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Do we have a technique in proving equality that says if $A$ is contained in $B$ and the cardinality of $A$ is equal to the cardinality of $B$ then $A=B$?

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The set of integers and the set of even integers make a counterexample. Because both have the same cardinality

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Here's a proof for finite sets (as mentioned, the claim is false for infinite sets; take $A=\mathbb{N}$ and $B=\mathbb{Q}$).

Suppose $A\subset B$, but $A\neq B$. Write $B=A\cup (B-A)$. Since $\#B$ is finite and $A\cap(B-A)=\emptyset$, we have $\#B=\#A+\#(B-A)$. Now, since $A\neq B$, $B-A$ is nonempty, so $\#(B-A)>0$ and hence $\#B>\#A$.

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    Absolutely. Sometimes I find contrapositive/contradiction proofs more intuitive to start out with though ("what happens if its not true?"), then convert to a direct argument if necessary.2012-11-20
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If $A$ is Dedekind-finite (which means finite in the context of most modern set theory), and $B\subseteq A$ then $A=B$ if and only if $|A|=|B|$.

In fact this is an equivalent definition for Dedekind-finite sets. Namely, $A$ is Dedekind-finite if and only if it is true that if $B\subseteq A$ and $|A|=|B|$ then $A=B$.

The natural numbers are not Dedekind-finite. For example, $\mathbb N\setminus\{0\}\neq\mathbb N$, but both have the same cardinality.