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Consider a parameterization $$c(t) = (x(t), y(t)) $$

For every $t$, you have some $(x,y)$ output. This will be some curve in the plane. My textbook seems to imply that only a single parameter is needed for a curve because curves are 1-dimensional. Now consider a parametrization

$$ g(u,v) = (x(u,v), y(u,v))$$ For every $(u,v)$ you have some $(x,y)$ output. This will be some surface in the plane. My textbook seems to imply that 2 parameters are needed for a surface because surfaces are 2-dimensional.

So my question is, does a single parameter always result in a curve? Why can't the curve fold up on itself to form a 2-dimensional surface in the plane? And if we go beyond the plane, why can't a parameterization $c(t)$ give you a surface in 3-space (adding in a $z(t)$)or a volume in 3-space? If I just scribble my pen on a sheet of paper, I create a bunch of lines. But couldn't my scribbling form a surface if it was fine enough? I feel like this is getting into the "philosophy of continuous mathematics." Limits, continuous number lines, and whatnot as opposed to "discrete mathematics". Likewise, why can't $g(u,v)$ describe a volume? Does a single parameter always give a curve, 2 a surface, and 3 a volume?

DWade64
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    "Why can't the curve fold up on itself to form a 2-dimensional surface in the plane?" Interesting question. But ask yourself. If you have many line segments available and you put these line segments next to each other, will you get an area? How thick do you think a line actually is (strictly from a mathematical standpoint?) – imranfat Apr 19 '18 at 16:19
  • In you title, I think you mean $y(t)$ not $y(x)$ – saulspatz Apr 19 '18 at 16:22
  • @imranfat You can have a curve that fills up an area, like the Peano curve – saulspatz Apr 19 '18 at 16:24
  • @saulspatz Thank you, it has been fixed – DWade64 Apr 19 '18 at 16:24
  • @saulspatz. But can such a curve be described in parametric form, as suggested in the title? – imranfat Apr 19 '18 at 16:25
  • @imranfat Yes, otherwise it wouldn't be a curve. A curve in the plane is a continuous function $\mathbf R^1\to\mathbf R^2$ by definition. Of course, space-filling curves are wildly non-differentiable. – saulspatz Apr 19 '18 at 16:32
  • @saulspatz. Interesting for sure. – imranfat Apr 19 '18 at 19:06
  • Have you heard about fractals? There're some mathematical shapes with fractional dimension. – Ng Chung Tak Apr 19 '18 at 19:14
  • @NgChungTak I have but I've never really understood them (hopefully one day I'll get to know that side of mathematics). It seems like space-filling curves and fractals have something in common – DWade64 Apr 20 '18 at 14:36

3 Answers3

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There are space-filling curves, such as the Hilbert curve. However, such curves are never differentiable.

José Carlos Santos
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The underlyng reason is that we can't have continuous bijection from $\mathbb{R}^{2} \to \mathbb{R}$.

user
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As you mentioned $$c(t) = (x(t), y(t))$$ is a curve which is a one dimensional manifold embedded in tow dimensional plain.

If you want to cover a surface with a curve you need to go to infinite time interval and take limit which is not quite practical.

For example covering a disk of radius one with $$c(t) = ( e^t \cos(t), e^t \sin (t))$$ is not possible in finite time.

Mohammad Riazi-Kermani
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