Author |
Topic: Skinned Balls (Read 1273 times) |
|
william wu
wu::riddles Administrator
Gender:
Posts: 1291
|
|
Skinned Balls
« on: Oct 17th, 2003, 2:39pm » |
Quote Modify
|
In some metric space (M,d), consider the set B'(x,a) = { y[in]M : d(x,y) [le] a }, where x[in]M,a[in][bbr]. (A "ball with skin".) Is B'(x,a) an open or closed set? Or is it both open and closed? Or is it neither open nor closed? Background Information Note 1: Metric Spaces A metric space (M,d) is a set M and a function d : M x M [to][bbr] such that a. positvity: d(x,y)[ge]0 for all x,y [in]M b. nondegeneracy: d(x,y) = 0 iff x = y c. symmetry: d(x,y) = d(y,x) for every x,y[in]M d. triangle inequality: d(x,y) [le] d(x,z) + d(z,y) for all x,y,z[in]M An example of a metric space would be the set [bbr]n with the Euclidean distance function d(x,y) = sqrt{[sum]i=1 to n(xi - yi)2} where x = (x1,x2,...,xn) and y = (y1,y2,...,yn). Note 2: Open vs. Closed Sets In a metric space (M,d), a set U [subset] M is said to be open if for each x[in]U, there exists [epsilon]>0 such that B(x,[epsilon]) = { y[in]M | d(x,y) < [epsilon] } [subset] U. (One could think of B(x,[epsilon]) as an [epsilon]-radius ball "with no skin", centered about x.) A set V[in]M is closed if its complement (the set {M \ V}) is open. Edit: 11:54 AM 10/21/2003 Clarified the wording of the problem so it doesn't suggest that all sets are either exclusively open or closed (... although as far as the solution goes, this doesn't matter ...)
|
« Last Edit: Oct 21st, 2003, 11:55am by william wu » |
IP Logged |
[ wu ] : http://wuriddles.com / http://forums.wuriddles.com
|
|
|
Icarus
wu::riddles Moderator Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863
|
|
Re: Skinned Balls
« Reply #1 on: Oct 18th, 2003, 6:28pm » |
Quote Modify
|
So, you're learning topology. Great! It'll improve your understanding of analytical topics considerably. Besides which, I think it's a fun subject in its own right. This one is not particularly hard, but I think I know where you are heading. As a hint: These have a commonly used name, and it's not "skinned". The balls without skins also have a common name, not "with no skin".
|
|
IP Logged |
"Pi goes on and on and on ... And e is just as cursed. I wonder: Which is larger When their digits are reversed? " - Anonymous
|
|
|
Barukh
Uberpuzzler
Gender:
Posts: 2276
|
|
Re: Skinned Balls
« Reply #2 on: Oct 19th, 2003, 1:54am » |
Quote Modify
|
I want to learn topology too. Icarus, could you please recommend a good book on the subject? Preferably one which is not very tough reading. Thanks...
|
|
IP Logged |
|
|
|
Icarus
wu::riddles Moderator Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863
|
|
Re: Skinned Balls
« Reply #3 on: Oct 19th, 2003, 8:52pm » |
Quote Modify
|
Alas, I cannot help you there. I have been away from the mathematical community for some time. The only book on point set topology (the basis version you need to start with) that I am familiar with is the textbook I first learned it from 20 years ago. That book, "Topology, A First Course" by James R. Munkres, published by Prentice-Hall, may not even be in print anymore. Besides which, while not a bad textbook it never struck me as a particularly outstanding one either. You might ask William what he is studying. Pietro may know of a good one, since his mathematical education is far more recent than mine.
