Wimpy Banach-Tarski paradox - should it break my intuition?
in [Math]
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From: Frederick Williams on 17 Feb 2010 12:47 "David C. Ullrich" wrote: > The fact that it's impossible to define the volume of a subset of R^3 > in such a way that > > (i) volume(E) is defined for _every_ set E in R^3 > (ii) if you obtain F from E by a rigid motion then volume(F) = > volume(E) > (iii) if E and F are disjoint then the volume of the union is the sum > of the volumes. Suppose one considers "volumes" of sets in R^n instead of R^3. For which n do (i,ii,iii) hold? -- .... A lamprophyre containing small phenocrysts of olivine and augite, and usually also biotite or an amphibole, in a glassy groundmass containing analcime.
From: A on 17 Feb 2010 13:06 On Feb 17, 12:47 pm, Frederick Williams <frederick.willia...(a)tesco.net> wrote: > "David C. Ullrich" wrote: > > The fact that it's impossible to define the volume of a subset of R^3 > > in such a way that > > > (i) volume(E) is defined for _every_ set E in R^3 > > (ii) if you obtain F from E by a rigid motion then volume(F) = > > volume(E) > > (iii) if E and F are disjoint then the volume of the union is the sum > > of the volumes. > > Suppose one considers "volumes" of sets in R^n instead of R^3. For > which n do (i,ii,iii) hold? > > -- > ... A lamprophyre containing small phenocrysts of olivine and > augite, and usually also biotite or an amphibole, in a glassy > groundmass containing analcime. The only notion of volume for subsets of R^n satisfying those three conditions, for any positive n at all, is the one in which all subsets of R^n have zero volume. Suppose n is positive; then the volume of a single-point set {x} in R^n is equal to the volume of any other single- point set {y} in R^n, by condition ii above. So the volume of a single- point set is either nonzero, in which case the volume of any infinite set of points is infinite, hence volume fails to be defined for infinite subsets of R^n, contradicting condition i above; or the volume of a single-point set is zero, in which case, by condition iii above, the volume of every subset of R^n (being the union of its single-point subsets) is also zero. (Conditions i and iii above are too strong to give us a good notion of volume.) For anyone who is following this thread and who is confused or surprised by it--this stuff is the subject of measure theory, which is essential to the theory of integration, so every graduate student (and many undergraduates) in mathematics is required to take a course in measure theory that covers these issues with defining the notion of volume, as well as the construction of a reasonable, well-behaved notion of volume for MANY (but not ALL) subsets of R^n, and this is the notion of "volume" that is used in rigorously constructing the integral. Try Googling "Lebesgue measure" for more about this.
From: A on 17 Feb 2010 13:08 On Feb 17, 1:06 pm, A <anonymous.rubbert...(a)yahoo.com> wrote: > On Feb 17, 12:47 pm, Frederick Williams > > > > > > <frederick.willia...(a)tesco.net> wrote: > > "David C. Ullrich" wrote: > > > The fact that it's impossible to define the volume of a subset of R^3 > > > in such a way that > > > > (i) volume(E) is defined for _every_ set E in R^3 > > > (ii) if you obtain F from E by a rigid motion then volume(F) = > > > volume(E) > > > (iii) if E and F are disjoint then the volume of the union is the sum > > > of the volumes. > > > Suppose one considers "volumes" of sets in R^n instead of R^3. For > > which n do (i,ii,iii) hold? > > > -- > > ... A lamprophyre containing small phenocrysts of olivine and > > augite, and usually also biotite or an amphibole, in a glassy > > groundmass containing analcime. > > The only notion of volume for subsets of R^n satisfying those three > conditions, for any positive n at all, is the one in which all subsets > of R^n have zero volume. Suppose n is positive; then the volume of a > single-point set {x} in R^n is equal to the volume of any other single- > point set {y} in R^n, by condition ii above. So the volume of a single- > point set is either nonzero, in which case the volume of any infinite > set of points is infinite, hence volume fails to be defined for > infinite subsets of R^n, contradicting condition i above; or the > volume of a single-point set is zero, in which case, by condition iii > above, the volume of every subset of R^n (being the union of its > single-point subsets) is also zero. > > (Conditions i and iii above are too strong to give us a good notion of > volume.) > > For anyone who is following this thread and who is confused or > surprised by it--this stuff is the subject of measure theory, which is > essential to the theory of integration, so every graduate student (and > many undergraduates) in mathematics is required to take a course in > measure theory that covers these issues with defining the notion of > volume, as well as the construction of a reasonable, well-behaved > notion of volume for MANY (but not ALL) subsets of R^n, and this is > the notion of "volume" that is used in rigorously constructing the > integral. Try Googling "Lebesgue measure" for more about this. Oops--I misread condition iii as saying that the volume of ANY union (possibly an infinite union) of subsets should be the sum of the volume of the subsets; the actual condition iii written above only states that the volume of a FINITE union of subsets should be the sum of the volumes of the subsets, which is a much better and more appropriate condition. So never mind the objections I made my last post.
