For these sets: $S \subseteq \mathbb{R}\times \mathbb{R}$, Are there $A,B\subseteq \mathbb{R}$ so that $S=A\times B$ For the next given sets $S \subseteq \mathbb{R}\times  \mathbb{R}$, Are there $A,B\subseteq \mathbb{R}$ such that $S=A\times B$ ?
A) $S=\left \{\left ( x,y \right ):x\in \mathbb{N}  \right \}$ 
B) $S=\left \{\left ( x,y \right ):0<y\leq1   \right \}$
C) $S=\left \{\left ( x,y \right ):x<y  \right \}$
D) $S=\left \{\left ( x,y \right ):x^2+y^2=1  \right \}$
I think that C, D it's true because there can't be an  empty set but in A, B not necessarily true, but i don't know how to show it.
Thanks.
 A: Proposed sets, $A, B$:
$(a)\quad A = \mathbb N, \; B = \mathbb R$.
$(b)\quad A =\mathbb R, \;B = (0, 1] \subseteq \mathbb R$.
$(c)\quad$ Here, whether $(x, y) \in S$ depends on both $x$ and $y$, relative to one another. $x$ can be any element whatsoever, but whatever it is, $y$ must be greater. We can not decide, based only on two proposed sets $A, B$, whether an ordered pair belongs to $S$ unless we know that $x \lt y$. For example, while $(1, 2) \in S, (2, 3)\in S$, $(2, 2)\notin S$. So there are no sets $A, B$ such that $A\times B = S$
$(d)\quad$ This presents a problem. $x^2 + y^2 = 1$ defines a circle with radius $1$ centered at the origin. For $(x, y) \in S$, even though $x$ ranges from $0$ to $1$, as does $y$, they do not do so independently of the other. Once we fix $x$, $y$ is determined, and vice-versa. So there are not $A, B$ such that $x \in A, y \in B$ and such that $A\times B = S$.
A: Remember the definition of cartesian products.
$$
A\times B = \{\,(a,b) : a\in A, b\in B\,\}
$$
So for each of the questions you are looking for sets $A,B\subseteq \mathbb R$ such that $S=A\times B$ is the set of all ordered pairs with first entry in $A$ and second entry in $B$.
In A) the first entry is always in $\mathbb N$ while the seconds entry can be any number in $\mathbb R$. Is $S=\mathbb N \times \mathbb R$ in A)?
In D) the set $S$ is a circle of radius $1$ around the origin. In particular $(0,1)$ and $(1,0)$ are in $S$. If we assume $S=A\times B$ here, we can conclude $0\in A$, $1\in B$ from $(0,1)\in S$ and we can conclude $1\in A, 0\in B$ from $(1,0)\in S$. Then also $(1,1)\in A\times B=S$, since we found $1\in A, 1\in B$. But $1^2+1^2\neq 1$, so $(1,1)\notin S$, a contradiction. Thus there are no such sets $A$ and $B$ and $S$ is not a cartesian product in D).
Try to find similar arguments for the other question.
