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Question about the Banach–Tarski paradox.

  1. Aug 7, 2011 #1
    I was reading a little bit about the Banach–Tarski theorem. Is this similar to a line segment of length 1 having the same points as a square with side lengths of 1. And then also a cube with sides of length 1. So then I should be able to take a square and pick out all the points and construct a cube with the same side length. And I should be able to construct as many cubes as I want from that square just by picking out points and constructing my cube. Is this related to the Banach–Tarski theorem or am I crazy.
  2. jcsd
  3. Aug 7, 2011 #2
    Hi cragar! :smile:

    The Banach-Tarski paradox is a little bit stronger. When you have a line, you can indeed form a square out of this line. However, what we do there is we take every point separately and map it to a point in the square. So we have to cut our line into an infinite number of pieces and then reassemble it.
    Banach-Tarski is a lot stronger: it says we can cut our sphere into [itex]\mathbf{5}[/itex] pieces, and then reassemble it to form two balls. You can't do this with the line: you can't take 5 line pieces and reassemble it to form a square!! But you can do it with a ball.
  4. Aug 7, 2011 #3
    so we can do it with a ball but not a line or square.
  5. Aug 7, 2011 #4
    Indeed, it has been proven that we can't do this in one or two dimensions.
  6. Aug 7, 2011 #5
    thanks for your answers by the way. Ok lets say I have a sphere of radius 1, can i view the points as infinitesimally small cubes? Then from these cubes I could construct 2 other spheres of radius 1. You said that the theorem cuts the sphere into 5 parts. Why cant I just say when I pick my little cubes from the sphere that I do it 2, 3, or how ever many ways and I put these cubes in a box. So I have an infinite amount of cubes in each box and I might have 3 boxes, then I use these 3 boxes to construct 2 other spheres of radius 1, And the boxes represent my finite area partitions of the original sphere.
  7. Aug 7, 2011 #6


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    The two key qualitative facts of the Banach-Tarski paradox are:
    • The motions are simple -- it uses Euclidean translations and rotations (volume-preserving operations) on finitely many objects
    • The sets involved are so "complicated" that the notion of volume doesn't have any meaning for them (they're called non-measurable sets)

    (aside: there are lots of "measures" -- notions like "how many", "length", "area", and "volume" are all different sorts of measures)

    The first point is rather important -- without it (or something similar), there's no reason to believe that such an argument would preserve measure. As you point out, it's a rather simple matter to take the points of a line and rearrange them into a square -- but the way you do it gives us no reason to think that it should preserve measure*

    Previous pseudo-paradoxes that properly use measure-preserving transformations had other factors against them that make it easy for people to mentally brush off the use of non-measurable sets and simply ascribe any poor behavior of measure to the ways in which the argument is complicated.

    The Banach-Tarski (pseudo-)paradox is significant because there is pretty much no room to rationalize things away -- it really does a good job of forcing people to acknowledge non-measurable sets and just how badly the idea of measure behaves in their presence.

    (Of course, this acknowledgement leads some people to adopt versions of set theory in which non-measurable sets don't exist)

    *: well, we have reason to believe the counting measure is preserved, and it is. ([itex]+\infty[/itex] for both a line and for a square)
  8. Aug 7, 2011 #7
    I guess I need to read more about the theorem and measure.
  9. Aug 7, 2011 #8
    Maybe read my blog post about it: https://www.physicsforums.com/blog.php?b=2993 [Broken]
    It might help...
    Last edited by a moderator: May 5, 2017
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