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I've been thinking over this problem this weekend and although my investigations on it has led to other interesting theorems I am still nowhere close to solving it.

For any set of disjoint unordered pair of points (in $\mathbb{R}^2$), call it $S$, there is a set of simple continuous curves satisfying the following properties:

  1. The set of end points for each curve is a member of $S$ and vice-versa
  2. If two curves can intersect, then the intersection must be an end-point of one of the curves

I feel like I was able to give a proof for this when $S$ was countable using mathematical induction. Basically used that fact the plane remains connected when $|S|=1$ and if the plane is connected after adding a number of curves, it is possible to create an extra curve keeps it connected. So by induction if $S$ is countable then the constructed curves satisfy the given properties. I've never used mathematical induction in this manner so I'm not sure if this is a valid proof.

Anyway, I can not think of a way to go about proving or giving a counterexample for an uncountable $S$. I'm stuck even when I let $S$ be the closed unit square (with interior) which I suspect fails the conjecture. What if $S$ is nowhere dense? Will that always satisfy the conjecture?

genepeer
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    Not sure what you are asking. Do you require one curve for each unordered pair of points $\in S$. If yes, the statement is false. Counter examples are the complete graph $K_5$ and the complete bipartite graph $K_{3,3}$. They are non-planar graphs (i.e graph that cannot be embedded in the plane). In fact, Kuratowski has proved a theorem which assert a graph is planar iff it does not contains $K_5$ or $K_{3,3}$ as sub-graph. – achille hui Mar 26 '13 at 09:44
  • I just realized I left out the $S$ forms a partition of the plane. For some reason, I wrote "distinct unordered pairs" when I meant "disjoint unordered pairs". – genepeer Mar 26 '13 at 13:58

3 Answers3

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By the Brouwer fixed-point theorem, "every continuous function $f$ from a convex compact subset $K$ of a Euclidean space to $K$ itself has a fixed point". Because $[0,1]$ is a compact, convex subspace of $\mathbb R$, your bijection is impossible.

Indeed, if you are generalising up to $K=[0,1]\times[0,1]$, it's still impossible.

Ian Coley
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  • I guess I didn't think about your general investigation as much as the second part of your question. Where'd you come up with that map $f$ from your initial investigation? – Ian Coley Mar 26 '13 at 06:42
  • Actually, sorry but the fixed-point theorem doesn't apply since $f$ need not be continuous. The thinking behind it was if there happened to be set of curves that work for $S$=unit square. Then I could define $f$ as the set of ordered end-points. Since the end-points share the same curve $f(f(x))=x$. So if I could show there's a fixed-point even when $f$ is discontinuous, then I would have a contradiction because end-points are distinct. – genepeer Mar 26 '13 at 06:51
  • So if $S$ is defined as ordered pairs, then I guess what we want is two spaces $S_1$ and $S_2$ such that every path goes from $S_1$ to $S_2$. Are we assuming those spaces are the same, e.g. paths from one copy of the unit interval to the other? – Ian Coley Mar 26 '13 at 07:00
  • Yes, the two spaces are the same. – genepeer Mar 26 '13 at 07:04
  • So I got an $f$ such an $x$ that doesn't have a fixed point. The second part of the question isn't any help to the first after all. Back to square one. – genepeer Mar 26 '13 at 07:13
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Let $S=\{c_n=\{x_n,y_n\} \mid n \geq 1\}$ be a set of disjoint pairs of points in $\mathbb{R}^2$. For convenience, let $C_0= \bigcup\limits_{n \geq 1}c_n$.

Notice that $\mathbb{R}^2 \backslash C_0$ is path connected, because between any points $x,y \in \mathbb{R}^2 \backslash C_0$ there are uncountably many disjoint linear piecewise paths whereas $C_0$ is countable. Let $X_1$ be the range of a path between $x_0$ and $y_0$ in $\mathbb{R}^2 \backslash C_0$.

By induction, suppose we have built $X_n$ such that $X_n$ is the union of paths connecting the pairs $c_k$, for $1 \leq k \leq n$ as required. Again, $\mathbb{R}^2 \backslash (X_n \cup C_0)$ is path connected, so you can introduce $X_{n+1}$ as the union of $X_n$ and the range of a path connecting $x_{n+1}$ and $y_{n+1}$ in $\mathbb{R}^2 \backslash (X_n \cup C_0)$.

Finally, set $X= \bigcup\limits_{n \geq 1} X_n$. By construction, $X$ is a countable disjoint union of ranges of paths connecting the pairs $c_n$. Moreover, the union is increasing, so two curves don't intersect (otherwise, they would interset in some $X_n$, impossible by construction).

Added: For a counterexample when $S$ is uncountable, take $S= ...$ (under edition).

