# Challenge 10b: Temperatures of the Earth

1. Dec 11, 2013

### Office_Shredder

Staff Emeritus
For the sake of this challenge, you can assume that the world behaves nicely (everything is continuous, differentiable etc, the world is a sphere blah blah blah) but figurative bonus points if you assume fewer things. If you're looking for an easier challenge check out Challenge 10a.

The Challenge: Show that there exists a curve dividing the earth into two pieces such that for every point p on the curve, the temperature at p and at -p (the point on the opposite side of the earth) are equal, or construct a temperature function in which no such dividing curve exists.

Last edited: Dec 11, 2013
2. Dec 11, 2013

### D H

Staff Emeritus
I'm game, but I skipped lunch for a meeting.

I'm going to have a ham sandwich. Even though the bread is a bit lopsided, I'm probably going to get a bit OCD and try to cut it exactly in half.

3. Dec 11, 2013

### Office_Shredder

Staff Emeritus
Re-reading my question, I realized I forgot to mention that you only get bonus points for assuming the temperature is bad if you prove the curve exists. No bonus points for making a horribly discontinuous function and noticing it's horribly discontinuous.

4. Dec 11, 2013

### jbunniii

Here is a start which probably doesn't go very far, but it recasts the problem in terms of topology instead of temperature functions.

We identify the earth with the unit sphere $S$, in other words, the set of all points $(x,y,z)$ such that $x^2 + y^2 + z^2 = 1$. We denote the temperature at each point $p = (x,y,z) \in S$ as $t(p)$. We assume that $t$ is continuous. We also define the difference between the temperature at a given point compared with its opposite point on the other side of the world: $d(p) = t(p) - t(-p)$. We note that $d(-p) = -d(p)$.

We partition $S$ into three regions, $S^+$, $S^0$, and $S^-$, which are, respectively, the regions where $d$ is positive, zero, and negative, respectively.

Challenge 10(a) shows us that $S^0$ is nonempty.

If $S^+$ and $S^-$ are both empty, then $d=0$ everywhere on the sphere, so any curve that divides the sphere into two pieces will work, and we're done.

So suppose that one of $S^+$ or $S^-$ is nonempty. Then both are nonempty, because of the relation $d(-p) = -d(p)$. So in this case the sphere is divided into three disjoint nonempty regions. Now $S^+$ and $S^-$ are open sets, because they are the inverse images $d^{-1}((0,\infty))$ and $d^{-1}((-\infty,0))$ of open sets, and $d$ is continuous. Similarly, $S^0$ is a closed set because it is the inverse image $d^{-1}(\{0\})$.

So now we need to argue that the boundary between $S^+$ and $S^-$ contains a simple closed path. This will be the fun part, but I'll have to think about how to continue from here. At this point I'm not even sure it has to be true.

5. Dec 12, 2013

### Staff: Mentor

Consider the set of points that are path-connected to a specific point p0 with d(p0)>0 such that d is positive everywhere on the path. It has a non-empty interior, its complement has a non-empty interor as well. Its boundary can still look odd, but it forms a closed cycle around p0.

6. Dec 12, 2013

### jbunniii

I agree with all of this...
And this seems plausible, but I'm struggling with how to prove it. Let's say that $P$ is a connected component of $S^+$. The boundary of $P$ need not consist only of a closed cycle around its exterior. There might also be "holes" in the interior consisting of isolated points or non-closed curves of points in $S^0$, and these would also be part of the boundary of $P$, but they might not contain any closed cycles.

I guess I'm not sure how to define the boundary of the exterior of $P$. This is the part of the boundary that presumably must contain a cycle. Maybe it would be $\text{closure}(P) \cap \text{closure}(\text{interior}(P^c))$, but there's probably some weird configuration where this wouldn't give me what I want.

I'm also not sure in general how to go about proving that a given set of points contains a closed curve. If you give me some random closed curve on the sphere, how do I prove it's a closed curve, assuming we don't have a parameterization for it? Maybe I should have paid more attention in algebraic topology. :tongue:

7. Dec 12, 2013

### Staff: Mentor

Ah, good point.
There is no unique way to do this, unless we define -p0 (as obvious example) to be "outside" and everything not connected to -p0 (here via a path in the region d<0) as "inside".

8. Dec 12, 2013

### jbunniii

Right, I was thinking of the exterior as the portion that contains $-p_0$. I agree that topologically this just looks like another "hole" since we're on the surface of the sphere. I guess we just need that the boundary between $P$ and some connected component of $\text{interior}(P^c)$ must contain a loop. I don't know if this has to be true. The boundary $\text{closure}(P) \cap \text{closure}(\text{interior}(P^c))$ has certain nice properties: it is compact and it contains no isolated points, but the same can be said of the Cantor set...

