Find two sets, P and Q, such that:
1) Both P and Q are contained within the square in with vertices (1, 1), (1, -1), (-1, -1) and (-1, 1). (In other words all (x, y) in P or Q must satisfy , .)
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) is empty. (P and Q have no point in common.)
No, sorry. When I said "Both P and Q are contained in the square in ...", I meant that every point of P and Q are in .
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 includes the entire line segment (0, y) for y between -1 and 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 and let C be the line from to (1, 1), y= . Let .
Let X be the line from (-1 ,1) to (0, .5). Let Y be the set set . Let Z be the line from to (1, -1). Let . 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).
The curves I came up with are:
while you came up with:
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.
That's nice!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).
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.