# Connectedness puzzle

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• Jan 28th 2010, 02:31 AM
HallsofIvy
Connectedness puzzle
Here's a puzzle I mentioned in another thread but it might be of interest here. I came up with it many years ago based on a math article I had read.

Find two sets, P and Q satisfying the following:
1) They are both subsets of the square in $\displaystyle R^2$ defined by
$\displaystyle -1\le x\le 1$ and $\displaystyle -1\le y\le 1$

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

3) P and Q are both connected sets.

4) P and Q are disjoint.

The key is that while P and Q are required to be connected, they cannot be path-connected.
• Jan 28th 2010, 10:00 AM
Opalg
Nice problem! I didn't think that was possible at first, but then I came up with this idea. For convenience in writing the formulas, it works on the unit square $\displaystyle [0,1]\times[0,1]$ rather than on the square from –1 to 1.

Let P consist of the graph of the function $\displaystyle \textstyle y= \frac13\cos(\pi/x) + \frac13\ (0<x\leqslant1)$ together with the line segment from $\displaystyle \bigl(0,\tfrac23\bigr)$ to $\displaystyle (0,1)$, and let Q consist of the graph of the function $\displaystyle \textstyle y= \frac13\cos(\pi/x) + \frac23\ (0<x\leqslant1)$ together with the line segments from $\displaystyle (0,0)$ to $\displaystyle \bigl(0,\tfrac13\bigr)$ and from $\displaystyle \bigl(1,\tfrac13\bigr)$ to $\displaystyle (1,1)$. Then P and Q are disjoint, connected (but not path-connected), P contains (0,1) and (1,0), and Q contains (0,0) and (1,1).
• Jan 31st 2010, 01:29 PM
HallsofIvy
Basically, yes, though my solution uses sin(1/x) rather than cosine but is basically the same as yours. I'm impressed! I am sure I spent much longer working it out than you did! (But not surprised. You may have noticed that I seldom respond to any question you have already answered- you have already said it much better than I could!)

Let A be the straight line from (-1, 1) to (0, 1/2), let $\displaystyle B= \{(x,y)|0< x\le \frac{1}{\pi}, y= .9 sin(1/x)- .05\}$ (the "-.05" is to shift it downward slightly and the ".9" is to make sure that, even shifted, it stays above y= -1), and let C be the line from $\displaystyle (\frac{1}{\pi},-.05)$ to (1, -1). Let $\displaystyle P= A\cup B\cup C$.

Let X be the straight line from (-1, -1) to (0, -1/2), let $\displaystyle Y= \{(x, y)|0< x\le \frac{1}{\pi}, y= .9 sin(1/x)+ 0.5\}$ (the "0.5} is to shift it upward slightly and the ".9" is to make sure that, even shifted, it stays below y= 1), and let Z be the line from $\displaystyle (\frac{1}{\pi}, .05)$ to (1, 1). Let $\displaystyle Q= X\cup Y\cup Z$.

It is easy to see that P and Q satisfy (1) and (2) of the requirements. It is also easy to see that they satisfy (4). For every x, the corresponding y of one set is different from the corresponding y in the other set.

Each is clearly made up of three connected sets and two of those sets overlap at $\displaystyle (\frac{1}{\pi}, -.05)$ and $\displaystyle (\frac{1}{\pi}, .05)$ for P and Q, respectively. Finally, the closures of B and Y include an interval of the y axis that intersects A and X.