Connectedness of parts used in the Banach–Tarski paradox












11












$begingroup$


A quote from the Wikipedia article "Axiom of choice":




One example is the Banach–Tarski paradox which says that it is
possible to decompose the 3-dimensional solid unit ball into finitely
many pieces and, using only rotations and translations, reassemble the
pieces into two solid balls each with the same volume as the original.




I know that at least some of the parts (called pieces here) must be non-measurable sets. I wonder if each of them can be chosen to be a path-connected set (otherwise it's really misleading to call them pieces, I think).






measure-theory set-theory axiom-of-choice connectedness paradoxes







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    If I recall correctly, one of these pieces is a singleton.
    $endgroup$
    – Asaf Karagila♦
    May 29 '14 at 23:43










  • $begingroup$
    @AsafKaragila A singleton is a path-connected set, right? Just use a constant function that maps $[0,1]$ to its only point. My question is can we make each part to be path-connected.
    $endgroup$
    – Vladimir Reshetnikov
    May 30 '14 at 17:11






  • 1




    $begingroup$
    Yes, I know, which is why I didn't post an answer. I simply don't know the answer. It sounds a bit unlikely, though. In any case, historically Banach-Tarski didn't specify how many pieces; von Neumann claimed he can do it in $9$, and some time later Sierpinski claimed he can do it in $8$; and only some more time later Robinson proved that you can do it in $5$, but you can't do it in $4$. (All this appears in Halbeisen's Combinatorial Set Theory which includes a chapter on the Banach-Tarski paradox.)
    $endgroup$
    – Asaf Karagila♦
    May 30 '14 at 17:13










  • $begingroup$
    There was an exposition of this result in American Mathematical Monthly. I can't recall the title or year.
    $endgroup$
    – DanielWainfleet
    Feb 22 '17 at 4:09
















11












$begingroup$


A quote from the Wikipedia article "Axiom of choice":




One example is the Banach–Tarski paradox which says that it is
possible to decompose the 3-dimensional solid unit ball into finitely
many pieces and, using only rotations and translations, reassemble the
pieces into two solid balls each with the same volume as the original.




I know that at least some of the parts (called pieces here) must be non-measurable sets. I wonder if each of them can be chosen to be a path-connected set (otherwise it's really misleading to call them pieces, I think).






measure-theory set-theory axiom-of-choice connectedness paradoxes







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    If I recall correctly, one of these pieces is a singleton.
    $endgroup$
    – Asaf Karagila♦
    May 29 '14 at 23:43










  • $begingroup$
    @AsafKaragila A singleton is a path-connected set, right? Just use a constant function that maps $[0,1]$ to its only point. My question is can we make each part to be path-connected.
    $endgroup$
    – Vladimir Reshetnikov
    May 30 '14 at 17:11






  • 1




    $begingroup$
    Yes, I know, which is why I didn't post an answer. I simply don't know the answer. It sounds a bit unlikely, though. In any case, historically Banach-Tarski didn't specify how many pieces; von Neumann claimed he can do it in $9$, and some time later Sierpinski claimed he can do it in $8$; and only some more time later Robinson proved that you can do it in $5$, but you can't do it in $4$. (All this appears in Halbeisen's Combinatorial Set Theory which includes a chapter on the Banach-Tarski paradox.)
    $endgroup$
    – Asaf Karagila♦
    May 30 '14 at 17:13










  • $begingroup$
    There was an exposition of this result in American Mathematical Monthly. I can't recall the title or year.
    $endgroup$
    – DanielWainfleet
    Feb 22 '17 at 4:09














11












11








11


2



$begingroup$


A quote from the Wikipedia article "Axiom of choice":




One example is the Banach–Tarski paradox which says that it is
possible to decompose the 3-dimensional solid unit ball into finitely
many pieces and, using only rotations and translations, reassemble the
pieces into two solid balls each with the same volume as the original.




I know that at least some of the parts (called pieces here) must be non-measurable sets. I wonder if each of them can be chosen to be a path-connected set (otherwise it's really misleading to call them pieces, I think).






measure-theory set-theory axiom-of-choice connectedness paradoxes







share|cite|improve this question









$endgroup$




A quote from the Wikipedia article "Axiom of choice":




One example is the Banach–Tarski paradox which says that it is
possible to decompose the 3-dimensional solid unit ball into finitely
many pieces and, using only rotations and translations, reassemble the
pieces into two solid balls each with the same volume as the original.




