Can any subset of $\mathbb{R}^2$ be expressed in form $A\times B$, where $A$ and $B$ are subsets of $\mathbb{R}$? This is a very elementary question I'm a little confused about. 

Can any subset of $\mathbb{R}^2$ be expressed in form $A\times B$, where $A$ and $B$ are subsets of $\mathbb{R}$? 

I'm thinking that it might not necessarily be so. For instance, if we consider the intervals $[0,2]$ and $[3,5]$ in $\mathbb{R}$, and throw out a single point, say $(1,4)$, out of $[0,2]\times [3,5]$ then there do not exist subsets $A$ and $B$ of $\mathbb{R}$ such that $A\times B = ([0,2]\times [3,5]) \setminus \{(1,4)\}$. Because if they were equal, surely $A$ would be equal to $[0,2]$ and $B$ would be equal to $[3,5]$. But as there is no way to put the constraint that $1$ cannot pair up with $4$, $A\times B$ will always be $[0,2]\times [3,5]$.
Am I thinking in the right direction? Thanks.
 A: Yes you are thinking in the right direction. The answer is no. Here is a simple counter example: Consider the set $S = \{(1,2), (2,1) \}$. If there were sets $A,B \subseteq \mathbb R$ with $S = A \times B$ then $1,2 \in A$ but also $1,2 \in B$ so that $A \times B$ would have to contain $(1,1) $ and $(2,2)$ and therefore $A \times B$ is strictly greater than $S$. 
A: If a subset $\mathrm X$ of $\mathbb R^2$ can be written as a product $\mathrm A \times \mathrm B$, then it is true that $\mathrm{pr_1}(\mathrm X) = \mathrm A$ and $\mathrm{pr_2}(\mathrm X) = \mathrm B$.
Then take the example of the unit disc in $\mathbb R^2$, it is not equal to $[-1,1]^2$ so it is not a product of two subset of $\mathbb R$.
A: For finite sets $A$ and $B$, the cardinality of $A\times B$ is the product of the two separate cardinalities. So for any set $S$ of $p$ points in the plane, $p$ being prime, the only way for $S$ to be a product set is for either the $x$-coordinates or the $y$-coordinates to be all the same.
A: No. You can think of sets of the form $A \times B$ as 'rectangles' (although they are really only rectangles when $A,B$ are intervals). Not all sets are rectangles, for example, take the square and rotate it $45^\circ$.
To further illustrate, with the set $A \times B$, you can pick the element $a \in A$ and $b \in B$ 'independently' to get $(a,b) \in A \times B$, but this is not always the case. (For example, if I have the line $\{(x,y) | x = y\}$ and I choose $x=1$, then I must have $y=1$, whereas if I have the square $\{ (x,y) | x \in [0,1], y \in [0,1] \}$, then if I choose $x=1$, I can still pick any $y \in [0,1]$.)
A: Take the unitary open disc, for example: $$D=\{(x,y)\in \mathbb{R}^2\mid x^2+y^2<1\}.$$
It is a good exercise to show that $D$ is not a product of any two subsets of $\mathbb{R}$. Here is a simple solution.
Suppose that $D=A\times B$, for some subsets $A,B\subseteq \mathbb{R}$. Note that $(x,0)\in D$ for any $x\in (-1,1)$. So $(-1,1)\subset A$ and  $0\in B$. For the same reason, $0\in A$ and $(-1,1)\subset B$. Then $(-1,1)\times (-1,1)\subseteq A\times B=D$ and so take $\sqrt{2}/2< a,b < 1$ such that $(a,b)\in A\times B$ with $1< a^2+b^2$. Contradiction!
A: Let $S:= \{ (x,y) \in \mathbf{R}^2 : x < y \}$, then I'm not sure if $S$ can ever be expressed as a Cartesian product of two subsets of $\mathbf{R}$. 
