How to fit an elephant (2011) 50 points by Kristine1975 9 months ago | hide | past | web | favorite | 23 comments

 I can't help but feel like a complex number is two parameters (real&imag / mod&arg) - so really this is 8 parameters.
 This is a good point. If we're posing Occam's Razor-like arguments against models for having "too many free parameters", then we should probably really be comparing something more precise, like the models' Kolmogorov complexities.Otherwise, it's just too easy to hide a lot of complex machinery inside a "single parameter". In fact, from this perspective it's arguable that an arbitrary real number is actually a (countably) infinite set of parameters, since it takes that many bits to uniquely specify any real number.
 Cantor showed that these sets have the same cardinality. You can represent a complex (in the form of two reals) by interleaving digits or using a space-filling curve for example.
 > Cantor showed that these sets have the same cardinality. You can represent a complex (in the form of two reals) by interleaving digits or using a space-filling curve for example.But this (set) isomorphism between R and C is not continuous. Indeed one can show that there exists no continuous epimorphism f: R^n -> R^m, where m > n, since for every such continuous map f the image f(R^n) has a measure of 0 with respect to the Borel measure in R^m.
 > there exists no continuous epimorphism f: R^n -> R^m, where m > nReally? Then what is a https://en.wikipedia.org/wiki/Space-filling_curve?
 Right from that article you linked: "A non-self-intersecting continuous curve cannot fill the unit square because that will make the curve a homeomorphism from the unit interval onto the unit square (any continuous bijection from a compact space onto a Hausdorff space is a homeomorphism). But a unit square has no cut-point, and so cannot be homeomorphic to the unit interval, in which all points except the endpoints are cut-points."
 Yeah, a non-self-intersecting map cannot, but OP didn't specify that, only "epimorphic" which certainly can.Moreover OP's argument specifically proves too much, because space-filling curves (as described in the article) have a range with positive Borel measure.
 Sure, but did we need continuity? Also, if you want to be awkward, you can get around this by using the discrete topology, I don't think we needed the metric structure of R^n.
 > Sure, but did we need continuity? Also, if you want to be awkward, you can get around this by using the discrete topologyThis is indeed possible - but this is clearly not the topology that "ordinary people" and physicists mean when talking about continuity of functions from R^n to R^m.
 So, it's all actually just one parameter?
 Pairing functions only work on countable sets. This is funny because Cantor is the same person who proved real numbers are uncountable, and that there is no pairing function between 1 real number and naturals, let alone 2.
 I'm claiming a "pairing function" between single reals and pairs of reals. They have respective cardinalities 2^N0 and 2*2^N0=2^N0 where N0<2^N0 is the cardinality of the naturals.
 You're right. I was wrong. I found an explanation of how to make interleaving method work. https://math.stackexchange.com/a/183383
 He could have used an octonion, then it was 1 parameter.
 The only thing missing here is a little anecdote about how Von Neumann, when challenged on this, did it in his head and started rattling off the parameters. For arbitrary animals.
 Well, if you're familiar with sums, then the "shortcut" isn't all that much faster; it's mostly a function of how quickly you can grok the salient points:The fly travels 3/2 times the speed of a train, so every bounce the fly travels 3/5 of the remaining track and leaves 1/2 * 2/5 = 1/5 track to travel, so we just compute the geometric sum 3/5 * \sum 1/5^r = 3/5 * 1/(1 - 1/5) = 3/5 * 1/(4/5) = 3/5 * 5/4 = 3/4.  One nice thing about this sum is that it encodes a bit more insight about the fly's flight path than the shortcut method.