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For all sets A, B and C, if $A-(B \cup C) = \emptyset$ ; then $A-C\subseteq B$. Is it true? If it is, how to prove it? I think it's true...

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  • $\begingroup$ Remember, if $A\setminus B=\emptyset$, then $A\subseteq B$. $\endgroup$ – frabala Feb 27 '14 at 22:33
  • $\begingroup$ (From template of comments) After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?. $\endgroup$ – frabala Feb 28 '14 at 4:39
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Notice that $$U-V=U\cap V^c$$ and $$(U\cup V)^c=U^c\cap V^c$$ hence we have $$A-(B\cup C)=A\cap(B\cup C)^c=A\cap(B^c\cap C^c)=(A\cap C^c)\cap B^c=(A-C)\cap B^c=\emptyset$$ hence $$A-C\subset B$$

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  • $\begingroup$ we cant use this $\endgroup$ – michael Feb 27 '14 at 22:37
  • $\begingroup$ What do you can use so? $\endgroup$ – user63181 Feb 27 '14 at 22:38
  • $\begingroup$ definitions like x belongs A-C, then x is in A, not in C $\endgroup$ – michael Feb 27 '14 at 22:42
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$A\setminus (B\cup C)=\emptyset\Rightarrow A\subseteq B\cup C\Rightarrow A\setminus C\subseteq B$. Done.

Another proof:

Because $A\setminus (B\cup C)=\emptyset$, we have that for all $x\in A$ it can't hold that $x\notin B\cup C$. So, for $x\in A$ we have $x\in B\cup C$ (by the way, this means that $A\subseteq B\cup C$). But if also $x\notin C$, meaning $x\in A\setminus C$, then $x\in (B\cup C)\setminus C= B$. So, $A\setminus C\subseteq B$.

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  • $\begingroup$ the question should be like is "-" not "\" $\endgroup$ – michael Feb 27 '14 at 22:39
  • $\begingroup$ @michael If you like it better that way, it's ok. I see you changed it. $\endgroup$ – frabala Feb 27 '14 at 22:45
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If you want to go for a purely logical proof then:

$x \in A \setminus C \implies x \in A $ and $x \not \in C ----(1)$

But $A \setminus (B \cup C) = \emptyset \implies $ There is no $x$ such that $x \in A$ and $x \not \in (B \cup C)$

By $(1)$, $ x \in A \implies x \in (B \cup C). $ But since $x \not \in C, \; x \in B \implies A \setminus C \subseteq B$

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