From: David C. Ullrich on 18 Feb 2010 06:29 On Wed, 17 Feb 2010 17:47:31 +0000, Frederick Williams <frederick.williams2(a)tesco.net> wrote: >"David C. Ullrich" wrote: > >> The fact that it's impossible to define the volume of a subset of R^3 >> in such a way that >> >> (i) volume(E) is defined for _every_ set E in R^3 >> (ii) if you obtain F from E by a rigid motion then volume(F) = >> volume(E) >> (iii) if E and F are disjoint then the volume of the union is the sum >> of the volumes. > >Suppose one considers "volumes" of sets in R^n instead of R^3. For >which n do (i,ii,iii) hold? Or rather, for which n does there _exist_ a _definition_ of "volume" with these properties... I'm really not sure about this - someone who knows should reply. My impression is that there _does_ exist a finitely additive measure defined on all subsets of R^n, invariant under euclidean motions (and with a positive finite value on the cube) for n = 1 and 2. Of course BT shows it's impossible for n >= 3. The wikipedia article on BT seems to say as much: "In fact, the Banach-Tarski paradox demonstrates that it is impossible to find a finitely-additive measure (or a Banach measure) defined on all subsets of a Euclidean space of three (and greater) dimensions that is invariant with respect to Euclidean motions and takes the value one on a unit cube. In his later work, Tarski showed that, conversely, non-existence of paradoxical decompositions of this type implies the existence of a finitely-additive invariant measure."
From: Rupert on 18 Feb 2010 07:05
On Feb 18, 10:29 pm, David C. Ullrich <ullr...(a)math.okstate.edu> wrote: > On Wed, 17 Feb 2010 17:47:31 +0000, Frederick Williams > > <frederick.willia...(a)tesco.net> wrote: > >"David C. Ullrich" wrote: > > >> The fact that it's impossible to define the volume of a subset of R^3 > >> in such a way that > > >> (i) volume(E) is defined for _every_ set E in R^3 > >> (ii) if you obtain F from E by a rigid motion then volume(F) = > >> volume(E) > >> (iii) if E and F are disjoint then the volume of the union is the sum > >> of the volumes. > > >Suppose one considers "volumes" of sets in R^n instead of R^3. For > >which n do (i,ii,iii) hold? > > Or rather, for which n does there _exist_ a _definition_ of "volume" > with these properties... > > I'm really not sure about this - someone who knows should reply. > My impression is that there _does_ exist a finitely additive > measure defined on all subsets of R^n, invariant under euclidean > motions (and with a positive finite value on the cube) for n = 1 and > 2. Of course BT shows it's impossible for n >= 3. > Yes, this is correct. It is all discussed in Stan Wagon's book. It arises from the fact that the isometry groups of R^1 and R^2 are solvable, but the isometry group of R^n when n>=3 has a free subgroup of rank two. > The wikipedia article on BT seems to say as much: > > "In fact, the Banach-Tarski paradox demonstrates that it is impossible > to find a finitely-additive measure (or a Banach measure) defined on > all subsets of a Euclidean space of three (and greater) dimensions > that is invariant with respect to Euclidean motions and takes the > value one on a unit cube. In his later work, Tarski showed that, > conversely, non-existence of paradoxical decompositions of this type > implies the existence of a finitely-additive invariant measure." |