How to prove that an equivalence relation $E$ over a set $A$ satisfying $|A|^3=|A|$ will satisfy $|E|=|A|$?

Question

Given set $A$, $|A| = a$ and $a^{3} = a$. Let $E$ be an equivalence relation over $A.$

Prove that $|E| = a$

My Attempt -

We know that there isn't any finite set $A$ such that $|A|^{3} = |A|$, therefore $A$ is an infinite set.

By definition equivalence relation $E$ over and infinite set $A$ -

$E$ is an equivalence relation $ \iff I_A \subseteq E$ (let $I_A$ be the identity relation over set $A$), and all equivalence relation $E$ over $A$ uphold $E \subseteq A \times A$ (that part am i not sure of).

Therefore, $I_A \subseteq E \subseteq A \times A$, translating to cardinals of infinite sets - $a = |I_A| \le |E| \le |A \times A| = a$.

$\Rightarrow |E| = a$

Am i correct ? are my mathematical assumptions right ? Would love to hear your thoughts.

Answer 1

If $A$ has $0$ or $1$ element we have $|A|^3=|A|$, so $A$ can be finite. Your argument works for the finite case as well.

Answer 2

Your proof is essentially correct. Essentially because it doesn't use the assumption that $A$ is infinite, which as pointed out is incorrect since $0^3=0$ and $1^3=1$ as well.

But the proof works just fine for these two finite cases as well.