# How to prove :: $A \subset B, B \subset C, C \subset A \Rightarrow B = C$

Let $A,B \text{ and } C$ are three sets then if $A \subset B, B \subset C, C \subset A \Rightarrow B = C$

How could we prove this ?

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"Axioms" are not meant to be proven. –  Ｊ. Ｍ. Dec 22 '10 at 13:53
As you say sir. –  Damir Dec 22 '10 at 13:57
@Anthony: both $\subset$ and $\subseteq$ usually mean subset or equal to, and $\subsetneq$ is used to denote proper inclusion. –  Asaf Karagila Dec 22 '10 at 14:37
@Anthony: Strictly speaking, $\subset$ should indeed mean proper inclusion, in analogy to $<$ meaning strictly less than. However, in this world this probably won't be accepted any more. –  Hendrik Vogt Dec 22 '10 at 15:46
@Hendrik: I'm a TA in an introductory course in set theory, as I told my students on the first class: Some people use this notation and other use that notation. If you want to be absolutely clear use $\subseteq$ when the inequality is weak and $\subsetneq$ when it is strong. And since strong $\implies$ weak anyway, use the weak one when you're not certain. –  Asaf Karagila Dec 22 '10 at 16:15

This is effectively asking to prove $B \subseteq C \land C \subseteq B \implies B = C$. The usual way to prove this is to use the Axiom of Extensionality - i.e. take an element $b \in B$ and show that it is in $C$. Then show that $c \in C \implies c \in B$. Extensionality now tells you that the two sets are identical.

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But it's $\subset$ not $\subseteq$. –  Damir Dec 22 '10 at 14:23
Well, if it's proper set inclusion then the problem is even easier because then LHS is always false, and "False $\implies$ Whatever" is always true. You can for example look at the truth table for implication –  kahen Dec 22 '10 at 14:30
Incidentally this also demonstrates why it's a BIG problem that people use $\subset$ and $\subseteq$ interchangably. When I take over the world ("Of course!") my first decree shall be that all mathematicians that use $\subset$ when they don't mean proper set inclusion shall be put in labour... eh I mean happy camps ;-) –  kahen Dec 22 '10 at 14:37
kahen, I think that this is a classic case of confusing the notations of subset and proper subset. –  Asaf Karagila Dec 22 '10 at 14:38
$\huge\subsetneq$ –  Nate Eldredge Dec 22 '10 at 20:29
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Verbosely:

Say that $A \subseteq B \subseteq C \subseteq A$. Then in particular,

$x \in A \Rightarrow x \in B$ by the first inclusion but then by the second we have $x \in B \Rightarrow x \in C$.

Contracting, $x \in A \Rightarrow x \in C$, but the rightmost inclusion tells us that $x \in C \Rightarrow x \in A$ so that $x \in A \Leftrightarrow x \in C$. By the axiom of extensionality we obtain that $A = C$.

Now by the second inclusion, $x \in B \Rightarrow x \in C$, but since $C=A$, we must have $x \in B \Rightarrow x \in A$, so that with the first inclusion $x \in A \Leftrightarrow x \in B$ and again by the axiom of extensionality we have that $A=B$. Now $A=C$, and so $A=B=C$.

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