9. Dec 12, 2013

### jbunniii

Let us consider a single great circle $C$. We may assume that $C$ does not consist only of points in $S^0$, otherwise we're done.

So $C$ contains a pair of opposite points $p$ and $q$, one in $S^+$ and the other in $S^-$. By challenge 10(a), $C$ also contains at least two points in $S^0$.

Moreover, the points in $C \cap S^0$ cannot all be isolated points of $S^0$. For suppose this were true. Then we could modify $C$ slightly to make detours around the isolated points of $S^0$, resulting in a closed curve that connects $p$ and $q$ but contains no points of $S^0$. This is impossible by the intermediate value theorem. Therefore $C$ contains at least one limit point (hence at least two limit points, the other being the opposite point) of $S^0$.

We can do the same thing for every great circle that passes through $p$ and $q$. If we orient the sphere so that $p$ and $q$ are at the north and south pole, then every "longitudinal great circle" (or whatever you call it) contains a limit point of $S^0$. Thus we have uncountably many limit points of $S^0$, at least one at every longitude angle.

So that's promising: certainly if we didn't have a limit point of $S^0$ at every longitude angle, it would not be possible to draw a loop that places $p$ and $q$ into disjoint regions. But the converse is not at all obvious.

Now $S^0$ is a closed subset of the compact surface $S$, hence $S^0$ is compact. Let $\epsilon > 0$. Cover $S^0$ by placing an $\epsilon$-ball centered at $x$, for every $x \in S^0$. By compactness, a finite subcollection of these balls covers $S^0$.

Now it should be possible to argue (but the devil is in the details) that there must be a "chain" of finitely many overlapping $\epsilon$-balls, each centered on a point of $S^0$, that separates the region containing $p$ from the region containing $q$. Otherwise, there would be a path from $p$ to $q$ which does not touch $S^0$, and that's impossible. We can find such a chain for any $\epsilon$, and maybe if we make the right limiting argument... I'll think about this more tonight as time allows. [Refinement to consider: I think I can replace $S^0$ with "the set of limit points of $S^0$" in this paragraph and the last one.]

Last edited: Dec 12, 2013
10. Dec 12, 2013

### Boorglar

We can parameterize the sphere's surface with two parameters $θ, ψ$ (longitude and latitude, for example) on the domain $S = [0,2\pi)×[-\pi,\pi]$. Then we have a temperature function $T:S→ℝ$ defined on every point, with value $T(θ,ψ)$ for a given longitude and latitude.

I will assume that $T$ is continuous and has continuous partial derivatives everywhere. The difference between the temperature at point P and the temperature at point -P will be denoted by $D(θ,ψ)$ where P has longitude θ and latitude ψ. Since $D$ is a difference of continuously differentiable functions, it is also continuously differentiable. An additional assumption that I make is that $\frac{∂D}{∂ψ} ≠ 0$ everywhere on the sphere. (This assumption doesn't seem necessary, but it allows me to use the Implicit Function Theorem with no additional work).

From the solution of 10a, we know there is a point $(θ_0,ψ_0)$ in $S$ such that $D(θ_0,ψ_0)=0$. Therefore, by the implicit function theorem, there exists an open region R around $(θ_0,ψ_0)$ such that in this region, there is a continuous function $ψ(θ)$ such that $D(θ,ψ(θ))=0$.

My aim is to show that this argument can be applied repeatedly so that the function $ψ(θ)$ exists and is continuous for every θ in [0, 2π). Otherwise, there must be a least upper bound $α$ to the set of all θ such that $ψ(θ)$ exists. But this means $D(θ,ψ(θ)) = 0$ for all $θ < α$ which are sufficiently close. By continuity, $D(α,ψ(α)) = 0$ too. Applying the implicit function theorem at that point shows that there are larger values for which the condition is also satisfied, contradicting the choice of the least upper bound.

This means that there is a continuous function ψ which exists for all θ and such that the point $(θ, ψ(θ))$ has the same temperature as the point opposite to it. The curve $(θ,ψ(θ))$ is continuous and divides the Earth in two (the set of points with ψ larger than ψ(θ) and those with ψ smaller than ψ(θ)).

11. Dec 13, 2013

### Staff: Mentor

This assumption is wrong, and in general you cannot find a function $ψ(θ)$ such that $D(θ,ψ(θ))=0$. Simple counterexample: Take a temperature that depends on θ only for some range of θ,ψ.
Alternatively, take something that looks like f(x)=x^2.

It is not sufficient to show that for each θ there is a ψ where you can find a function - you also have to show that all those functions are nicely connected.