Seirios
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  • This one is promising. One thing: I'm thinking of making $S_4 = {{(q,0),(q,1/2) | q \in \mathbb{Q}\cap (1,2) }}$. To avoid pairing $(1,0)$ with two points: $(1,1)$ in $S_1$ and $(1,1/2)$. Since $S$ is countable, I suspect there is a way to use the uncountable irrational "gaps" to construct the paths. I'll let you know what I find out. – genepeer Apr 16 '13 at 19:27
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    I found a possible solution to the counterexample, so I finally tried to prove the property is true. – Seirios Apr 17 '13 at 13:41
  • I guess I could write my paper on this for my senior project next year :) – genepeer Apr 17 '13 at 17:15
  • This is similar to my argument but I can't justify using induction on a countably infinite case. To the best of my knowledge, induction is applicable only to finite sets. Are you sure this is a valid proof? – genepeer Apr 17 '13 at 23:36
  • $X$ is not built by induction: the $X_n$'s are. Here, $X$ is a kind of limit, possible because the sequence $(X_n)$ is increasing. – Seirios Apr 18 '13 at 05:54
  • See Asaf's answer here on Finite Choice: http://math.stackexchange.com/a/64239/50955 He constructs $f_n$ just like you did but says he cannot set $f = \cup f_n$ to give a proof of the axiom of countable choice. I hope you see where my confusion lies. – genepeer Apr 18 '13 at 06:47
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    There is a problem only when you don't assume the axiom of choice: Asaf Karagila said that you need the principle of dependent choice to define $\cup f_n$. – Seirios Apr 18 '13 at 07:44
  • I was hoping there may be a proof that doesn't require the axiom of choice since I'm not very familiar with it. But the more I think of it, the more this problem seems like the sort that must require some principle of choice. – genepeer Apr 18 '13 at 17:38
  • Do you think we can prove the $S$ is uncountable case by using the Well-Ordering Theorem and transfinite induction? – genepeer Apr 22 '13 at 17:37
  • The result is no longer true when $S$ is uncountable; I added a counterexample. – Seirios Apr 22 '13 at 19:56
  • The problem is that whenever I think of an $S$ that could be a counterexample, I can't prove that it is. Which leads me to believe it has a solution that requires axiom of choice. E.g. $S = {{ (x,0), (x-1,0)} | x \in (0,1) }$. – genepeer Apr 22 '13 at 20:20
  • (Using complex plane for the sake of brevity.) The $S$ I just proposed also has an easy solution: $$f_x(\theta) = \begin{cases} xe^{i\theta}-1 & \theta \in [0, \pi] \ (x+.5)e^{i\theta}-.5 & \theta \in [\pi, 2\pi] \end{cases}$$ But I can't prove if $S={{e^{i\theta},e^{i(\theta+\pi)}} | \theta \in [0,\pi) }$ has a solution or not. – genepeer Apr 22 '13 at 22:02
  • I proved a counterexample and I posted it as an answer. What do you think? – genepeer Apr 23 '13 at 18:59
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Proof of a counterexample when $S$ is uncountable. Using the complex plane, let $S=\{\{e^{i\theta}, e^{i(\theta+\pi)}\}~|~\theta \in [0,\pi) \}$.

Suppose $\mathcal{F}$ is a set of simple curves that satisfy the hypothesis.

  1. By hypothesis, if two curves intersect, then the point of intersection is an endpoint for one of the curves.
  2. There are at most two curves which only intersect the circle at endpoints. All other curves must cross the circle at least once. This has been modelled in following image:

I'll assume this model but it can also be proven when there are only $0$ or $1$ which don't cross the circle. Without loss of generality, there are uncountably many curves $\mathcal{F}_1$ that join "red points" and cross $B_1$. Similarly, for these points of intersections, $I \subset B_1$, let $\mathcal{F}_2$ be set of uncountable curves that join them to $B_2$ and crossing $R_2$ (without loss of generality again).

$\mathcal{F}' = \mathcal{F}_1 \cup \mathcal{F}_2$.

By intermediate value theorem, the curves in $\mathcal{F}'$ cross the $x$-axis. For each point $p \in I$, there are two distinct point, $x_1$ and $x_2$, corresponding points on the $x$-axis (one for a curve in $\mathcal{F}_1$ and another for $\mathcal{F}_2$). Since the curves don't cross, there is no other intersection point between $x_1$ and $x_2$. Let $x_p$ be the midpoint of $x_1$ and $x_2$. This forms a bijection between $x_p$ and $p$. Each $x_p$ is isolated and any set of isolated points is countable.$^1$ This is a contradiction as there is also bijection from each function in $\mathcal{F}_1$ to a given $p$.

Therefore, no such $\mathcal{F}$ exists for the given $S$.

$^1$ One of the interesting theorems I proved in my earlier investigations. Glad I could finally use it somewhere.

genepeer
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