[HN Gopher] The Point of the Banach-Tarski Theorem
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The Point of the Banach-Tarski Theorem
Author : ColinWright
Score : 30 points
Date : 2023-01-22 21:05 UTC (1 hours ago)
(HTM) web link (www.solipsys.co.uk)
(TXT) w3m dump (www.solipsys.co.uk)
| CJefferson wrote:
| I always feel part of the confusion with Banach-Tarski is that
| lots of words don't use their "natural definitions", which makes
| the proof more surprising. People (not this article) often
| talking about "cutting" a sphere, which is really misleading.
|
| This result is, in many ways, quite similar to the idea I can
| "cut" the integers into the odd integers and even integers (but
| with many more fine details).
|
| This is still a nice article, which explains the actual result
| well.
| azeemba wrote:
| As the author points out that Banach-Tarski theorem is an example
| of hard-to-accept result that comes out of the easy-to-accept
| axiom of choice.
|
| There is a popular quote that related to this:
|
| > The axiom of choice is obviously true, the well-ordering
| principle obviously false, and who can tell about Zorn's lemma?
|
| From https://en.wikipedia.org/wiki/Axiom_of_choice
|
| Axiom of choice, the well-ordering principle and Zorn's lemma are
| equivalent statements (any one proves the other two). But each
| has a very different "believability" feel to it.
| whatshisface wrote:
| Here is a potentially daft question that I nonetheless would
| appreciate if someone could answer. Is it possible to deny the
| axiom of choice for the purposes of measures while accepting it
| for vector spaces? I am wondering if you could say, "there are
| two kinds of sets, ones equipped with a choice function and ones
| without it, and measurable sets are of the latter kind."
| joerichey wrote:
| If you accept that _all_ vector spaces have a (Hamel) basis,
| you can then prove the Axiom of Choice:
| http://www.math.lsa.umich.edu/~ablass/bases-AC.pdf
|
| This means if you want to deny the Axiom in some cases, you
| will also have to allow for the existence of vector spaces
| without a basis.
| contravariant wrote:
| Not in a meaningful way I think. I mean you could weaken it to
| 'all finite vector spaces have a basis', but I think regular
| induction is enough to prove that, you don't need the axiom of
| choice.
| moloch-hai wrote:
| I asked a mathematician about what a wacky conclusion it is. He
| said that whenever you allow infinity, you get results like that.
| It relies on uncountably-infinite division of an object, which
| corresponds to no real-world experience anywhere in the universe.
| Real objects have, you know, atoms.
|
| We use real numbers a lot, but we are careful never to rely on
| their more extreme properties anywhere it would matter. In
| practice, in fact, we use floating-point numbers, not reals, when
| doing actual calculations, and use numerical analysis to stay
| well clear of nonsensical results. If you tried to rely on BT in
| a real calculation, you would find a lot of NaNs and Infs.
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