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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.2010-12-22
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    As you say sir.2010-12-22
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    @Damir Is this a homework assignment?2010-12-22
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    I know @Asaf; the reason for that comment is that the [previous version](http://math.stackexchange.com/posts/15188/revisions) of the question said that it was.2010-12-22
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    @J.M. whoops... I only saw that now.2010-12-22
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    Don't you mean $\subseteq$ rather than $\subset$?2010-12-22
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    @ Anthony Labarre: It is $\subset$.2010-12-22
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    @Anthony: both $\subset$ and $\subseteq$ usually mean subset or equal to, and $\subsetneq$ is used to denote proper inclusion.2010-12-22
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    Thanks for clarifying, Asaf, I'm used to different conventions.2010-12-22
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    @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.2010-12-22
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    @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.2010-12-22
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    @Asaf: I'm telling my students the same `:-)`2010-12-22
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    Aren't this also implies $A = B$ ?!2010-12-23

2 Answers 2

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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 $\subseteq$.2010-12-22
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    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](http://en.wikipedia.org/wiki/Truth_table#Logical_implication)2010-12-22
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    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 ;-)2010-12-22
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    kahen, I think that this is a classic case of confusing the notations of subset and proper subset.2010-12-22
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    $\huge\subsetneq$2010-12-22
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    @kahen: +1 for NC reference ^_^2011-07-05
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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 $A=B=C$.