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Prove that $A \subset B$ if and only if $A \setminus B = \emptyset$.

What is the correct and mathematically strict way to prove the above? (slightly different than Prove that if $A \setminus B = \emptyset$, then $A \subseteq B$ )

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    $\begingroup$ The symbol $\subset$ means inclusion or proper inclusion? $\endgroup$ – ajotatxe Nov 28 '14 at 15:39
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    $\begingroup$ The symbol $\subset$ must mean regular inclusion, same as in your linked answer, because otherwise it would be false. So the only difference between your question and your linked question is that yours is an equivalence, so you need to prove the converse as well. $\endgroup$ – Dustan Levenstein Nov 28 '14 at 15:41
  • $\begingroup$ It's not true if the inclusion is strict. Consider $A = B$. $\endgroup$ – brick Nov 28 '14 at 15:42
  • $\begingroup$ Nice question. In their book books.google.pt/… they mention "$A \subset A$ holds true for any $A$". So I suppose they use the sign $\subset$ instead of $\subseteq$, right? $\endgroup$ – Konstantinos Nov 28 '14 at 15:45
  • $\begingroup$ My answer to the question you link to answers this exact question, in exactly the same way as GDumphart's answer does. $\endgroup$ – Marnix Klooster Nov 28 '14 at 20:42
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\begin{align} A \subset B \ \ :\Leftrightarrow \ \ & \forall x \ : \ (x \in A) \rightarrow (x \in B) \\ \Leftrightarrow \ \ & \forall x \ : \ \lnot (x \in A) \lor (x \in B) \end{align}

\begin{align} A \backslash B = \emptyset \ \ :\Leftrightarrow \ \ & \forall x \ : \ x \notin A \backslash B \\ \Leftrightarrow \ \ & \forall x \ : \ \lnot (x \in A \backslash B) \\ \Leftrightarrow \ \ & \forall x \ : \ \lnot ((x \in A) \land \lnot (x \in B)) \\ \Leftrightarrow \ \ & \forall x \ : \ \lnot (x \in A) \lor \lnot(\lnot (x \in B)) \\ \Leftrightarrow \ \ & \forall x \ : \ \lnot (x \in A) \lor (x \in B) \\ \Leftrightarrow \ \ & A \subset B \ \ \ \end{align}

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If $A\subset B$, $\forall x\in A\Rightarrow x\in B$, so $A\setminus B=\emptyset$. I think this trivial.

Now, suppose $A\setminus B=\emptyset$. If $\exists x\in A$ but $x\notin B$, $x\in A\setminus B$ by the definition of set minus, which is a contradiction. So $x\in B$ and $A\subset B$.

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