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bonanova
Daedalian Member

 Posted: Mon Sep 22, 2008 10:50 pm    Post subject: 1 Your eccentric, rich Uncle Wilbur died and left you an uncountably infinite number of 0's, 1's, 2's, 3's, 4's, 5's, 6's, 7's, 8's, and 9's. You arrive at the warehouse with 10 crates - one for each number. But you're told the will stipulated the digits were not be stored three- dimensionally. Crazy old Wilbur wasn't going to make this easy. Instead, you are handed 10 infinite planes to load them onto. Again, one plane for each number. Now the will also stipulates that [1] there can be no contact [crossings or tangencies] among any of the individual digits, and [2] if, for any number, you can't get all the digits onto a plane, that number would be donated to charity. Which [if any] of the numbers will you be able to load up and take home?_________________ Vidi, vici, veni.
marcusI
Daedalian Member

 Posted: Tue Sep 23, 2008 12:55 am    Post subject: 2 My guess is: You can take them all. Heres how: Start at the center of each plane and tile them with the numbers using the fibonacci sequence as a pattern. This is the same as the seeds on a sunflower. Since this has the same cardinality as an infinite set of whole numbers you should be able to fit them all. I'm not sure I understand stipulation #1, but I think this might meet it.
bonanova
Daedalian Member

 Posted: Tue Sep 23, 2008 1:07 am    Post subject: 3 Clarification of [1]: a line segment that draws part of one digit may not intersect or be tangent to [may not touch] a line segment which draws part of another digit. Aren't the [whole numbers countable?]_________________ Vidi, vici, veni.
ralphmerridew
Daedalian Member

 Posted: Tue Sep 23, 2008 10:07 am    Post subject: 4 marcusI: That won't work. You'll only get countably many digits onto the plane that way. I think this depends on the exact font used. Also, "uncountably infinite" is too vague; there are many uncountable infinities. Does that mean "with the cardinality of the real numbers"? Perhaps using digits made this confusing. Suppose instead he'd left an uncountably infinite set of plus signs and an uncountably infinite set of minus signs.
bonanova
Daedalian Member

 Posted: Tue Sep 23, 2008 10:29 am    Post subject: 5 Yes. Cardinality of the reals._________________ Vidi, vici, veni.
ralphmerridew
Daedalian Member

 Posted: Tue Sep 23, 2008 12:02 pm    Post subject: 6 Exactly how is each digit written? Could you post an image of each? (If "1" is to be written as a single downstroke, or as a single downstroke with the small diagonal stroke, yes. If it is written with a serif at the bottom, no.)
bonanova
Daedalian Member

 Posted: Tue Sep 23, 2008 5:37 pm    Post subject: 7 Assume the simplest sans serif representation; the appearance of the OP will do. Or give a necessary / sufficient condition, whichever is easier to do. Sounds like you have it._________________ Vidi, vici, veni.
Zag
Tired of his old title

 Posted: Tue Sep 23, 2008 6:21 pm    Post subject: 8 I have trouble understanding why the font matters. Maybe I misunderstood something. Anyway, my thought was to make a fractal-based design with them. Either of two styles ought to work. 1. Arrange a bunch of them in some pattern, say, an isosceles triangle with three copies of the digit along the bottom and ten along each of the larger sides. Then take the three in the middle of each of the long sides, shrink them to 1/3 their size in each dimension, duplicate them twice (so now you have a row of 9 smaller ones), and put them back. The nine should just fill the space you removed the three from. Now, from the middle three of the shrunken ones as the base, duplicate the original triangle at 1/3 scale. Repeat infinitely. Also, realize that the three you originally used as a base were really just part of the side of an even bigger triangle, which is 3-to-1 scale of the triangle you just made, and so on. The other pattern is easier to describe in colors. You can replace the colors with groups of numbers oriented in certain directions. Choose any point on the plane and send out three rays that are 120 degrees apart, separating the plane into 3 infinite pie slices. Paint one red, another blue, and another yellow. In between the blue and the yellow, paint a red wedge which is 1/10 the amount of the smaller of the blue and yellow wedges, the space for it stolen from the larger of the two. Between the red and blue, paint a yellow wedge, and in between the red and yellow, paint a blue wedge. Repeat infinitely. Anywhere two colors meet, paint a wedge of the third color between them, according to the "one-tenth" formula.
ralphmerridew
Daedalian Member

 Posted: Wed Sep 24, 2008 2:05 am    Post subject: 9 Are you familiar with the difference between "countably infinite" and "uncountably infinite"?
Zag
Tired of his old title

