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Math Help - Connectedness puzzle

  1. #1
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    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 R^2 defined by
    -1\le x\le 1 and -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.
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  2. #2
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    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 [0,1]\times[0,1] rather than on the square from 1 to 1.

    Let P consist of the graph of the function \textstyle y= \frac13\cos(\pi/x) + \frac13\ (0<x\leqslant1) together with the line segment from \bigl(0,\tfrac23\bigr) to (0,1), and let Q consist of the graph of the function \textstyle y= \frac13\cos(\pi/x) + \frac23\ (0<x\leqslant1) together with the line segments from (0,0) to \bigl(0,\tfrac13\bigr) and from \bigl(1,\tfrac13\bigr) to (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).
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  3. #3
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    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 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 (\frac{1}{\pi},-.05) to (1, -1). Let P= A\cup B\cup C.

    Let X be the straight line from (-1, -1) to (0, -1/2), let 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 (\frac{1}{\pi}, .05) to (1, 1). Let 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 (\frac{1}{\pi}, -.05) and (\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.
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