12. Dec 19, 2013

### Citan Uzuki

Such a dividing curve need not exist. Consider the following: Let $A\subseteq \mathbb{R}^2$ be defined by:
$$A = \{(x, \sin(\frac{1}{x})): x\in [-\frac{1}{\pi}, 0)\cup (0,\frac{1}{\pi})\} \cup \{(x, 0): x \in [-\frac{\pi}{2}, -\frac{1}{\pi}] \cup [\frac{1}{\pi}, \frac{\pi}{2}]\} \cup \{(0, y): y\in [-1,1]\}$$. A is a compact topologist's sine curve between $-1/\pi$ and $1/\pi$ extended by straight horizontal lines to $-\pi/2$ and $\pi/2$. Now, let $f:\mathbb{R}^2 \rightarrow S^2$ be given by $f(\theta, \phi) = (\cos \theta \cos \phi, \sin \theta \cos \phi, \sin \phi)$ (here $\theta$ and $\phi$ represent latitude and longitude). Let $B = f(A) \cup -f(A)$. B is a topologist's sine curve on the sphere and its mirror image joined by horizontal lines along the equator. It is clear that B divides the sphere into two connected components -- one to the north of B and one to the south. Further, these components are clearly mirror images of each other. Let g be a smooth function on S^2 such that g and all its derivatives are zero on B and g is positive on the complement of B (this is possible since B is compact). Finally, let h be defined by:
$$h(x) = \begin{cases}g(x) & \mbox{if }x\mbox{ is north of }B \\ 0 & \mbox{if }x \in B \\ -g(-x) & \mbox{if }x \mbox{ is south of }B \end{cases}$$

h is a smooth function on $S^2$, and the set of points on $S^2$ where h(x) = h(-x) is precisely B. No curve dividing $S^2$ into two pieces can exist in B, because each path component of B is homeomorphic to either $[0, 1]$ or $\mathbb{R}$. Q.E.D.

13. Dec 20, 2013

### Office_Shredder

Staff Emeritus
Nice solution! It takes heavy advantage of the definition of curve. The natural followup question is, does there necessarily exist $B \subset S^2$ such that p(x) = p(-x) for all $x\in B$, and $S^2 \setminus B$ has two (or more?) connected components?

Last edited by a moderator: Dec 20, 2013
14. Dec 20, 2013

### jbunniii

Yes, if the temperature function is continuous.

Recall that I defined the difference function $d(p) = t(p) - t(-p)$.

Case 1: If $d(p) = 0$ everywhere, then we may take $B$ to be the equator. Then the upper and lower hemispheres give us the two connected components of $S^2 \setminus B$.

Case 2: If $d(p)$ is nonzero for some point $p$, then $d(-p)$ has the opposite sign. Thus both $P = d^{-1}((0,\infty))$ and $N = d^{-1}((-\infty,0))$ are nonempty. Also, Challenge 10(a) ensures that $d^{-1}(\{0\})$ is nonempty. We'll call this $B$. Thus the sphere $S^2$ is partitioned into three disjoint nonempty sets $P$, $N$, and $B$. Moreover, $P$ and $N$ are open, as they are the inverse images of open sets. This means that $S^2 \setminus B = P \cup N$ is the disjoint union of two open sets, so it has at least two connected components.

Last edited by a moderator: Dec 20, 2013
15. Dec 20, 2013

### jbunniii

Citan Uzuki - nice solution, by the way. I was vaguely thinking over the past few days that a boundary like the topologist's sine curve would probably be a counterexample, thus derailing my argument. But I was tired of thinking about this problem so I didn't pursue it. :tongue:

I was also thinking there might be an even a more pathological boundary $B$, one that is not even connected, let alone path connected, but somehow scattered in such a way that there is no path from $P$ to $N$ that does not touch $B$, so the intermediate value theorem is not contradicted. But I don't know enough weird topological constructions to come up with such an example, nor could I prove that one does not exist.

Last edited: Dec 20, 2013
16. Dec 20, 2013

### jbunniii

OK, it appears (I haven't read the proof yet) that if $B$ is a totally disconnected, compact subset of $\mathbb{R}^2$ (or presumably of $S^2$), then the complement of $B$ is connected.

http://math.stackexchange.com/quest...ally-disconnected-compact-subset-of-the-plane

Since our $d^{-1}(\{0\})$ is the inverse image of a closed set, it is closed, hence compact. So this shows in some sense that the most pathological sort of counterexample to the original problem is ruled out: $B$ cannot be totally disconnected.

17. Dec 21, 2013

### Staff: Mentor

Nice solution, Citan Uzuki.