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I can't help but feel like a complex number is two parameters (real&imag / mod&arg) - so really this is 8 parameters.
by oddeyed 9y ago
I can't help but feel like a complex number is two parameters (real&imag / mod&arg) - so really this is 8 parameters.
- Y_Y 9y agoCantor 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.
- Bromskloss 9y agoSo, it's all actually just one parameter?
- deleted 9y ago[deleted]
- wolfgke 9y ago> 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.
- waqf 9y ago> there exists no continuous epimorphism f: R^n -> R^m, where m > n Really? Then what is a https://en.wikipedia.org/wiki/Space-filling_curve https://en.wikipedia.org/wiki/Space-filling_curve?
- dahart 9y agoRight 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."
- waqf 9y agoYeah, 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.
- Y_Y 9y agoSure, 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.
- wolfgke 9y ago> Sure, but did we need continuity? Also, if you want to be awkward, you can get around this by using the discrete topology This 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.
- deleted 9y ago[deleted]
- dahart 9y agoPairing 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. https://en.m.wikipedia.org/wiki/Countable_set https://en.m.wikipedia.org/wiki/Countable_set
- Y_Y 9y agoI'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.
- dahart 9y agoYou're right. I was wrong. I found an explanation of how to make interleaving method work. https://math.stackexchange.com/a/183383 https://math.stackexchange.com/a/183383
- Aardwolf 9y agoHe could have used an octonion, then it was 1 parameter.
- xelxebar 9y agoThis 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.