Let's say we have given two sets: $A$ and $B$.

I thought about two possible statements:

1st: $A = B$ iff $|A| = |B|$ and $(\forall x \in A: x \in B)$ and $(\forall y \in B: y \in A)$

2nd: $A = B$ iff $|A \setminus (A \cap B)| = 0$ and $|B \setminus (A \cap B)| = 0$

Now I am curious if this statements are true and how I can prove them (or are they already fundamental/trivial).

Also are there other possible true statements?

  • 1
    $\begingroup$ $A=B\iff A\setminus B=B\setminus A=\emptyset$ $\endgroup$
    – Lehs
    Jan 18, 2018 at 16:19
  • $\begingroup$ Thats also true. So obviously the second one (from my question) is a little bit redundant, right? $\endgroup$ Jan 18, 2018 at 16:45
  • 1
    $\begingroup$ Instead of $|X| = 0$ you can write $X = \varnothing$. $\endgroup$ Jan 18, 2018 at 20:04
  • $\begingroup$ @PaŭloEbermann: Formally no you cannot, because that creates an infinitely recursive definition. $\endgroup$
    – Ben Voigt
    Jan 18, 2018 at 21:33

2 Answers 2


Obviously $|A| = |B|$ is redundant because of $(\forall x \in A: x \in B) \Rightarrow A \subseteq B$ and $(\forall x \in B: x \in A) \implies B \subseteq A$, therefore $\implies$ $A \subseteq B \land B\subseteq A \iff A=B$


I would just keep it on:$$A=B\text{ if and only if }\forall x\in A[x\in B]\wedge\forall x\in B[x\in A]$$which is the axiom of extensionality.

The demand $|A|=|B|$ is redundant.

The second statement is correct, but rests on cardinality. Let's stay close to the basics.


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