Can $\{(x,y) \mid x^2 + y^2 < 1\}$ can be written as the cartesian product of two subsets of $\mathbb{R}$?

Question

We consider the set $$S := \{(x,y) \mid x^2 + y^2 < 1\}.$$ The exercise in Munkres asks whether it is possible to write this set as the cartesian product of two subsets of $\mathbb{R}$. We, naturally, consider restricting ourselves to the interval $(-1,1)$. If $|x|, |y| \geq 1$, then $x^2 + y^2 \geq 1$, so we require $|x|, |y| < 1$. Hence, the set $(-1,1)$ should work. However, if I take, say, $x, y = \frac{3}{4}$, I get $$\left(\frac{3}{4}\right)^2 + \left(\frac{3}{4}\right)^2 = 2 \cdot \frac{9}{16} = \frac{9}{8} > 1,$$ so this construction fails.

I cannot figure out how to prove rigorously (and generally) that no such construction would work because it's possible that there is some kind of restrictions, maybe to $\left(-\frac{1}{2}, \frac{1}{2}\right)$, for example, that I am missing. Any help on this would be appreciated.

Answer 1

Suppose $S = A \times B$ with $A,B \subset \mathbb{R}$.

Since $(t,0)^T,(0,t)^T \in S$ for any $t\in (-1,1)$ we must have $t \in A$ and $t \in B$ and so we must have $(t,t)^T \in S$, and we have $t^2+t^2 = 2t^2$.

We can choose any $t$ satisfying $|t| \ge {1 \over \sqrt{2}}$ to get a contradiction.

Answer 2

Since $(0,0.99)$ and $(0.99,0)$ are in the set, $0.99\in A,B$, where $A,B$ are two sets whose cartesian product contains the given set.

But, $(0.99,0.99)$ is not in the given set, so $A\times B$ is in fact, not equal to the set. So, it is not possible.