 Posted: Wed Sep 24, 2008 3:18 am    Post subject: 10 I think so. I had thought that both of these fractal patterns would result in an uncountably infinite number of elements, but I could easily be confused on that score. Each new element you add causes an infinite number more elements to be added (and each of those, etc.) -- won't that make it uncountable? I certainly don't see any way to map them to the integers, or even the rationals, but I do see how to map them to the reals. Anyway, fractals are always fun to bring up, and the infinitely fringed edge (it has a real name, but I forget it) is crazy to contemplate.
Griffin
Daedalian Member

 Posted: Wed Sep 24, 2008 4:00 am    Post subject: 11 If the numbers have any "thickness" to them (i.e. they take up non-zero area), then you're stuck fitting only a countable number of numbers into the plane. This is because any two dimensional region contains a rational point (meaning a point with rational x coordinate, and rational y coordinate). Since there are only a countable number or rational points, there will be only a countable number of numbers. However, if the numbers are infinitely thin (i.e. made out of simple curves and line segments), then it might be possible. For zero 0, it is possible (imagine an infinite continuum of zeros shrinking down to a point). For the other numbers I'm not so sure ...
bonanova
Daedalian Member

 Posted: Wed Sep 24, 2008 8:30 am    Post subject: 12 Yes; assume zero thickness lines._________________ Vidi, vici, veni.
ralphmerridew
Daedalian Member

 Posted: Wed Sep 24, 2008 10:09 am    Post subject: 13 Do all digits have to be congruent, or only similar?
ralphmerridew
Daedalian Member

 Posted: Wed Sep 24, 2008 10:30 am    Post subject: 14 A tip about countability: The union of a countable number of countable sets is countable. As for Zag's first method: - On Step 1, there are finitely many digits. - On each step, he adds a finite number of digits. Therefore, after a countably infinite number of steps, there are countably many digits. An example of how to place uncountably many symbols on the plane: - Use minus signs. - Number them from 0 to 1 over the reals. - Place the minus sign labeled r at (0, r) - (1, r).
bonanova
Daedalian Member

Posted: Wed Sep 24, 2008 11:11 am    Post subject: 15

 ralphmerridew wrote: Do all digits have to be congruent, or only similar?

Only similar.
_________________
Vidi, vici, veni.
Thok
Oh, foe, the cursed teeth!

 Posted: Fri Sep 26, 2008 7:36 am    Post subject: 16 Here's an argument that T can't be tiled this way. It can be modified to a proof that 3,4,6,8,9 can't work also, by noticing that each such numeral has a subset that looks like a T and mimicing the below proof. A "T" can be defined by four points (the center, and the three endpoints). So, let's think of a "T" as an element of R 8 given by those 4 points. I claim that given any "T" in this form, there is a small open neighborhood of R 8 around this representation such that any other "T" in this neighborhood crosses the original "T" (this is intuitively obvious since if I move the center and the endpoints of a "T" very little, then the resulting figure crosses the original "T". I could make this rigorous if you want.) This means that any nontouching set of T's in R 2 gives rise to a set of disjoint open neighborhoods in R 8 , and a standard topology theorem says there any set of disjoint neighborhoods in R n is countable, which means that we could only have a countable number of nontouching T's. If the numbers are required to be similar, I think the above proof can be modified to works on 2 and 5 also. 0, 1, and 7 are trivial to place uncountably many of: just consider the "thick" form of the number and decompose as appropriate. (This doesn't work for 2 and 5 because the resulting decomposition gives nonsimilar shapes.)
lostdummy
Daedalian Member

 Posted: Fri Sep 26, 2008 9:06 am    Post subject: 17 I'm not sure that I follow this "countable infinity". If we have infinite plane with grid drawn on it, where each grid cell is big enough to put/draw one digit, and if we look at one diagonal of that grid - is there countable or uncountable number of cells in that infinite diagonal? If it is uncountable number, then we can presume that font with only horizontal/vertical lines is used for digits (like on calculators, and even examples like "T" satisfy that), and we can put each digit in next cell of one diagonal. And as I mentioned at start, I'm not sure that I follow how number of cells in infinite diagonal at infinite plane can be countable. Also, from previous post I understood that "union of countable sets is countable, even if it is union of infinite number of such sets". If it is so, than I dont see how you can have "uncountable infinity" with digits. If we look at each single digit as set of one digit, it is countable set. So even if we add infinite number of such sets, resulting infinite set will still be countable?
bonanova
Daedalian Member

