[wolfgke]

> 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]

> 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?

[dahart]

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]

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]

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]

> 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.