Quality humour [Matthew's Journal]

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A little more visual humour here. The joke is my fault, but the artistry is by

covertmusic :-)

Joke by emperor, art by covertmusic

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There are 14 comments on this entry. (Reply.)

Obviously this is only true for a Platonic lemon, real lemons aren't solid so Bnack & Tarski's trick won't work.

Otherwise I'd have perfected this for lemon and moved on to the gin and tonic components.

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Mmmm gin.

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I thought that. But *fermions* are, right? An individual electron wave function is isotropic, so (I hear) you should be able to do the BT trick (assuming you have a instantaneously-teleport-points-of-space-o-mat). So shouldn't you be able to copy each particle in the lemon onto a a particle in each of the others? That's glossing over a few details, but it sounds good, right?

(OK, you can't actually cut stuff up like that, but then you wouldn't be able to do it for a platonic lemon either.)

(Did I out-pedant Feynman?)

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Lemons do not good Gin & Tonics make- take the thought from your mind. Limes, I say, Limes, young aardvark, that is the way of g&t.

dagonet

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I was going to quibble with the maths here (shouldn't two lemmas be more proof than one, not less?) but decided that that really wouldn't make any sense.

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That's not the joke.

Zorn's lemma is equivalent to the Axiom of Choice which is equivalent to the Banach-Tarski paradox which says you can decompose a sphere into two equivalent spheres of the same volume.

Or, indeed, lemons.

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Anyway, not my joke, I just did the graphics. :)

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AC is well known to imply BT, but are you sure they're actually equivalent (i.e. that BT implies AC as well)?

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It scarcely sounds likely, does it? I don't see how you could even begin.

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AC is part of the proof of BT.

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That's not the question...

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It sucks, doesn't it? "What's yellow and implied by the axiom of choice? THE BANANACH-TARSKI PARADOX" is nowhere near as funny as ""What's yellow and equivalent to the axiom of choice? THE BANANACH-TARSKI PARADOX", but isn't true. And the same applies to the "Cardinine trichotomy theorem". I support just saying it anyway. (After all, only weirdos use enough choice to get b-t but no more, I guess? :))

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*thinks* I assume aleph-C choice is used in the proof, so ZF set theory, plus axiom of choice for set of real-line cardinality, but not choice for any higher-order sets, would let you prove B-T. If so, it *can't* imply complete AC, right?

Does it imply choice for aleph-C sets? If you take B-T to be "given two shapes with certain constraints in R^3, there exists a 1-1 mapping between the points in them"?

The obvious way to consider it is, assume B-T. Take an infinite set of pairs. Does that give you a choice function? I bet there isn't, because B-T doesn't seem to give a way to "pick an answer" that AC and Well-ordering do. (And any sort of embedding of the problem set in R^3 pretty much solves the question before you start and hence is impossible, as once it's embedded, you can choose things by choosing the one nearest the origin, etc)

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Sorry, I was probably a bit out of it, I should have explained more clearly. I'm well steeped in large number of axiom-of-choice–pun–variants. I thought this was very cute.

But something in me niggled at the details. "Zorn's Lemma => Banach Tarski", but "Zorn's Lemon (cut and reassemble) --> Two lemons" so there's a pun. But it feels like there ought to be a disproof at the end, but B-T isn't literally false, and the mathematical equivalent to two Zorn's Lemons is two Zorn's lemmas, and duplicating a mathematical statement is automatic and if anything reaffirms it... I knew I couldn't explain :)

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Quality humour

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October