Wimpy Banach-Tarski paradox - should it break my intuition?
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From: Frederick Williams on 18 Feb 2010 09:34 "David C. Ullrich" wrote: > > 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. That is what I thought. Thanks. But now consider the area of a _surface_ embedded in R^3. I seem to recall that there is something that looks like a truncated "faceted" cylinder that has no area: one computes it in one obvious way and it's one value, one computes in another equally obvious way and it's some other value. Sorry that's all very vague. I wonder what the definition of surface is. Something like continuum of dimension two, I suppose. > 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." -- ..... A lamprophyre containing small phenocrysts of olivine and augite, and usually also biotite or an amphibole, in a glassy groundmass containing analcime.
From: Gerry Myerson on 18 Feb 2010 16:41 In article <4B7D5004.702F280(a)tesco.net>, Frederick Williams <frederick.williams2(a)tesco.net> wrote: > I wonder what the definition of surface > is. Something like continuum of dimension two, I suppose. A surface is a compact, connected topological space every point of which has a neighborhood homeomorphic to an open disk or a half-plane. -- Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email)
From: Frederick Williams on 18 Feb 2010 18:01 Gerry Myerson wrote: > A surface is a compact, connected topological space > every point of which has a neighborhood homeomorphic > to an open disk or a half-plane. Thank you. And given one of *those* what does it mean for it to have an area? -- ..... A lamprophyre containing small phenocrysts of olivine and augite, and usually also biotite or an amphibole, in a glassy groundmass containing analcime.
From: Frederick Williams on 18 Feb 2010 18:44 master1729 wrote: > > nonsense , thats not even an answer , only an insult. If the cap fits... -- ..... A lamprophyre containing small phenocrysts of olivine and augite, and usually also biotite or an amphibole, in a glassy groundmass containing analcime.
From: Gerry Myerson on 18 Feb 2010 21:10
In article <4B7DC6B7.C7761E3A(a)tesco.net>, Frederick Williams <frederick.williams2(a)tesco.net> wrote: > Gerry Myerson wrote: > > > A surface is a compact, connected topological space > > every point of which has a neighborhood homeomorphic > > to an open disk or a half-plane. > > Thank you. And given one of *those* what does it mean for it to have an > area? About as much as it means for the color blue to have an area. The definition I gave is purely topological, and area is not a topological concept. Now there's a theorem that says that any surface can be embedded in R^4. Once you've embedded your surface, you could use concepts of area in Euclidean spaces to define an area for it. That's about as close as this number-theorist is going to get to giving you a useful answer. Maybe someone else can take up where I've left off. -- Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email) |