1) Both P and Q are contained within the square in $\displaystyle R^2$ with vertices (1, 1), (1, -1), (-1, -1) and (-1, 1). (In other words all (x, y) in P or Q must satisfy $\displaystyle -1\le x\le 1$, $\displaystyle -1\le y\le 1$.)

2) P contains the diagonally opposite points (1, 1) and (-1, -1) and Q contains the diagonally opposite points (-1, 1) and (1, -1).

3) P and Q are both connected sets.

4) $\displaystyle P\cap Q$ is empty. (P and Q have no point in common.)

Must the sets P and Q lie on the same surface of the plane $\displaystyle R^2$? If not then draw a straight line between (1,1) and (-1,-1) = P and a path that loops underneath from (-1,1) and (1,-1) = Q

Can I bend the space? If so, bend the plane R^2 so that (1,1) and (-1,-1) become the same point. Let (1,1) be P and let the line between (-1,1) and (1,-1) be Q

No, sorry. When I said "Both P and Q are contained in the square in $\displaystyle R^2$ ...", I meant that every point of P and Q are in $\displaystyle R^2$.

This uses sin(1/x) so that the two sets can "dodge" one another very rapidly! Also, it is of importance that the closure of the set $\displaystyle \{(x, y)| y= sin(1/x), x>0\}$ includes the entire line segment (0, y) for y between -1 and 1.

Last edited by HallsofIvy; Mar 8th 2013 at 08:06 AM.

No, sorry. When I said "Both P and Q are contained in the square in $\displaystyle R^2$ ...", I meant that every point of P and Q are in $\displaystyle R^2$.

This uses sin(1/x) so that the two sets can "dodge" one another very rapidly! Also, it is of importance that the closure of the set $\displaystyle \{(x, y)| y= sin(1/x), x>0\}$ includes the entire line segment (0, y) for y between -1 and 1.

A connected set is a set which cannot be partitioned into two nonempty subsets such that each subset has no points in common with the set closure of the other.

Nice. My solution was almost the same, still using the topologist's sine curve and requiring three "subsets" for each of P an Q instead of your two. Let A be the line from (-1, -1) to (0, -.5). Let B be the set $\displaystyle \{ (x, y)| y= .95 sin(1/x)+ .01, 0\le x\le \frac{1}{\pi}\}$ and let C be the line from $\displaystyle (1/\pi, .01)$ to (1, 1), y= . Let $\displaystyle P= A\cup B\cup C$.

Let X be the line from (-1 ,1) to (0, .5). Let Y be the set set $\displaystyle \{(x, y)| y= .95 sin(1/x)- .01, 0\le x\le \frac{1}{\pi}\}$. Let Z be the line from $\displaystyle (1/\pi, -.01)$ to (1, -1). Let $\displaystyle Q= X\cup Y\cup Z$. The point of the "+.01" and "-.01" is to keep the curves from crossing. The point of the ".95" is to keep the y values between -1 and 1. Of course, P and Q are connected because the closure of the "topologist's sine curve" contains the entire interval (0, -1) to (0, 1).

Nice. My solution was almost the same, still using the topologist's sine curve and requiring three "subsets" for each of P an Q instead of your two. Let A be the line from (-1, -1) to (0, -.5). Let B be the set $\displaystyle \{ (x, y)| y= .95 sin(1/x)+ .01, 0\le x\le \frac{1}{\pi}\}$ and let C be the line from $\displaystyle (1/\pi, .01)$ to (1, 1), y= . Let $\displaystyle P= A\cup B\cup C$.

Let X be the line from (-1 ,1) to (0, .5). Let Y be the set set $\displaystyle \{(x, y)| y= .95 sin(1/x)- .01, 0\le x\le \frac{1}{\pi}\}$. Let Z be the line from $\displaystyle (1/\pi, -.01)$ to (1, -1). Let $\displaystyle Q= X\cup Y\cup Z$. The point of the "+.01" and "-.01" is to keep the curves from crossing. The point of the ".95" is to keep the y values between -1 and 1.

Mine has simpler formulas, while yours looks more like crossing lines.
I'll leave it to any other observer to indicate what they like best.

Of course, P and Q are connected because the closure of the "topologist's sine curve" contains the entire interval (0, -1) to (0, 1).

That's nice!
On the one hand we have 2 curves that are each connected, that cross each other, but that do not have any point in common.
But on the other hand these 2 curves are connected to each other.