Cartesian Product of large circle and small circle I am bit confused with cartesian product. Suppose I have a set in R2 space and another set in R2 space so would it be right to say that cartesian product of those is in R4 space??
Following to this what is cartesian product of a small circle and a large cirlce?? If we go by previous question(if it is true??) then the geometric answer would be some shape in R4 so how to analyze this.
I hope I have clarified my question carefully. Please look at it and share your views and suggestions.
Thanks,
 A: Suppose the first circle is parametrized by $x(t) = (r \cos t, r \sin t)$, and the second is parametrized by $y(t) = (R \cos t, R \sin t)$. The product is a $2$-dimensional surface sitting in $\mathbb R^4$, parametrized by $f(s, t) = (x(s), y(t)) = (r \cos s, r \cos s, R \cos t, R \sin t)$. 
We can come up with a transformation that brings this surface into $\mathbb R^3$. What I'm about to show is in general not what people do, but I guess one example of extreme tedium would not hurt. The underlying idea I use here is essentially the same idea as "charts".
It is easy to check that $f$ restricted to $(s, t) \in [0, 2\pi)^2$ is one-to-one. Let $\tilde f$ be this restriction. It follows that $\tilde f$ is continuous, and it has an inverse. Let $S$ be the image of $\tilde f$ (which is equal to the image of $f$). We have $\tilde f^{-1}$ defined on $S$. $f^{-1}$ is continuous except at points corresponding to $s = 0$ or $t = 0$, i.e., $f(\{0\} \times [0, 2\pi) \cup [0, 2\pi) \times \{0\})$ is the set of discontinuities of $\tilde f^{-1}$.
Let $g(s, t) = ((R + r\cos s)\cos t, (R + r\cos s)\sin t, r\sin s)$, and let $T$
be the image of this function.
($T$ is a torus.)
It is straightforward to check that $\tilde g$, the restriction of $g$ to $[0, 2\pi)^2$, is one-to-one. (Note that we need $R > r$.)
So $\tilde g^{-1}$ is defined on $T$. We have similar discontinuities of $\tilde g^{-1}$ as those of $\tilde f^{-1}$.
Define $\varphi: S \to T$ by $\varphi = \tilde g \tilde f^{-1} = g\tilde f^{-1}$. Because $\tilde g$ and $\tilde f^{-1}$ are one-to-one, $\varphi$ is also one-to-one, and it has an inverse $\varphi^{-1} = f \tilde g^{-1}$.
$\varphi$ is continuous at points on $S$ that do not correspond to $s = 0$ or $t = 0$.
We want to show that in fact $\varphi$ is continuous over the whole $S$.
Now note that $f$ and $g$ are periodic in both parameters, with both periods equal to $2\pi$. So the restriction of domain in fact could have been anything of the form $[a, a + 2\pi) \times [b, b + 2\pi)$, and the restrictions of $f$ and $g$ on this domain will still have inverses.
In particular, we can define $\tilde f_1 = f|_{[-\pi, \pi) \times [0, 2\pi)}$, $\tilde f_2 = f|_{[0, 2\pi) \times [-\pi, \pi)}$ and $\tilde f_3 = f|_{[-\pi, \pi)^2}$, then derive the same $\varphi = g\tilde f_1^{-1} = g\tilde f_2^{-1} = g\tilde f_3^{-1} = g\tilde f^{-1}$. Each construction of $\varphi$ gives continuity at different places, and all of them combined give continuity over the whole $S$.
To see that $\varphi^{-1}$ is also continuous, define $\tilde g_i$ similarly, and derive continuity over the whole $T$ from the four constructions of $\varphi^{-1}$: $\varphi^{-1} = f\tilde g^{-1} = f\tilde g_1^{-1} = f\tilde g_2^{-1} = f\tilde g_3^{-1}$.
Every occurrence of the word "continuous" above can in fact be replaced by "smooth", and $\varphi$ is a diffeomorphism.
A: The cartesian product of two circles (of any size), is what is called a 'duo-cylinder' or bi-cylinder prism.  
The surface is described by $\operatorname{max}(\frac 1{r_1}\sqrt(x^2+y^2), \frac 1{r_2}\sqrt(w^2+z^2))=1$.
The surface might be described by taking a longish cylinder, and bending it in the fourth dimension, such that the two circular ends meet.  Likewise, one takes two other cylinders and wrap then.  The two then join.
A second way one might make one is to imagina a cylinder, and rotate it around a circular mid-height.  This causes the top and base to join in a circular motion (which represents the longish cylinder being rotated), and the original round surface, to become the second face, the vertical lines become circles, and one can see that the circumference becomes a stack of circles, wrapped into a circular ring (around the original cylinder).
If one is looking for the cartesian product of the surfaces of two circles, ie their circumferences: it goes like this.  Take a rectangle, the length and bredth being the circumferences of the circle.  Roll this up into a cylinder.  Now, in the fourth dimension, roll the cylinder up so that the ends meet.  You have what is variously called a 'torus' in four dimensions.  The Polygloss name for this is a bi-glomolatric prism.
A: Think about a cylinder. Choose an interval $A=[a,b)$ and a circle (disk)
$D = \{ (y,z): y^2 + z^2\le r^2 \}$. Then  $A \times D$ is a 3D set $(x,y,z)$, with $x \in [a,b]$ and $(y,z) \in D$. We say, that $a=0$, and
$b=2 \pi$. 
Imagine now that the cylinder is a slinky and joint the two (top and bottom) circles. They will then construct a torus. 
Technically speaking we can think of a map from $A=[0,2 \pi)$ into a circle by the function
\begin{eqnarray}
f : [0,2 \pi)  &\to& B \subset \mathbb{R}^2 \\
      t   &\to& (R \cos t, R\sin t)
\end{eqnarray} 
with $R>r$.
This map is one-to-one, smooth ... etc. 
So, the set $( \cos t, \sin t)$ as coordinates belongs in $\mathbb{R}^2$, but
they are really a one-D manifold, parametrized 
by $t$.
Moreover consider a polar coordinate representation defined by $f$ above.
We write the torus as
\begin{equation}
  \tau = B \times D \in \mathbb{R}^4
\end{equation}
However, if we in polar coordinates the set $B$ is represented by
the couple $B=(\theta, R)$  where $R$ is the fixed radius and
$\theta \in A$. So the product $\tau = B \times D = A \times \{R\} \times D$,
where $A$ is 1D, and $D$ is 2D. The singleton $\{ R \}$ (the large radius)
becomes just a label, and indeed the object is a 3D object embedded in
$\mathbb{R}^4$, with a fixed $R$, to choose among many.
