Prove that $A = B$ if and only if $A \subseteq B$ and $B \subseteq A$ I want to prove that $A = B$ iff $A \subseteq B$ and $B \subseteq A$. I'm unsure of how to approach this problem. It seems really easy but I have no idea. Help would be greatly appreciated.
 A: Hints:
1) If $A=B$, clearly $A \subseteq B$ and $B \subseteq A$.
2) Suppose that $A \subseteq B$ and $B \subseteq A$. If $A \not= B$, then there exists $x \in A\setminus B$(or $x \in B\setminus A$), however $A \subseteq B$, clearly $x\in B$. This is a contradiction. So $A=B$.
A: When you have to show something of "Property one holds if and only if property two holds", you approach it in the following way (and this is especially true for introductory level problems, like this one):


*

*You remember that in order to prove an "if and only if" statement, you need to show two implications hold. First you assume that property one holds, and you show that property two holds, and then the other way around.

*You open up your notes, book, or whatever, and you write down on a piece of paper the definitions for both property one and property two. In this case, you write down explicitly what does it mean that $A\subseteq B$ and what does it mean that $A=B$.

*You finish up the proof in the two big steps, each requiring about two smaller steps to prove.
A: The forward implication really asks you to prove that $A \subseteq A$. That's true because for any $x \in A$, it's true that $x \in A$.
The reverse implication, that $A \subseteq B$ and $B \subseteq A$ implies $A = B$, is actually an axiom of set theory called the "axiom of extensionality." Therefore the "proof" would simply amount to quoting the axiom.
I doubt that this answer is actually what's expected here, but I think it's highly likely that other "proofs" will involve some form of circular reasoning, or else "definitions" of set equality that are logically deficient in some way. 
I have seen books that define two sets to be equal when they have the same elements. However it does not follow from this "definition" of equality that any property that is true of the first set must necessarily be true of the second, since "equal," having been defined, cannot be taken with its ordinary meaning of "being identical." Without the axiom of extensionality, it does not follow from the fact that two sets have the same elements that they are identical in other respects.
A: I start from a suggestion of user180040
Suppose that A=B.  Then suppose x $\in$A.  It follows that x $\in$B.  Why?  Because 
anytime we have an equality, we can replace any instance of one side of the equality by 
the other side of the equality wherever it appears in any statement.  Thus, we can infer 
that x $\in$A implies x $\in$B.  Since it holds that "if x $\in$A, then x$\in$B", by 
definition, we then obtain that A$\subseteq$B.  Similarly, we can prove that 
B$\subseteq$A.  Thus, both A$\subseteq$B and B$\subseteq$A.  So, we can infer that if 
A=B, then both A$\subseteq$B and B$\subseteq$A. End of first part.
Now suppose that A$\subseteq$B and B$\subseteq$A.  By definition of $\subseteq$ and since 
we have that A$\subseteq$B, we have that 
"if x $\in$A, then x $\in$B".  Similarly, we can infer that "if x $\in$B, then x 
$\in$A."  So, we can infer that x $\in$A if and only if x $\in$B.  Since we've proved the 
antecedent of the axiom of extensionality, we can infer the consequent that A=B.  
Thus we can infer that, if A$\subseteq$B and B$\subseteq$A, then A=B.  End of second part.
Since we've proved that "if A=B, then A$\subseteq$B and B$\subseteq$A" and "if A$\subseteq$B and B$\subseteq$A, then A=B" we can infer that
A=B iff A⊆B and B⊆A.
A: $
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\notag \\ #1 \quad & \quad \text{"#2"} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
$Here is how I would prove this, simply and succinctly.  We start at the most complex side of the statement, expand the definitions, and simplify, as follows:
$$\calc
A \subseteq B \;\land\; B \subseteq A
\calcop{\equiv}{definition of $\;\subseteq\;$, twice}
\langle \forall x :: x \in A \Rightarrow x \in B \rangle \;\land\; \langle \forall x :: x \in B \Rightarrow x \in A \rangle
\calcop{\equiv}{logic: merge the $\;\forall x \;$ quantifications into one}
\langle \forall x :: (x \in A \Rightarrow x \in B) \;\land\; (x \in B \Rightarrow x \in A) \rangle
\calcop{\equiv}{logic: simplify mutual implication to equivalence}
\langle \forall x :: x \in A \;\equiv\; x \in B \rangle
\calcop{\equiv}{definition of $\;=\;$, i.e., set extensionality}
A = B
\endcalc$$
A: For the forward implication, assume $A=B$. Then for any $x \in A$, $x \in B$ and thus $A \subseteq B$. Similarly, if $x \in B$, then $x \in A$ and thus $B \subseteq A$. Now using similar tactics approach the reverse implication.
