HallsofIvy said:
Find two sets P and Q satisfying
I) P and Q are completely contained in the square, in R2, with vertices (1, 1), (1, -1), (-1, -1), and (-1, 1).
II) P contains the ponts (1, 1) and (-1, -1) while Q contains (-1, 1) and (1, -1).
III) P and Q are disjoint.
IV) P and Q are both connected sets.
P and Q cannot be path-connected, because any continuous path from (1, 1) to (-1, -1) would have to intersect one from (-1, 1) to (1, -1). So the problem has to be about the topological definition of connectedness rather than the geometric notion of path-connectedness, and we must look for sets that are connected but not path-connected.
[sp]Let $X = [-1,1]\times[-1,1]$, and define subsets of $X$ by $$UR = \{(x,y)\in X : x>0,\ y > \tfrac12\sin\tfrac1x\},$$ $$LR = \{(x,y)\in X : x>0,\ y < \tfrac12\sin\tfrac1x\},$$ $$UL = \{(x,y)\in X : x<0,\ y > \tfrac12\sin\tfrac1x\},$$ $$LL = \{(x,y)\in X : x<0,\ y < \tfrac12\sin\tfrac1x\},$$ $$UA = \{(x,y)\in X : x=0,\ y > 0\},$$ $$LA = \{(x,y)\in X : x=0,\ y < 0\},$$ (the names of the sets are meant to indicate Upper Right, Lower Left, etc., and A denotes $y$-Axis). These sets are all disjoint, and each of them is connected.
Let $P = UR\cup LL\cup UA$, $Q = LR\cup UL\cup LA.$ Then properties I), II), III) certainly hold. To see that IV) also holds, suppose that $U$ and $V$ are disjoint open sets with $P\subset U\cup V$. Since the three component parts of $P$ are connected, each of them must lie entirely within one of the sets $U$, $V$. In particular, $UA\subset U$ say. Then $U$, being open, must contain a neighbourhood extending each side of the positive $y$-axis and therefore contains points in both $UR$ and $LL$. But those sets are both connected, and it follows that $U$ must contain the whole of $P$, so that $V$ is disjoint from $P$. That shows that $P$ is connected; and a similar argument shows that so also is $Q$.[/sp]
