# Challenge 17: T and other letters

1. May 29, 2014

### micromass

Consider the letter T (written as such: thus we have two line segments).

1) Prove that it is impossible to to place uncountably many copies of the letter T disjointly in the plane $\mathbb{R}^2$.

2) Prove that it is impossible to place uncountably many homeomorphic copies of the letter T disjointly in the plane.

3) For which letters in the alphabet is (2) possible? The letters are written as follows

A B C D E F G H J I J K L M N O P Q R S T U V W X Y Z

4) Let $U\subseteq \mathbb{R}^3$ be shaped like an umbrella: it is a disc (with boundary) with a perpendicular segment attached to its center. Prove that uncountably many copies of $U$ cannot be placed disjointly in the space $\mathbb{R}^3$.

5) What happens to 4 if the line segment is attached to the boundary of the disc instead?

(Source: Pugh's Real mathematical analysis)

Please list any sources that you have used to solve this question. Using google or other search engines is forbidden. Wikipedia is allowed but not its search engine. Wolfram alpha and other mathematical software is allowed.

Points will be given as follows:
1) First person to post a correct solution to one of the above points will receive 3 points (So if you solved 3 and 4, you will receive 6 points).

2) Anybody to post a (original) correct solution to one of the above points will receive 1 point.

3) Anybody posting a (nontrivial) generalization of some of the above questions will receive 2 points

4) The person who posts a solution with the least advanced mathematical machinery will receive 1 extra point for his solution (for example, somebody solving this with basic calculus will have an "easier" solution than somebody using singular homology), if two people use the same mathematical machinery, then we will look at how complicated the proof is.

5) The person with the most elegant solution will receive 1 extra point for his solution (I decide whose solution is most elegant)

Please thank this post if you think this is an interesting challenge (no, I'm not doing this to get more thanks, I just want to see if I should post more things like this and the "thanks" system is the easiest way for this).

Private messages with questions, problem suggestions, etc. are always welcome!

2. May 29, 2014

### economicsnerd

If we're allowed to take earlier questions as given lemmata...

My answer for part (3) (taking the statement of (2) as given) is CDG(I)JLMNOSUVWZ. [The letter "I" makes the list if your font makes it a straight line, doesn't if your font gives it serifs.] Every omitted letter has a subset homeomorphic to T, so that it's not possible by part (2). Every included letter is homeomorphic to a subset of the unit circle $\mathbb S = \{x\in\mathbb R^2: \enspace ||x||_2=1\}$. Then we're done because $\{\alpha\mathbb S\}_{\alpha\in\mathbb R_{++}}$ is an uncountable collection of pairwise disjoint sets each homeomorphic to $\mathbb S$.

3. May 29, 2014

### Staff: Mentor

(1) first consider the problem if rotations are not allowed:
Consider a "T" with height h and width w at an arbitrary position in R^2. There is no way to put another T with a distance of less than $x=min\left(h,\frac{w}{2}\right)$ between the T (measured between equivalent points). Therefore, every T occupies a finite space of at least $\pi\,x^2$, and there is no way to place an uncountable number of them in R^2.

If rotations are allowed, consider the distances between the joints: the minimal distance is 0, but to get a second T closer than x/100 it has to look very close to an inverted T (which gives it a cross-like shape), and we cannot add another T as close as x/100*. Therefore, in a radius of x/100 around each joint of a T, there is at most one other T. Following the same argument as above (just with 2 T in each circle), there is no way to place an uncountable number of them in R^2.

(4) can be shown in the same way.

(5) If the line segment is attached to the boundary it is possible. We can shift the shape by $\alpha \vec{r}$ for every real $\alpha$, where $\vec{r}$ is orthogonal to the line between "end of line segment" and "center of disk" and in the same plane as the line segment.

Just to write that in an explicit way, a trivial extension of (2): This is impossible for every shape that has a subset homeomorphic to T.

*no formal proof, but it should be obvious if you draw it.

4. May 29, 2014

### micromass

Extra question: what if the T's in (1) (and others) do not have the same dimensions? (of course, this follows from (2) once that is proven).

5. May 29, 2014

### The_Duck

I'll consider disks instead of T's, but I think it shouldn't be too hard to make an argument that T's and disks are equivalent in this question (Edit: after thinking about it, maybe it is hard).

