Suppose that $A ⊆ A × A$. Show that $A =∅$ [duplicate]

Suppose that $A ⊆ A × A$. Show that $A =∅$ Hint: Using axiom of regularity.

My proof.

Suppose that $A ⊆ A × A$ and $A\not=∅$. let $x∈A$. Since $A ⊆ A × A$, $x∈A×A$ which implies $(x,∅)∈A×A \Rightarrow x∈A \mid x∈∅$ which is a contradiction $\Rightarrow A \not⊆ A × A$. But this contradicts our supposition, hence $A=∅$.

I don't know how to use axiom of regularity here.Also, please suggest how do i improve my proof if its correct. If not correct suggest the proof itself.

Solution 1:

The step

x∈A×A which implies (x,∅)∈A×A

is wrong. I don't know why you think this is true because you didn't write any justification in your proof.

If $x∈A×A$ then $x$ is an ordered pair of elements of $A$ (definition of $A\times A$), so we can write $x=( y,z)$ for some elements $ y,z$ of $A$.

At this point you probably need to apply the definition of an ordered pair and identify $x$ as a particular set. Can you do that?

Solution 2:

If $x\in A\subseteq A\times A$, then by definition $x$ has the form $\{\{y\},\{y,z\}\}$ for some $y,z\in A$. So $x$ is an element of $A$ which is also a subset of $2^A$ (power set of $A$). This is only fulfilled by the empty set $A=\emptyset$.