I know that at least some of the parts (called pieces here) must be non-measurable sets. I wonder if each of them can be chosen to be a path-connected set (otherwise it's really misleading to call them pieces, I think).






measure-theory set-theory axiom-of-choice connectedness paradoxes




measure-theory set-theory axiom-of-choice connectedness paradoxes






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked May 29 '14 at 23:40









Vladimir ReshetnikovVladimir Reshetnikov

24.3k4120232




24.3k4120232








  • 3




    $begingroup$
    If I recall correctly, one of these pieces is a singleton.
    $endgroup$
    – Asaf Karagila♦
    May 29 '14 at 23:43










  • $begingroup$
    @AsafKaragila A singleton is a path-connected set, right? Just use a constant function that maps $[0,1]$ to its only point. My question is can we make each part to be path-connected.
    $endgroup$
    – Vladimir Reshetnikov
    May 30 '14 at 17:11






  • 1




    $begingroup$
    Yes, I know, which is why I didn't post an answer. I simply don't know the answer. It sounds a bit unlikely, though. In any case, historically Banach-Tarski didn't specify how many pieces; von Neumann claimed he can do it in $9$, and some time later Sierpinski claimed he can do it in $8$; and only some more time later Robinson proved that you can do it in $5$, but you can't do it in $4$. (All this appears in Halbeisen's Combinatorial Set Theory which includes a chapter on the Banach-Tarski paradox.)
    $endgroup$
    – Asaf Karagila♦
    May 30 '14 at 17:13










  • $begingroup$
    There was an exposition of this result in American Mathematical Monthly. I can't recall the title or year.
    $endgroup$
    – DanielWainfleet
    Feb 22 '17 at 4:09














  • 3




    $begingroup$
    If I recall correctly, one of these pieces is a singleton.
    $endgroup$
    – Asaf Karagila♦
    May 29 '14 at 23:43










  • $begingroup$
    @AsafKaragila A singleton is a path-connected set, right? Just use a constant function that maps $[0,1]$ to its only point. My question is can we make each part to be path-connected.
    $endgroup$
    – Vladimir Reshetnikov
    May 30 '14 at 17:11






  • 1




    $begingroup$
    Yes, I know, which is why I didn't post an answer. I simply don't know the answer. It sounds a bit unlikely, though. In any case, historically Banach-Tarski didn't specify how many pieces; von Neumann claimed he can do it in $9$, and some time later Sierpinski claimed he can do it in $8$; and only some more time later Robinson proved that you can do it in $5$, but you can't do it in $4$. (All this appears in Halbeisen's Combinatorial Set Theory which includes a chapter on the Banach-Tarski paradox.)
    $endgroup$
    – Asaf Karagila♦
    May 30 '14 at 17:13










  • $begingroup$
    There was an exposition of this result in American Mathematical Monthly. I can't recall the title or year.
    $endgroup$
    – DanielWainfleet
    Feb 22 '17 at 4:09








3




3




$begingroup$
If I recall correctly, one of these pieces is a singleton.
$endgroup$
– Asaf Karagila♦
May 29 '14 at 23:43




$begingroup$
If I recall correctly, one of these pieces is a singleton.
$endgroup$
– Asaf Karagila♦
May 29 '14 at 23:43












$begingroup$
@AsafKaragila A singleton is a path-connected set, right? Just use a constant function that maps $[0,1]$ to its only point. My question is can we make each part to be path-connected.
$endgroup$
– Vladimir Reshetnikov
May 30 '14 at 17:11




$begingroup$
@AsafKaragila A singleton is a path-connected set, right? Just use a constant function that maps $[0,1]$ to its only point. My question is can we make each part to be path-connected.
$endgroup$
– Vladimir Reshetnikov
May 30 '14 at 17:11




1




1




$begingroup$
Yes, I know, which is why I didn't post an answer. I simply don't know the answer. It sounds a bit unlikely, though. In any case, historically Banach-Tarski didn't specify how many pieces; von Neumann claimed he can do it in $9$, and some time later Sierpinski claimed he can do it in $8$; and only some more time later Robinson proved that you can do it in $5$, but you can't do it in $4$. (All this appears in Halbeisen's Combinatorial Set Theory which includes a chapter on the Banach-Tarski paradox.)
$endgroup$
– Asaf Karagila♦
May 30 '14 at 17:13




