Why do "path diagonals" of a rectangle need to intersect? 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.
 A: 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.
A: (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.
