A Geometry Puzzle

[HallsofIvy]

Find two sets, P and Q, in \(\displaystyle R^2\), such that
1) P and Q are both contained in the square with vertices at (1, 1), (1, -1), (-1, -1) and (-1, 1).
2) P contains the diagonally opposite points (1, 1) and (-1, -1) and Q contains (1, -1) and (-1, 1).
3) P and Q are disjoint.
4) P and Q are both connected sets.

 

[Nehushtan]

Easy. Take a portion of the topologist’s sine curve, rotate it through \(\displaystyle \frac{\pi}4\) radians, call it \(\displaystyle P\). Take \(\displaystyle Q\) to be a portion of the line \(\displaystyle y=-x\).

 

[daon2]

Easy. Take a portion of the topologist’s sine curve, rotate it through \(\displaystyle \frac{\pi}4\) radians, call it \(\displaystyle P\). Take \(\displaystyle Q\) to be a portion of the line \(\displaystyle y=-x\).

Nice. I thought of a similar, uglier solution. Multiply the topologists' sine curve by 1/2 and adjoin line segments {-1} x [0,1], [-1,0) x {0}, (0,-1] x {0}, {1} x [0,-1]. This is P.

For Q, Take the line segments [-1,0] x {-1}, {0} x [-1,1], [1,0] x {1}