 Posted: Fri Sep 26, 2008 9:31 am    Post subject: 18 The positive integers 1, 2, 3, .... are a countable, infinite set. Any set whose elements can be matched up with 1, 2, 3, ... is also countable, for example the set of unit squares in the plane. Here's how to see that: Consider the first quadrant of the plane x>=0, y>=0. Place a 1 in the lower left box [between 0 and 1 for x and y]. Place 2 and 3 in the boxes above and to the right of box 1. Place 3, 4, 5 in the next diagonal row of boxes, and so on. You cover the quadrant that way. Similarly cover the other three quadrants and you've covered the plane. So the unit squares are a countably infinite set._________________ Vidi, vici, veni.
lostdummy
Daedalian Member

 Posted: Fri Sep 26, 2008 11:10 am    Post subject: 19 In that case, how is it possible to have "uncountable infinite set of digits" ? Since we cant have fractional number of digits, we can clearly map each digit to some positive integer. For example, label each digit as 1st, 2nd, 3rd... that would suggest that ANY infinite set of whole items (like digits) is countable? Similary, is union of any two countable sets also countable? In that case, since single digit is countable, adding single digit to any countable digits set will still result in countable digit set (even after infinite adds). And since we must start any set with single digit, it could follow (same as above) that ANY set of digits is countable set? In other words, it does not seem to me that we can have "uncountable infinite number of 5's". And if we can have only countable infinite number of 5's, we can put them along [countable infinite number of] diagonal cells in infinite plane, and solve the problem ;p
ralphmerridew
Daedalian Member

 Posted: Fri Sep 26, 2008 11:20 am    Post subject: 20 The set of reals in the closed interval U = [0,1] is uncountable. Proof: Suppose it were. Then there exists a function f(N)->U with range U. Define a_i: : a_i = 7 if the ith digit in the decimal expansion of f(i) is between 0 and 4 inclusive. : a_i = 3 if the ith digit in the decimal expansion of f(i) is between 5 and 9 inclusive. Let K be the number in U such that the ith digit in the decimal expansion of K is a_i. K is not equal to f(i) for any i. Therefore, the range of f is not U. Contradiction; the reals are uncountable.
lostdummy
Daedalian Member

 Posted: Fri Sep 26, 2008 11:38 am    Post subject: 21 My previous post was not stating that any set can not be uncountable. I was wondering if set of whole items (like digits) can be uncountable? One easy attempt at explaining uncountable digit set would be to say "create set of digits from decimal representation of reals in interval [0,1]". And if we have uncountable number of reals, and each real has one or more digits, if should result in uncountable number of digits ... or not. Because I dont think that it is valid assumption - in order to "create set of digits from all reals in interval" we need to be able to count "all reals in interval", and since that is not possible (by definition of uncountable infinity), it is not possible to create such set of digits. And again, no matter how we got set of digits, it seems clear that we can map that set to set of natural numbers as I mentioned above (1st, 2nd...) , so it still seems to me that any set of digits is countable.
ralphmerridew
Daedalian Member

 Posted: Fri Sep 26, 2008 11:48 am    Post subject: 22 The size of a set has nothing to do with what kind of elements it holds. If I haven't made any mistakes: Consider the set of paths { 0 <= x <= 1 | (x-.3, x+.8) to (x, x+1) to (x, x) }
Zag
Tired of his old title

 Posted: Fri Sep 26, 2008 12:18 pm    Post subject: 23 I still don't buy that you can't make an uncountable number of elements using fractals. rm, in your example with minus signs, you basically said "Make a minus sign for every real number between 0 and 1 and arrange them so they don't overlap. Since reals are uncountable, this set will be, too." OK. I wish to make a 0 for every real number [0, 1), and arrange them so they don't overlap. Here is my arrangement: Start a single 0 which will represent then real number 0.0000... Around the inside of it, draw nine more smaller zeroes, and leave space for an undrawn tenth one. The drawn ones represent the real numbers 0.1000... to 0.9000... Inside each of these (including the undrawn one), duplicate the drawing. For every real number in the range [0,1), I could point to the 0 that represents its decimal representation. Similarly, for every 0 I've drawn, there is a unique real number. This isn't fundamentally different from saying that there exists a decimal representation of every real number in the range (upon which your proof of the uncountability of the real numbers in a range rests). How do you construct all those decimal representations? By starting with "0." and duplicating it ten times, adding a different digit from 0 to 9 at the end of each. Then duplicate each of those 10 times, etc. Why is this any different? It results in all the reals in that range, which is an uncountable number, doesn't it?
ralphmerridew
Daedalian Member

 Posted: Fri Sep 26, 2008 12:53 pm    Post subject: 24 You only draw zeroes corresponding to terminating decimal expansions. Which 0 corresponds to 1/3 or pi-3?
Zag
Tired of his old title

Posted: Fri Sep 26, 2008 7:13 pm    Post subject: 25

 ralphmerridew wrote: You only draw zeroes corresponding to terminating decimal expansions. Which 0 corresponds to 1/3 or pi-3?

