is that it does not matter where the origin is
 
7:53 AM
This question had 7 edits during the first 7 minutes. I wonder it that was a record. math.stackexchange.com/posts/96583/revisions
 
u mean r-cos-theta will work... I dont think so
 
has this got something with what qiaochu said that the closure of the open ball of some radius in a metric space need not be the closed ball?
@coure2011 Easy: If your circle's center is at say $(x,y)$ then shift that point $(a,b)$ to $(a- x, b-y)$
now work with everything at the origin
when you get your answer at the end, add $(a,b)$ to it
 
Problem is that I dont know the Cx,Cy center points
I am not sure how to find it
 
Well then how do you even know your initial point?
 
its a real world problem for me to solve
 
7:55 AM
@tb I think I get it: I am tacitly assuming that the closure of the open ball is a closed ball
@coure2011 I don't understand your question, why don't you post it on math.stackexchange.com?
I don't really get the question, you are telling me the full details
type out everything here in full
 
@BenjaminLim No, it's not that (that's a remark one should be aware of, though). If $G_1 = (-\infty, 1)$ and $G_2 = (0,\infty)$ and $E_1 = (0,1)$ then...
 
math.stackexhange.com is only for experts not for simple persons like me
 
@coure2011 I am n00b and I'm on there
 
If I will post there I will get 5-10 answers each with complex equations without explaining what is what and who is who
 
(so my complaint is on a more elementary level)
 
7:58 AM
@tb just choose the radius of the ball to be smaller?
 
exactly. While $\overline{B_r(x)}$ can be a proper subset of the closed ball with radius $r$ around $x$ it is always contained in it.
 

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