$begingroup$
Yes, I know, which is why I didn't post an answer. I simply don't know the answer. It sounds a bit unlikely, though. In any case, historically Banach-Tarski didn't specify how many pieces; von Neumann claimed he can do it in $9$, and some time later Sierpinski claimed he can do it in $8$; and only some more time later Robinson proved that you can do it in $5$, but you can't do it in $4$. (All this appears in Halbeisen's Combinatorial Set Theory which includes a chapter on the Banach-Tarski paradox.)
$endgroup$
– Asaf Karagila♦
May 30 '14 at 17:13












$begingroup$
There was an exposition of this result in American Mathematical Monthly. I can't recall the title or year.
$endgroup$
– DanielWainfleet
Feb 22 '17 at 4:09




$begingroup$
There was an exposition of this result in American Mathematical Monthly. I can't recall the title or year.
$endgroup$
– DanielWainfleet
Feb 22 '17 at 4:09










1 Answer
1






active

oldest

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6












$begingroup$

Dekker and de Groot proved (Decompositions of a sphere, Fund. Math. 43 (1956), 185–194) that the pieces can be made to be totally imperfect, and hence connected and locally connected. I don't know about path connected, though.



Although you didn't ask about this, one can also argue that "decompose and reassemble" implicitly suggests that the rearrangement can take place using continuous motions during which the pieces never intersect. That this is possible was proved by Trevor Wilson
(A continuous movement version of the Banach–Tarski paradox: A solution to de Groot's problem, J. Symb. Logic 70 (2005), 946–952). But I don't know if there is a common generalization of the Dekker–de Groot result above and Wilson's result.






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    6












    $begingroup$

    Dekker and de Groot proved (Decompositions of a sphere, Fund. Math. 43 (1956), 185–194) that the pieces can be made to be totally imperfect, and hence connected and locally connected. I don't know about path connected, though.



    Although you didn't ask about this, one can also argue that "decompose and reassemble" implicitly suggests that the rearrangement can take place using continuous motions during which the pieces never intersect. That this is possible was proved by Trevor Wilson
    (A continuous movement version of the Banach–Tarski paradox: A solution to de Groot's problem, J. Symb. Logic 70 (2005), 946–952). But I don't know if there is a common generalization of the Dekker–de Groot result above and Wilson's result.






    share|cite|improve this answer









    $endgroup$


















      6












      $begingroup$

      Dekker and de Groot proved (Decompositions of a sphere, Fund. Math. 43 (1956), 185–194) that the pieces can be made to be totally imperfect, and hence connected and locally connected. I don't know about path connected, though.



      Although you didn't ask about this, one can also argue that "decompose and reassemble" implicitly suggests that the rearrangement can take place using continuous motions during which the pieces never intersect. That this is possible was proved by Trevor Wilson
      (A continuous movement version of the Banach–Tarski paradox: A solution to de Groot's problem, J. Symb. Logic 70 (2005), 946–952). But I don't know if there is a common generalization of the Dekker–de Groot result above and Wilson's result.






      share|cite|improve this answer









      $endgroup$
















        6












        6








        6





        $begingroup$

        Dekker and de Groot proved (Decompositions of a sphere, Fund. Math. 43 (1956), 185–194) that the pieces can be made to be totally imperfect, and hence connected and locally connected. I don't know about path connected, though.



        Although you didn't ask about this, one can also argue that "decompose and reassemble" implicitly suggests that the rearrangement can take place using continuous motions during which the pieces never intersect. That this is possible was proved by Trevor Wilson
        (A continuous movement version of the Banach–Tarski paradox: A solution to de Groot's problem, J. Symb. Logic 70 (2005), 946–952). But I don't know if there is a common generalization of the Dekker–de Groot result above and Wilson's result.






        share|cite|improve this answer









        $endgroup$



        Dekker and de Groot proved (Decompositions of a sphere, Fund. Math. 43 (1956), 185–194) that the pieces can be made to be totally imperfect, and hence connected and locally connected. I don't know about path connected, though.



        Although you didn't ask about this, one can also argue that "decompose and reassemble" implicitly suggests that the rearrangement can take place using continuous motions during which the pieces never intersect. That this is possible was proved by Trevor Wilson
        (A continuous movement version of the Banach–Tarski paradox: A solution to de Groot's problem, J. Symb. Logic 70 (2005), 946–952). But I don't know if there is a common generalization of the Dekker–de Groot result above and Wilson's result.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 9 '18 at 0:53









        Timothy ChowTimothy Chow

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