Hmmm. But I'm not positing that they are actually drawn according to a process that takes time. They just are all there, in the same way that you can say that 1/3 has a decimal representation.

Or maybe not. Anyway, I thought it was worth a shot -- or at least to understand why they are different.
ralphmerridew
Daedalian Member

Posted: Fri Sep 26, 2008 9:25 pm    Post subject: 26

 Zag wrote: For every real number in the range [0,1), I could point to the 0 that represents its decimal representation. Similarly, for every 0 I've drawn, there is a unique real number.

Okay, point to the zero that corresponds to 1/3 or pi-3 (or point to any zero that corresponds to a number not of form i/10^k, where i and k are nonnegative integers).
Griffin
Daedalian Member

 Posted: Sat Sep 27, 2008 4:48 am    Post subject: 27 I agree with Thok's solutions for the most part (very clever proof!). However, 'm not sure how 3 has a "forbidden T" in it. 3 is possible, at least in my interpretation of it. Also, 2 might be possible depending on angles and such.
Trojan Horse
Daedalian Member

 Posted: Sat Sep 27, 2008 5:05 am    Post subject: 28 I guess both the 3 and the 2 depend "on angles and such". To my eyes, it looks like the 3 and the 2 (at least in the font used here) are curved more than in your picture; they are curved enough to stop your plans. But it's hard to tell.
Griffin
Daedalian Member

 Posted: Sat Sep 27, 2008 5:06 am    Post subject: 29 Also, can we pack an uncountable number of each capital letter into the plane? We can quickly eliminate A, B, E, F, H, K, P, Q, R, T, X, Y from Thok's forbidden T theorem. Of the remaining letters, only S seems impossible. It's interesting that Z is possible, but S and 5 isn't. I think it has something to do with how steep the angle is on the tips of the S-shape, relative to what is happening with the central portion.
Griffin
Daedalian Member

Posted: Sat Sep 27, 2008 5:10 am    Post subject: 30

 Trojan Horse wrote: I guess both the 3 and the 2 depend "on angles and such". To my eyes, it looks like the 3 and the 2 (at least in the font used here) are curved more than in your picture; they are curved enough to stop your plans. But it's hard to tell.

I agree with the 2, but with the 3, I think it is possible even with a lot more curvature. I'll try to do a picture:

Thok*
Guest

 Posted: Sat Sep 27, 2008 8:06 am    Post subject: 31 3 depends on how you write it: if you write it so there's a point with three line segments coming out of it, it will never work. (Yes, two of those three are "curvy" but the point is that a small movement of those line segments will still result in crossings.) If you draw a 3 as two curves touching at a point, then you could be able to tile then appropriately. A cleaner version of my proof would be to think of a letter being given by a four dimensional object (pick a point on the letter, then the four dimensions are a location of that point in R^2, a rotation around that point, and an expansion/dilation factor with respect to that point.) [Yes, I've ignored the possibility of reversing orientation, but the resulting union of 2 setsdoesn't effect the argument] Then you get a theorem of the form: if there aren't arbitrarily small transformations that don't cross, you can't stack uncountably many of that object into the plane. I think it's easy to show there's a lack of arbitrarily small transformations of S (choose the center point, obviously to make life easier), or if you fail to prove this, you'll likely get an uncountable set of S's.
bonanova
Daedalian Member

 Posted: Sat Sep 27, 2008 10:48 am    Post subject: 32 This seems like a sufficient condition, and perhaps it's necessary as well? [Find a point not on the digit, from which you can sweep a ray through 360 degrees where the ray intersects the digit no more than once at any time. If [and only if?] such a point exists, you can place uncountably infinite instances of the digit. The simplest example is 0, and an obvious counter example is 8. My answers would be 0, 1, 2, 3[maybe], 5[maybe], and 7.]_________________ Vidi, vici, veni.
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