Let \(\mathbb{R}\) denote the set of real numbers. For each of the following subsets of \(\mathbb{R} \times \mathbb{R}\), determine whether it is equal to the cartesian product of two subsets of \(\mathbb{R}\) .
\(\{(x, y) \mid x \text{ is an integer}\}\) .
\(\{(x, y) \mid 0 < y \leq 1\}\) .
\(\{(x, y) \mid y > x\}\) .
\(\{(x, y) \mid x \text{ is not an integer and } y \text{ is an integer}\}\) .
\(\{(x, y) \mid x^{2} + y^{2} < 1\}\) .
Solution
Yes. The set is \[ \mathbb Z\times\mathbb R. \]
Yes. It is \[ \mathbb R\times(0,1]. \]
No. If the set were \(A\times B\), then since \((0,1)\) and \((1,2)\) both satisfy \(y>x\), we would have \[ 0,1\in A,\qquad 1,2\in B. \] Hence \((1,1)\in A\times B\), contradicting \(1\not>1\).
Yes. It is \[ (\mathbb R-\mathbb Z)\times\mathbb Z. \]
No. If the open unit disk were \(A\times B\), then from \[ (0.9,0),\ (0,0.9) \] being in the disk we would obtain \(0.9,0\in A\) and \(0,0.9\in B\). Thus \((0.9,0.9)\in A\times B\), but \[ 0.9^2+0.9^2=1.62>1, \] so this point is not in the disk, a contradiction.