1. ## A topological puzzle

Find two sets, P and Q, such that:

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

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

2. ## Re: A topological puzzle

Spoiler:
Must the sets P and Q lie on the same surface of the plane $R^2$? 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

3. ## Re: A topological puzzle

Now that's the coolest thing I've run across for a while.

-Dan

4. ## Re: A topological puzzle

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

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 $\{(x, y)| y= sin(1/x), x>0\}$ includes the entire line segment (0, y) for y between -1 and 1.

5. ## Re: A topological puzzle

Well it's still cool.

-Dan

6. ## Re: A topological puzzle

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

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 $\{(x, y)| y= sin(1/x), x>0\}$ includes the entire line segment (0, y) for y between -1 and 1.
I finally got it!

According to mathworld:
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.

Spoiler:
The sets

$P = \{(-1,-1)\} ~\cup~ \{(x, y)~|~y= \frac 1 2 (1+x) + \frac 1 2 |1-x| \sin(\frac {\pi}{x+1}), -1 < x\le 1\}$

and

$Q = \{(-1,1)\} ~\cup~ \{(x, y)~|~y= - \frac 1 2 (1+x) + \frac 1 2 |1-x| \sin(\frac {\pi}{x+1}), -1 < x\le 1\}$

do the trick.

They are derived from the topologist's sine curve.

Here's Wolfram's plot.

7. ## Re: A topological puzzle

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 $\{ (x, y)| y= .95 sin(1/x)+ .01, 0\le x\le \frac{1}{\pi}\}$ and let C be the line from $(1/\pi, .01)$ to (1, 1), y= . Let $P= A\cup B\cup C$.

Let X be the line from (-1 ,1) to (0, .5). Let Y be the set set $\{(x, y)| y= .95 sin(1/x)- .01, 0\le x\le \frac{1}{\pi}\}$. Let Z be the line from $(1/\pi, -.01)$ to (1, -1). Let $Q= X\cup Y\cup Z$. 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).

8. ## Re: A topological puzzle

Originally Posted by topsquark

Well it's still cool.

-Dan
Yes, it is!

9. ## Re: A topological puzzle

Originally Posted by HallsofIvy
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 $\{ (x, y)| y= .95 sin(1/x)+ .01, 0\le x\le \frac{1}{\pi}\}$ and let C be the line from $(1/\pi, .01)$ to (1, 1), y= . Let $P= A\cup B\cup C$.

Let X be the line from (-1 ,1) to (0, .5). Let Y be the set set $\{(x, y)| y= .95 sin(1/x)- .01, 0\le x\le \frac{1}{\pi}\}$. Let Z be the line from $(1/\pi, -.01)$ to (1, -1). Let $Q= X\cup Y\cup Z$. 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.
Hmm, let's see.
The curves I came up with are:

Wolfram|Alpha

while you came up with:

Wolfram|Alpha

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.