[HN Gopher] The Point of the Banach-Tarski Theorem ___________________________________________________________________ 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. ___________________________________________________________________ (page generated 2023-01-22 23:00 UTC)