Is every subset of a product a product of subsets?

Is every subset of a product a product of subsets ?

i.e Let $E$ and $F$ two non empty sets and we define the Cartesian product $E \times F$.

Now given a non empty subset $A$ of $E\times F$, can we write $A$ as the product of two subsets of $E$ and $F$: i.e is there $E_1 \subset E$ and $F_1 \subset F$ such that $$A=E_1 \times F_1$$

My idea is that this statement is false, and some counter-example I thought of is $$\{(x,y) \in \mathbb{R}^2, \,\, x^2+y^2=1\}$$ $$\{(x,1/x), \,\, x\in \mathbb{R}^*\}$$ But I could not find a way to prove we can't write these two sets as product of two subsets of $\mathbb{R}.$

Keep it simple. A minimal counterexample is the following: $$A=\{a,b\},B=\{d,e\}$$ We have

$$A\times B=\left\{(a, d), (a, e), (b, d), (b, e)\right\}$$ And the subset $$E=\left\{ (a, e), (b, d)\right\}$$ is not the cartesian product of two sets.

Assume $\{(x,y): x^2+y^2=1\}=A\times B$ for some $A,B\subseteq\mathbb{R}$. Let $x\in A$. Then for every $y\in B$ we have $x^2+y^2=1$. However, there can be at most two real numbers $y$ which satisfy $x^2+y^2=1$, and hence we conclude that $|B|\leq 2$. Similarly, $|A|\leq 2$. But this is obviously a contradiction, because $\{(x,y): x^2+y^2=1\}$ is an infinite set.

No. If $A \subset X \times Y$ has the form $A = E_1 \times F_1$, then

$$E_1 = \{ x \in E \mid \exists y \in F : (x,y) \in A\}, F_1 = \{ y \in F \mid \exists x \in E : (x,y) \in A\} .$$

In other words, $E_1$ is the image $\pi_E(A)$ of $A$ under the projection $\pi_E : E \times F \to E$, similarly $F_1 = \pi_F(A)$.

Your first set, the unit circle, has both images $= [-1,+1]$, but $[-1,+1] \times [-1,+1]$ is bigger than your set.

For your second set, the hyperbola with two branches, you get images $\mathbb R^*$ which again does not fit.

An easier counterexample is to let $E=F=\{0,1\}$ and $A=\{\langle 0,0\rangle,\langle 1,1\rangle\}$. If $A=E_1\times F_1$ for some $E_1\subseteq E$ and $F_1\subseteq F$, then clearly $0\in E_1$ and $0\in F_1$, and $1\in E_1$ and $1\in F_1$. But then $E_1=E=F=F_1$, so $$E_1\times F_1=E\times F\ne A\,.$$

Alternatively, you can argue from cardinality: $|A|=2$, and the subsets of $E$ and $F$ have cardinalities $0,1$, and $2$, so if $E_1\times F_1=A$, then one of $E_1$ and $F_1$ must have one element, and the other must have two. But if $|E_1|=1$, the members of $A$ must all have the same first component, while if $|F_1|=1$, they must all have the second component, and neither of these is in fact the case.

This second argument needs only a tiny modification to show that if $E$ and $F$ both have at least two points, then $E\times F$ has a subset that is not a product: if $e_1$ and $e_2$ are distinct points of $E$, and $f_1$ and $f_2$ are distinct points of $F$, the subset $\{\langle e_1,f_1\rangle,\langle e_2,f_2\rangle\}$ of $E\times F$ cannot be a product for the same reason that $A$ above is not a product.

$$ S = \{ (x,y)\in \mathbb{R}\times \mathbb{R} \vert y = x\} $$ is a subset of $\mathbb{R}\times \mathbb{R}$.

But $S$ is not a cartesian product. To see this, notice that: $$(0,0)\in S$$ $$(1,1)\in S$$ But $$(0,1)\notin S.$$

Therefore $S$ is not a cartesian product. $\Box$

FYI, this $S$ is sometimes called the diagonal subset of $\mathbb{R}\times \mathbb{R}$, and I think this name should make some sense to you if you draw the graph of $S$. Some of the other answers that have already been posted also use diagonal subsets (of sets other than $\mathbb{R}$), so this answer is really no different from theirs. But it might be easier to visualize.