Can a continuous function from 2D to 3D be onto?

Are you ready to dive into the world of topology and analysis? Today, we explore an intriguing question: Can a continuous function from 2D to 3D be onto?

We take a deep dive into the concept of space-filling curves and the relationship between continuity and dimensionality. You'll learn about the limits and possibilities of continuous functions and discover the beauty and complexity of the mathematical world. This video is perfect for students looking to expand their understanding of mathematical concepts and deepen their appreciation for the subject. Get ready to be excited by the endless possibilities of math, and join us on this journey of discovery!

Music creds:
Home-Prod. Riddiman

[azurebrown]

You basically have explained the essence of the the seed of higher math. It took much study, for me to realize all higher math is us essentially building different types of numbers lines "onto" a space.

[AKfire]

The explanation is quite good. However, the place where you started explaining about 2D to 3D conversion, it seemed to me that it should have been explained in more detail. I couldn't comprehend what was going on there.. The animations are quite good.. especially the 3D one.

[patricius]

That's why circles can help us discovering physics
Circles give us the idea of periodic functions, because they are just a line ending in itself . Other shapes, like square, are not finite lines, thats why you cannot write them as an equation . Circle has an equation .
From the trigonometry we just barely started our way in the new game of physics, because the science of continuity, made us to discover field equations and the first steps in the the quantum world

[ffs]

A much easier example of such a mapping would be mapping two real numbers 0.(a1)(a2)(a3)... and 0.(b1)(b2)(b3)... into 0.(a1)(b1)(b2)(b2)(a3)(b3)... I know that the point was to demonstrate visuals, but the space-filling curve require limits, and it is not obvious how the things behave in the limit, but demonstrating that finite dimensions can be interlaced to create finite dimensions... a much more simple (and general!) concept. No need to collapse hyperplanes into lines etc.

[quantumgaming]

I remember watching 3blue1brown's video about this very subject (Hilbert Curves are called I believe?). I found it pretty cool as he also gave a somewhat use of them in real life, but this video it's so much better! I can't believe how cool this is.

[ilonachan]

for the 2d->3d case, couldn't you just use the 1d->2d map on one of the two input dimensions, and use the identity for the other? that'd also extrude the square into a cube without issue, right? and it gives us an almost trivial way to generalize to any case n->(n+1). Together with a generalization to any 1->n case, which you claim exists for n=3 and it _seems plausible_ that it'd extend, that allows us to just map from any dimension into any (higher) dimension in an onto way.

[thomasgrady]

3:11. Could you explain why the map which does the same as the 1D to 2D map at each x coordinate (essentially folding the plane like a hilbert curve) does not satisfy this criterion?

[hkayakh]

Bro why are you drawing the hilburt curve upside down?

[AnCoSt]

lovely video. so question: where am I going wrong with my argument? Since I know that the cardinalities of R, R^2, R^3, and indeed all finite n R^n, are all the same, that means we can establish a bjiection between R and any R^n. So - is it simply the "continuous" criterion that causes the problem?

[TheJara]

Nice videos of higher math, but bit too fast, may it is good idea to slow down and make little bit more detailed videos? please take more time to present slowly, longer time is better for these kind of high math...thanks...

[maxwibert]

Doesn't seem like you proved the punchline conclusion, that there does not exist a continuous one to one and onto map from 2d to 3d. You only gave one example of a continuous onto map from 2d to 3d which doesn't happen to be one to one. In fact, I don't even think that assertion is true. If you take f(x, y) = (H(x), y) where H(x) : R -> R^2 is the limit hilbert curve you described in your video. It's one-to-one because both x|->H(x) and y|->y are both one-to-one. It's onto because to get (a, b, c) in R^3, you just take y=c and whatever x gives H(x) = (a, b), which is guaranteed to exist since H is onto R^2. Thus f(x, y) = (a, b, c). Lastly it's continuous because in order to gaurantee that f(x, y) is within epsilon of f(w, z), you can choose w close enough to x (say within gamma) so that H(x) is within epsilon/2 of H(w), which you can do because H is continuous. Next, choose z within epsilon/2 of y. Then setting delta = min(gamma, epsilon/2) and we have that |(x, y) - (w, z)| < delta implies |f(x, y) - f(w, z)| = |(H(x), y) - (H(w), z)| <= |H(x) - H(w)| + |y - z| < epsilon. Thus f is continuous.

[TC]

You didn't prove that the Hilbert curve is continuous.

[atreidesson]

OK so that's a new channel, but it looks exactly like other quality math videos... which is almost definitely international.

[teeweezeven]

Ngl i came here because the thumbnail looks like bocchi from bocchi the rock

[MohammadHBakr]

If f is continuous and one-to-one
That means for every distinct image there's only one distinct input, meaning that you can't draw a horizontal line and make it cross the function twice
And here is the point without crossing we can't fill the space, so if f is a continuous function it's either one-to-one or onto, Not both.

[Manuel_Bache]

Didn't you say that a continuous function from 2d to 3d couldn't be onto?😊😊