I want to show that there is no uncountable set of disjoint disks each of finite radius. To prove this, suppose there is some set $S$ of disjoint disks. We will show that $S$ is countable. Consider the subset $S_d$ of $S$ consisting of disks that lie entirely within a distance $d$ of the origin. This set might be infinite. But consider the sub-subset $S_{d,r}$ of $S_d$ consisting of disks with a distance $d$ of the origin with radius at least $r$. As long as $r > 0$, $S_{d,r}$ consists of only a finite number of disks, because the total area of all the disks in $S_{d,r}$ must be at most $\pi d^2$.

I claim that $S_d$ is countable. Consider the sequence of sets
$$S_d^{(n)} = \bigcup_{i = 1}^n S_{d,r_i}$$
where $r_i$ is some sequence of positive real numbers that tends to $0$ as $i \to \infty$. Any disk in $S_d$ must appear in $S_d^{(n)}$ for some finite $n$. This gives us a way to count the disks in $S_d$. First we write down all the disks in $S_{d,r_1}$. Then we write down all the disks in $S_{d,r_2}$ that weren't in $S_{d,r_2}$. Then we write down all the disks in $S_{d,r_3}$ that we haven't written down yet. And so on. Every disk eventually gets written down in this way, and we only add a finite number of disks to the list in each step. So this puts the elements of $S_d$ in one to one correspondence with (a subset of) the positive integers. So $S_d$ is countable.

Now by a similar argument we can show that $S$ is countable. Consider the sequence of sets
$$S^{(n)} = \bigcup_{i = 1}^n S_{d_i}$$
where $d_i$ is a sequence of positive real numbers that tends to infinity. Every disk in $S$ eventually appears as an element of one of the $S^{(n)}$. We can use this to count the disks in $S$, but we have to remember that the $S^{(n)}$'s can be infinite (but we have established that they are countable). We count the elements of $S$ as follows. First we write down the first element of $S^{(1)}$. Then we write down the first element of $S^{(2)}$ and the second element of $S^{(1)}$. Then we write down the first element of $S^{(3)}$, the second element of $S^{(2)}$, and the third element of $S^{(1)}$. And so on. (We skip any element we have already written down). This procedure eventually writes down all the disks in $S$, so $S$ is countable.

This argument shows that any set of disjoint disks, all of finite radius, must be countable.

Last edited: May 29, 2014
6. May 29, 2014

### Staff: Mentor

A similar approach for the extended question 1, via contradiction:

If I refer to a T via a single point, I mean the "center" of a T (where the 3 lines meet).
Assume there is an uncountable set of "T"s of arbitrary size and orientation in R^2, then there is at least one unit square with an uncountable set of "T" inside (otherwise we can find a way to count all "T"). Take this unit square:

Making a "T" smaller (while keeping the central point) keeps the property of no overlap between the "T". We make a countable set of decreasing size classes an with a limit 0, and reduce each T to the nearest smaller size class. Based on post 3, every class has at most a finite number of "T". As a result, our uncountable set can be classified with a countable number of sets with a finite content. Contradiction.

7. May 29, 2014

### verty

I'll note for number 1 that it becomes possible if we remove a single point from T, the point where the two lines meet. As proof, center all T's at a single point and rotate each by an angle $0 ≤ Θ < {∏ \over 2}$, there are uncountably many of these angles and the only common point is the point where the lines meet. Note that T minus that point is not homeomorphic to a subset of the circle (I believe). Clearly there is more to it than just that.

8. May 30, 2014

### verty

Actually, I suppose it may be homeomorphic to a subset of the circle if we can just align the three line segments around a circle with some gaps between them. This is not a subject I know much about but it is interesting that that one point causes so much trouble.

9. May 30, 2014

### verty

Here is something interesting that I've noticed. Starting with one T, we can place a second T very close by if we bend the top bar of the T to become a right angle, so it looks like an up arrow. It can then fit in the right "armpit" of the T, as close as we like to place it. This up arrow has a right armpit as well where can place a narrower arrow. Each iteration will halve the angle of the top bar, so there are countably many arrows where each squeezes into a right armpit.

Now we can think of this a kind of divide-and-conquer situation. As mentioned in my earlier post, there are uncountably many angles $Θ$ such that $0 ≤ Θ < {∏ \over 2}$, but we can see that these $∏ \over 2$ radians have been divided and conquered by the arrows because each halves the angle, and the result is that we have countably many arrows rather than uncountably many.

And set theory tells us that the cardinality of the real numbers (or any uncountable set with this same cardinality) has cardinality $2^{\aleph_0}$. And the arrows, being countable, have cardinality $\aleph_0$.

