How to use the Axiom of Extension to prove the equality of sets?

Question

I have a problem understanding the Axiom of Extension.

The axiom states that $A = B \iff \forall x \ (x \in A \Leftrightarrow x \in B)$. In my understanding, this is a bi-conditional statement that means two things:

1. if two sets are the same, then they have the same elements, i.e., $(A = B) \Rightarrow \forall x \ (x \in A \Leftrightarrow x \in B)$; and
2. it two sets have the same elements, then they are the same, i.e., $\forall x \ (x \in A \Leftrightarrow x \in B) \Rightarrow (A=B)$

Now I want to use the axiom of extension to prove that $A = A$.

Proof: The statement $x \in A \Rightarrow x \in A$ is obviously true. Therefore, the statement $x \in A \Leftrightarrow x \in A$ is also true. By the Axiom of Extension, since we have $\forall x \ (x \in A \Leftrightarrow x \in A)$, we get $A = A$, i.e., $(x \in A \Leftrightarrow x \in A) \Rightarrow A=A$.

The problem is that I do not understand how to prove the other side of the implication, i.e., that $A = A \Rightarrow (x \in A \Leftrightarrow x \in A)$, because I am not sure what $A=A$ means, other than that both sets have the same elements.

So my question in essence is the following: Is it necessary to prove the other side of the implication? and is my understanding of the axiom correct?

Thank you very much.

Answer 1

I think there is a confusion in your mind between proving a bi-conditional statement and using this statement. .

Suppose you want to prove a bi-conditional statement, in that case, you actually have to show that the implication " works" in both directions.

But here you only want to use this biconditional, as a premise in your reasoning.

The structure of the proof you want to achieve is as follows :

(1) $LHS \iff RHS$

(2) $RHS$ is true, because...

(3) Therefore $LHS$ is true

with $LHS : A=A$

and $RHS : \forall (x) [ x\in A \iff x\in A]$

• Example. Suppose I want to prove that figure $F$ is a square.

I know that : F is a square iff F is a parallelogram with 1 right angle and 4 equal sides. ( Note that I do not prove this statement, I use it as a given and as a premise in my reasoning).

I prove the $RHS$ , namely that $F$ is a parallelogram with 1 right angle and 4 equal sides.

I conclude that the $LHS$ is true, namely that $F$ is a square.

Answer 2

View Point 1 :

You have Proved 1-Direction. The other Direction is almost same: Let $A=A$.
Then $\forall x : x \in A \implies x \in A$
& $\forall x : x \in A \impliedby x \in A$
Hence $\forall x : x \in A \iff x \in A$
Hence, $(A=A) \implies (\forall x : x \in A \iff x \in A)$

Overall, we have, $(A=A) \iff (\forall x : x \in A \iff x \in A)$

View Point 2 :

In certain treatments, The Axiom of Extension is not BiDirectional.
In wiki (Here) , it says:

In Zermelo–Fraenkel Axioms, the Implication is 1-Way.
$\forall A \forall B (\forall X (X \in A \iff X \in B) \implies A = B)$
The converse of this axiom follows from the Substitution Property of Equality.
$\forall A \forall B ( A = B \implies \forall X ( X \in A \iff X \in B ) )$
In this case, the Proof of $A=A$ will use the Substitution Property of Equality.

View Point 3 :

In certain treatments, there is no Primitive Equality.
Then that becomes the "Definition" of Equality, using new Axiom with Substitution Property.
In high level, $\forall X : P(X) \implies \forall Y : P(Y)$ is the Substitution Property.
In this case too, the Proof of $A=A$ will use the Substitution Property & the "Definition" of Equality.

View Point 4 :

In certain treatments, [ Eg : Logic & Structure by Dirk van Dalen ] , Identity (Equality) itself has multiple Axioms, which interact with the Axiom of Extension, giving rise to new ways to look at $A=A$.