# Cartesian Product involving the set $\mathbb{R}$ [closed]

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 $\mathbb{R}$:

$1)\enspace \{\, (x,y) \mid xy < 4 \,\}$,

$2)\enspace \{\, (x,y) \mid x^2+y^2 > 9\,\}$.

## closed as off-topic by user21820, Saad, ArsenBerk, Namaste, ScientificaOct 6 '18 at 10:21

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• Could you tell us what you have tried and achieved so far? – Suzet Aug 26 '18 at 20:08
• For part 1, I said that it could not be written as the cartesian product of two subsets of R because if x=3 and y=3, then xy=9 which is not less than 4. – amber patel Aug 26 '18 at 20:51

Both are not.

A set $C$ of the form $A \times B$ has the property that $$x \in \pi_1[C] \text{ and } y \in \pi_2[C] \implies (x,y) \in C$$ as well.

This fails for the $C = \{(x,y) : xy < 4\}$ as $3 \in \pi[C]$ as witnessed by $(3,1) \in C$ and $(1,3) \in C$ shows $3 \in \pi_2[C]$, but $(3,3) \notin C$ as $9 > 4$.

Also for $C= \{(x,y): x^2 + y^2 > 9 \}$ using $x= 1 = \pi_1(1,10) \in \pi_1[C]$ and $y=1 = \pi_2(10,1) \in \pi_2[C]$ etc.

None of the following is a Cartesian product of two sets of real numbers.$$\enspace \{\, (x,y) \mid xy < 4 \,\}$$

$$\enspace \{\, (x,y) \mid x^2+y^2 > 9\,\}$$

The first one is a disconnected set with two disjoint components and the second one is the set out side of a circle.

Proof by contradiction: suppose the first set is of the form $A\times B$. As $(-1,-1)$ is in our set, both $A,B$ should contain negative numbers.

If $a_0\in A$ is negative then for any non-negative $b$, the point $(a_0,b)$ satisfies the condition $a_0b<4$. So all nonnegative numbers should be in $B$. And for any positive $b_0$ every negative $x$ satisfies $xb_0<4$, and so every negative number should be in $A$. Further $(x,y)$ is in the set iff $(y,x)$ is in the set. This means $A,B$ both should be whole of real numbers, and the product would be the whole plane. However lot of points, for example $(2+u, 2+v), for u,v >0$ are missing in the first set.

Contradiction. The second example can also be proved using similar arguments.