But now I pull this trick. The starting T has two armpits, I can place 2 arrows, one each side. And for the next step there are 4 armpits, and so on. Although I'm halving the angle, I'm placing twice as many arrows each step. So if I place an arrow into every armpit at every step, is it not that there are now $2^{\aleph_0}$ arrows?

I know the mathematical answer, we can order the arrows by top bar angle and count them like we count the rationals, so there are countably many. But it is unfortunate that this part of math is so strange.

10. May 30, 2014

### micromass

At each stage $n$, you have $2^n$ arrows. So if you complete infinitely many steps, it makes sense that you have $\lim_{n\rightarrow +\aleph_0} 2^n$ arrows. Sadly, the operation $2^X$ is not continuous when the exponents are cardinals, so you can't just substitute $n=\aleph_0$ into the limit. However, when we look at $2^X$ as ordinals, then it is continuous. So we can deduce that there are $2^\omega$ arrows, where we consider ordinals. However, the ordinal $2^\omega$ is countable.

11. May 30, 2014

### verty

I said earlier that we can rotate the T's so that they share only a single point, so that if we remove that point, there are uncountably many T's. But suppose now that each T has at least one point with rational coordinates, I'll call these R-points. The T's are disjoint, so there are uncountably many of these R-points. If we sum the two coordinates, there are only countably many sums, so for at least one of the sums, there must be uncountably many R-points. But knowing the sum, we can calculate the second coordinate of these R-points by knowing the first. So there are countably many of these R-points, absurd.

So there are only countably many T's with an R-point, but uncountably many with no R-points. By placing the T's point-of-meeting on a point with an irrational coordinate and extending the T's legs to infinity in both directions, we see that there are lines in the plane with no R-point.

So there are so many reals that there are these crazy lines. This supports to some degree the idea that there are "many more" irrationals than rationals.

-- Hopefully this is all correct and I've not made any mistakes. I'll not post again unless it is in reply or to attempt to solve one of the challenge points.

12. May 30, 2014

### Staff: Mentor

@verty: I think this is just an argument that there are more real than rational numbers - but with exactly this as information as input.

$2^{\aleph_0}$ contains limits of the interval process, your set of T does not (not all).

13. May 30, 2014

### verty

My point was not to argue that because there are many reals, there are many real lines, my point was that there is some intuitive oddness to say that there are uncountably many angles for which no R-point is met in either direction. This was more about intuition than fact.

Around the year 1900, people were trying to show that math and logic were compatible in the sense that logic could be applied to answer all mathematical questions. It turned out that was not the case, even arithmetic was too large for that to happen. But there was a desire that math be logical, ordered, sensible. We can call this a geometric desire; non-Euclidean geometries were discovered by changing the axioms and I think we can view the logicist program as an attempt to rein in or tame the complexity of axiomatic reasoning, sort of like discovering America and then desiring to have a map created for future navigations.

So there was this desire for it all to make sense, but there is also the idea that math entities exist in a separate universe where there is only math. The point is this, if you look at any language, there are words that refer to things, snow refers to the cold white stuff on the ground, and so on. To understand snow, one understands the white stuff. And with math it should be the same: to understand the real plane, one must understand that mathematical thing with crazy lines. I can't say it better than this.

14. Jun 16, 2014

### Joffan

I definitely believe that S should also be omitted from the answer to (3), as copies of S cannot be nested arbitrarily closely together. I'm not sure about M (with true parallel edges).

G and J also depend on font, incidentally.

------------
Edit: I'm officially embarrassed about skip-reading and missing the homeomorphism aspect, as per micromass below. So S and M stay in the answer to 3, but at least I generated a new question :-).

Last edited: Jun 16, 2014
15. Jun 16, 2014

### micromass

But in question (3), I merely ask whether it is possible for those shapes homeomorphic to the letter. A line is homeomorphic to S, and it is possible for lines. So S should not be omitted in (3).

However, you raise a good question. Thus, I add:

6) For which letters in the alphabet is (1) possible (I allow the letters to be rotated and I allow decrease is size). What if I don't allow rotation/decrease in size/both?

16. Jun 17, 2014

### Staff: Mentor

Well, we restricted the operations we can do with the letters, so we don't have to consider A, B, E, ... again, they won't work.

Without serifs:

I, L, N, V, W and Z in those simple forms can be shifted (e.g. N: up+left; Z: up+right). No scaling or rotation necessary.

With different sizes (with or without rotation):
CDMOU work with a simple scaling process with a suitable center, as they can be a subset of the border of a star domain.
I don't remember a (common) font where J would be an issue. A point "inside" should work.
G depends on the font.
This just leaves S open, depending on the precise shape.