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Let $X$ be a rectangle in $\mathbb{R}^2$. Let $A,B,C,D$ denote its vertices.

Assume that there is a path $\alpha$ connecting $A$ and $C$ such that its image (besides endpoints) is completely inside of the interior of the rectangle. Similarly, assume that there is a path $\beta$ connecting $B$ and $D$ such that its image (besides endpoints) is completely inside of the interior of the rectangle.

So my question is: Do the paths need to intersect?

It seems to be almost obvious that the answer is yes; however, I was not able to find an easy proof using basic topology tools.

I tried using connectedness, but unfortunately I am not even sure how to justify that one such path splits a rectangle into 2 connected components... Actually, it does not even have to be true, since paths can have "loops," i.e. paths do not need to be injective. However, I even struggled with the injective case.

In other words, I am stuck. Thus, any help would be very appreciated.

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    $\begingroup$ iirc this is basically equivalent to the Jordan curve theorem, so it is seemingly obvious but also very hard. $\endgroup$ – mercio Sep 22 '16 at 19:34
  • $\begingroup$ Hm, good idea. I think one could use the Jordan curve theorem here if the paths were injective... $\endgroup$ – Pawel Sep 22 '16 at 19:46
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Let us assume for definiteness of notation that $X=[0,1]^2$. Suppose you have a pair of paths $\alpha,\beta:[0,1]\to X$ with $\alpha$ going from $A=(0,0)$ to $C=(1,1)$ and $\beta$ going from $B=(1,0)$ to $D=(0,1)$, and suppose they never intersect. Define $f:[0,1]^2\to\mathbb{R}^2\setminus\{(0,0)\}$ by $f(s,t)=\alpha(s)-\beta(t)$, and consider the restriction of $f$ to $\partial[0,1]^2\cong S^1$.

Note that the restriction of $f$ to $[0,1]\times\{0\}$ is a path from $A-B=(-1,0)$ to $C-B=(0,1)$ that is contained entirely in the set $\{p-B:p\in X\}=[-1,0]\times[0,1]$, and also does not pass through the origin. Since $[-1,0]\times[0,1]\setminus\{(0,0)\}$ is convex, this path is homotopic (rel the endpoints) in $\mathbb{R}^2\setminus\{(0,0)\}$ to the straight line path from $(-1,0)$ to $(0,1)$, by a straight line homotopy.

Similarly, the restriction of $f$ to $\{1\}\times[0,1]$ is homotopic rel its endpoints to the straight line path from $(0,1)$ to $(1,0)$, and the restriction to the other two edges of $\partial[0,1]^2$ are homotopic rel their end points to the straight line paths from $(1,0)$ to $(0,-1)$ and from $(0,-1)$ to $(-1,0)$. Putting it all together, we get that the restriction of $f$ to $\partial[0,1]^2$ is homotopic to a loop that traces the square with vertices $(0,\pm 1)$ and $(\pm 1,0)$ clockwise. This loop has winding number $-1$ around the origin, and in particular this means that $f|_{\partial[0,1]^2}:\partial[0,1]^2\to\mathbb{R}^2\setminus\{(0,0)\}$ is not nullhomotopic. But $f|_{\partial[0,1]^2}$ extends to a map on all of $[0,1]^2$ (namely, $f$), and so is nullhomotopic. This is a contradiction.

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  • $\begingroup$ I think when constructing your first straight line path, you meant "to the straight line path from $(-1,0)$ to $(0,1)$". $\endgroup$ – Pawel Sep 22 '16 at 21:14
  • $\begingroup$ Besides that, I think your proof works! Thank you very much! $\endgroup$ – Pawel Sep 22 '16 at 21:20
  • $\begingroup$ Oops, you're right. I've corrected that. $\endgroup$ – Eric Wofsey Sep 22 '16 at 22:12
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(this note is too long for a comment)

Regarding connectedness, it turns out there exists two sets such that:

  • Each set is connected
  • Each set is contained in the rectangle
  • Each set contains a pair of opposite corners of the rectangle
  • The two sets do not intersect

The basic idea to finagle the "crossing" is demonstrated by the two functions

$$ f(x) = \begin{cases} \sin(1/x) & x > 0 \\ -1 & x \leq 0 \end{cases} $$ $$ g(x) = \begin{cases} \sin(1/x) - 1 & x > 0 \\ 0 & x \leq 0 \end{cases} $$

The graph of each of these functions is a connected subset of the plane, and we have

  • $f(x) < g(x)$ for all $x \leq 0$
  • $f(x) > g(x)$ for all $x > 0$
  • $f(x) = g(x)$ never

Of course, neither function is continuous at $x=0$, so this phenomenon doesn't happen with paths.

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  • $\begingroup$ Thank you for this nice observation! $\endgroup$ – Pawel Sep 23 '16 at 2:20
  • $\begingroup$ Also, maybe a silly question, but what is your domain for $f$ and $g$? $\endgroup$ – Pawel Sep 23 '16 at 14:38
  • $\begingroup$ @Paquarian: Real numbers: any interval containing zero in the interior. $\endgroup$ – Hurkyl Sep 23 '16 at 16:52
  • $\begingroup$ Thanks! This is truly a great example! And it made me realize that the concept of "connectivity" is not as intuitive as I thought it was. $\endgroup$ – Pawel Sep 23 '16 at 21:00

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