# Can (x(t), y(t)) generate a surface? If so, can the surface be continuous?

Intuitively, the parametric equation $z = (x(t), y(t))$ seems to only be able to generate one-dimensional objects, i.e. curves.
However...

Let $x(t)$ be "the odd-indexed digits of the real number $t$", and let $y(t)$ be "the even-indexed digits of the real number $t$". So for example, $x(123.4567) = 13.57$ and $y(123.4567) = 2.46$.

Now let's say we're given the parametric equation $z = (x(t),\ y(t))\ \ \ \forall\ t \in \mathbb{R}$.
This equation, essentially, represents a bijective mapping between $\mathbb{R}$ and $\mathbb{R} \times \mathbb{R}$.

My questions are:

1. Can $z$ be called a (two-dimensional) "surface" (instead of a curve)? (Why?)

2. If so, is it correct to say the surface is continuous? (Why?)

-
+1 Nice question, Mehrdad. –  Babak S. Jan 18 '13 at 4:24
@BabakSorouh: Thanks! –  Mehrdad Jan 18 '13 at 4:36

You have to make sure that it is well-defined, which means you need to decide which representation to use for $t=0.499999....=0.5000...$. You'd also have to prove the function is continuous. The tricky point is at the same values of $t$ where it wasn't well-defined.
Even then, the function is not $1-1$. If you take $t_0=0.409090909090...$ and $t_1=0.500000....$ they get sent to the same value. What you have here is called a "space-filling curve."
That said, there is no meaning to "continuous surface" in this sense. A "curve" is not just the image of the map $(x(t),y(t))$, but also the path through that image. Under that view, this is still a continuous curve, not a continuous surface.