Can $A \times B$ give you a circle of radius $1$.

Question

I've heard that $A \times B$ cannot be a circle of radius $1$. However I think this is false as I know from $\sin^2 x+\cos^2 x=1$. So can we take the set $A=\{\sin \theta | \theta \in R \}$ and $B=\{\cos \theta | \theta \in R\}$ or does this not work?

Answer 1

Your construction gives $[-1,1] \times [-1,1]$ because the two sets you've written down are those two closed intervals (and the set product is every pair of things where the first is from the first interval and the second is from the second interval). However, there is a stupid way to do this ...

Let $A$ be a circle of radius $1$ and $B = \{0\}$. Then $A \times B$ is still a circle of radius $1$. It's elements are just written like "$(a,0)$" where $a$ is any point in a circle of radius $1$. But the second element is constant, so this is homeomorphic to $A$.

Answer 2

Suppose that $A \times B$ is a circle (i.e. a ring) of positive radius. Note that for any $a \in A$, the set $\{a\} \times B$ is bijective to $B$ via the map $(a,b) \mapsto b$. Now, if $a$ is the $x$-coordinate of the center point of our circle, then $\{a\} \times B$ has exactly two points and hence $B$ has exactly two points. On the other hand, if $a' \in A$ is a point where the tangent on our circle is parallel to the $y$-axis, then $\{ a' \} \times B$ only contains a single point and hence $B$ may only contain one element. Contradiction!

Answer 3

In your example $A$ and $B$ are both the set $[-1,1]$, so $A\times B$ is just a filled-in rectangle.

There's no way for a carteseian product to specify that you only want to pair elements that arise from the same $\theta$.

You can write $\{\langle x,y\rangle\mid \exists\theta: x=\cos\theta, y=\sin\theta \}$ for the unit circle, but that is not a cartesian product, because the conditions for $x$ and $y$ are not independent of each other.

Answer 4

Let $S^1 $ be the circle of radius 1. Suppose $S^1 = A \times B $. Then $(1,0) \in S_1$ and so $1 \in A$. Similarly $1 \in B$ and so $(1,1) \in A \times B$. But $\|(1,1)\| = \sqrt{2}$ hence $S^1 \neq A \times B$.