|
|
IP Logged |
"Pi goes on and on and on ... And e is just as cursed. I wonder: Which is larger When their digits are reversed? " - Anonymous
|
|
|
william wu
wu::riddles Administrator
Gender:
Posts: 1291
|
|
Re: Skinned Balls
« Reply #4 on: Oct 19th, 2003, 9:14pm » |
Quote Modify
|
I'm currently auditing honors real analysis, in which we cover some basic point set topology, but many other things too. So the textbook we're using is Elementary Classical Analysis by Marsden and Hoffman, which I'm not sure I can recommend. It's structure is awkward, in that theorems are not immediately proved after they are stated, but rather, all the proofs are tacked at the end of the chapter. I understand the motivation for this was to better convey the big picture on the first pass of the material. However, I still find myself flipping back and forth between the chapter material, and the back of chapters where the theorems are proved. Also, the book had a ridiculous amount of errata for a seventh printing (it even caused a protest of textbook choice at my university), but the recent ninth printing seems to have most of them fixed. It seems that neither my undergraduate nor my graduate institution offers a course dedicated solely to topology. It seems like something you learn a little bit about over many courses. I have an e-book titled "Algebraic Topology", by Allen Hatcher. It is 553 pages and 3.73 MB, deemed by the author as "a readable introduction to Algebraic Topology" (from the preface), and intended to be freely distributable. I've only read the first two pages, which are indeed very friendly and readable. There's also pretty pictures. http://www.math.cornell.edu/~hatcher/AT/ATpage.html As Icarus said, Pietro (math major) might know better; you could just e-mail or instant message him and tell him we referred you. Check out the "Members" link in the forum's top menu bar for lots of useful contact information.
|
« Last Edit: Oct 19th, 2003, 11:11pm by william wu » |
IP Logged |
[ wu ] : http://wuriddles.com / http://forums.wuriddles.com
|
|
|
Barukh
Uberpuzzler
Gender:
Posts: 2276
|
|
Re: Skinned Balls
« Reply #5 on: Oct 20th, 2003, 3:28am » |
Quote Modify
|
Icarus, William, thank you for this information. Icarus, just for the update: the second edition of Munkres's book was published in 2000.
|
|
IP Logged |
|
|
|
Icarus
wu::riddles Moderator Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863
|
|
Re: Skinned Balls
« Reply #6 on: Oct 20th, 2003, 10:06am » |
Quote Modify
|
Munkres is alright, but I suspect there are better choices. It sounds like Marsden did no better with his Hoffman than he did with Tromda (?). That being the book I took a similar course from. William's description sound familiar! One thing: AVOID Algebraic Topology until you have a strong grasp of point-set topology, and at least some understanding of abstract algebra. Algebraic Topology is a very powerful tool, but it comes in two forms: homotopy is easily understood but it is nearly impossible to prove things; homology is nearly impossible to understand, but proofs are easy!
|
|
IP Logged |
"Pi goes on and on and on ... And e is just as cursed. I wonder: Which is larger When their digits are reversed? " - Anonymous
|
|
|
Pietro K.C.
Full Member
Gender:
Posts: 213
|
|
Re: Skinned Balls
« Reply #7 on: Oct 21st, 2003, 10:21am » |
Quote Modify
|
Oh, you guys are too much! You remember me?! I've been away for so long because of university/work, and my name is the one that comes up?! That's so moving... I'm actually a 4th-year (out of 5) computer engineering major, but I've been taking the math courses as well, and I'm about halfway down that road. It's funny you should ask, because this very semester I'm also enrolled in a Real Analysis course, the first of three that math majors must take. The book we use in the course is really, really good - "Análise Real", by Elon Lages Lima - but it's only available in portuguese, unfortunately. I've also looked for other references, and one teacher of mine - who confirmed my theory that good books on topology are rare indeed - did name an author she said had published decent material. I've forgotten his name, but I have it here somewhere, so I'll be back later with it. About the riddle: I think it's interesting to point out that, unlike in ordinary usage, open and closed are not mutually exclusive in topology, nor are they logical complements - it is entirely possible for a set to be neither open NOR closed, or both open AND closed. The question posed by the riddle seems to imply that this is not the case, and may throw off beginners.
