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Math Help - A topological puzzle

  1. #1
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    A topological puzzle

    Find two sets, P and Q, such that:

    1) Both P and Q are contained within the square in R^2 with vertices (1, 1), (1, -1), (-1, -1) and (-1, 1). (In other words all (x, y) in P or Q must satisfy -1\le x\le 1, -1\le y\le 1.)

    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) P\cap Q is empty. (P and Q have no point in common.)
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    Senior Member MacstersUndead's Avatar
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    Re: A topological puzzle

    Spoiler:
    Must the sets P and Q lie on the same surface of the plane R^2? If not then draw a straight line between (1,1) and (-1,-1) = P and a path that loops underneath from (-1,1) and (1,-1) = Q

    Can I bend the space? If so, bend the plane R^2 so that (1,1) and (-1,-1) become the same point. Let (1,1) be P and let the line between (-1,1) and (1,-1) be Q
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    Forum Admin topsquark's Avatar
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    Re: A topological puzzle

    Now that's the coolest thing I've run across for a while.

    -Dan
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    Re: A topological puzzle

    No, sorry. When I said "Both P and Q are contained in the square in R^2 ...", I meant that every point of P and Q are in R^2.

    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 \{(x, y)| y= sin(1/x), x>0\} includes the entire line segment (0, y) for y between -1 and 1.
    Last edited by HallsofIvy; March 8th 2013 at 08:06 AM.
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    Forum Admin topsquark's Avatar
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    Re: A topological puzzle


    Well it's still cool.

    -Dan
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    Super Member ILikeSerena's Avatar
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    Re: A topological puzzle

    Quote Originally Posted by HallsofIvy View Post
    No, sorry. When I said "Both P and Q are contained in the square in R^2 ...", I meant that every point of P and Q are in R^2.

    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 \{(x, y)| y= sin(1/x), x>0\} includes the entire line segment (0, y) for y between -1 and 1.
    I finally got it!

    According to mathworld:
    A connected set is a set which cannot be partitioned into two nonempty subsets such that each subset has no points in common with the set closure of the other.

    Spoiler:
    The sets

    P = \{(-1,-1)\} ~\cup~ \{(x, y)~|~y= \frac 1 2 (1+x) + \frac 1 2 |1-x| \sin(\frac {\pi}{x+1}), -1 < x\le 1\}

    and

    Q = \{(-1,1)\} ~\cup~ \{(x, y)~|~y= - \frac 1 2 (1+x) + \frac 1 2 |1-x| \sin(\frac {\pi}{x+1}), -1 < x\le 1\}

    do the trick.

    They are derived from the topologist's sine curve.

    Here's Wolfram's plot.
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    Re: A topological puzzle

    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 \{ (x, y)| y= .95 sin(1/x)+ .01, 0\le x\le \frac{1}{\pi}\} and let C be the line from (1/\pi, .01) to (1, 1), y= . Let P= A\cup B\cup C.

    Let X be the line from (-1 ,1) to (0, .5). Let Y be the set set \{(x, y)| y= .95 sin(1/x)- .01, 0\le x\le \frac{1}{\pi}\}. Let Z be the line from (1/\pi, -.01) to (1, -1). Let Q= X\cup Y\cup Z. 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).
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    Re: A topological puzzle

    Quote Originally Posted by topsquark View Post

    Well it's still cool.

    -Dan
    Yes, it is!
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  9. #9
    Super Member ILikeSerena's Avatar
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    Re: A topological puzzle

    Quote Originally Posted by HallsofIvy View Post
    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 \{ (x, y)| y= .95 sin(1/x)+ .01, 0\le x\le \frac{1}{\pi}\} and let C be the line from (1/\pi, .01) to (1, 1), y= . Let P= A\cup B\cup C.

    Let X be the line from (-1 ,1) to (0, .5). Let Y be the set set \{(x, y)| y= .95 sin(1/x)- .01, 0\le x\le \frac{1}{\pi}\}. Let Z be the line from (1/\pi, -.01) to (1, -1). Let Q= X\cup Y\cup Z. 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.
    Hmm, let's see.
    The curves I came up with are:


    Wolfram|Alpha

    while you came up with:


    Wolfram|Alpha

    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.


    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).
    That's nice!
    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.
    Attached Thumbnails Attached Thumbnails A topological puzzle-crossing_topologist_curves_ils.png   A topological puzzle-crossing_topologist_curves_hoi.png  
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    Re: A topological puzzle

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