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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

by Y_Y 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.

Bromskloss 9y ago
So, 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 ago
    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."
    - waqf 9y ago
      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.
- Y_Y 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, 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 ago
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. https://en.m.wikipedia.org/wiki/Countable_set https://en.m.wikipedia.org/wiki/Countable_set
- Y_Y 9y ago
  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.
  - dahart 9y ago
    You'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

by Y_Y 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.

Bromskloss 9y ago

So, 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 ago
  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."
  - waqf 9y ago
    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.
Y_Y 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, 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 ago

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. https://en.m.wikipedia.org/wiki/Countable_set https://en.m.wikipedia.org/wiki/Countable_set

Y_Y 9y ago
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.
- dahart 9y ago
  You'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