|
|
IP Logged |
"I always wondered about the meaning of life. So I looked it up in the dictionary under 'L' and there it was --- the meaning of life. It was not what I expected." (Dogbert)
|
|
|
Icarus
wu::riddles Moderator Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863
|
|
Re: Skinned Balls
« Reply #8 on: Oct 21st, 2003, 4:53pm » |
Quote Modify
|
Pietro, we could hardly forget you! I hoped you would be more familiar because, although it is not your major, you clearly are well-educated mathematically. I should have considered the language problem. Perhaps Barukh is so cosmopolitan as to be well-versed in both English and Portugese in addition to his own language, but it seems unlikely. Concerning open and closed, I was wondering if that was the point of William's question: to receive one answer, then spring an example where the opposite is true, to see if the respondent would mistake this as contradicting the original answer (a sneaky, but effective, teaching technique). The answer of course is: the ball with skin is closed, the ball without skin is open. This is why they are usually called "closed" and "open" balls, respectively. To prove it, use the triangle inequality. Every topological space has at least two sets that are both open and closed: the empty set [emptyset] and the entire space. A topological space is said to be connected if these two are the only sets that are both open and closed. Connectedness is a very useful topological property. Things that are easily seen to be true locally can usually be shown to hold for the entire space if it is connected. On the subject of good text books, my favorite while in college was Peter S. Bartle's The Elements of Real Analysis (Wiley). This book was a standard in US colleges at the time for a first course in Real Analysis, and in my opinion well deserved to be. I think I learned more from it than from any other textbook. Of course, people of my mindset are rare, so you might not enjoy it as much. For topology, it restricts itself to the topology on subsets of [bbr][supn], and this further gets lost among the many analytical ideas it is bringing in, so I do not suggest it for the study of that worthy subject.
|
|
IP Logged |
"Pi goes on and on and on ... And e is just as cursed. I wonder: Which is larger When their digits are reversed? " - Anonymous
|
|
|
wowbagger
Uberpuzzler
Gender:
Posts: 727
|
|
Re: Skinned Balls
« Reply #10 on: Oct 22nd, 2003, 2:30am » |
Quote Modify
|
on Oct 21st, 2003, 6:39pm, william wu wrote: Actually, I thought the answer is [hidden] |
| That was my first thought, too. All I'm sure of, however, is that if you can prove the existence of a border element z, i.e. z is in B'(x,a), but not in B(x,a), then B'(x,a) cannot be open. Quite trivial, I know, but I'm not sure about the triviality of the proof of existence of such an element. I haven't dealt with this stuff for a long time, so my tries to conjure up an example for open B' (in spite of Icarus's "of course"-answer) haven't been crowned with success. It's good to see you're still around, Pietro.
|
« Last Edit: Oct 22nd, 2003, 3:31am by wowbagger » |
IP Logged |
"You're a jerk, <your surname>!"
|
|
|
Icarus
wu::riddles Moderator Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863
|
|
Re: Skinned Balls
« Reply #11 on: Oct 22nd, 2003, 8:47pm » |
Quote Modify
|
Since there seems to be some doubt here... B'(x, a) is always closed. It may also be open, depending on the metric space, but it is always closed. And B(x, a) is always open. Proof: let B = B'(x, a). For any y [in] M - B, d(x, y) = b > a. If z is in B(y, b-a), then b = d(x, y) [le] d(x, z) + d(y, z) < d(x, z) + b - a. Solving gives d(x,z) > a, so z is not in B. Thus for each y in M - B, there is a ball around y that is contained in M - B. I.e. M - B is open, and so B is closed. To show that B(x, a) is open: if y [in] B(x,a), then d(x,y) = b < a. By the triangle inequality, you can show that B(y, a - b) [subset] B(x, a). Thus B(x, a) is open. QED As a simple example of a metric space where B is both open and closed, just let M be the union of two disjoint closed balls in [bbr]3. Choosing x to be the center of one ball, and a the radius, we see by the proof that B is closed. But its complement in M is just another closed ball, so B is also open. As I said above, the only way to have the set both open and closed without it being the entire space (or empty) is for the space to be disconnected. Thus we need 2 disjoint balls to pull this off (by the way, they do not need to be closed or open balls in [bbr]3 - they'll always be both in M.
|
|
IP Logged |
"Pi goes on and on and on ... And e is just as cursed. I wonder: Which is larger When their digits are reversed? " - Anonymous
|
|
|
|