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$$B \cap C \subseteq A \implies (C-A) \cap (B-A) = \varnothing.$$

I don't think this is true because B and neither C are necessarily a subset of A. Only B intercept C is a subset of A.

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how come b intercept c is a subset of a implies that b and c are a subset of a? i can draw a diagram where only b intercept c is a subset of a and a = b intercept c. –  hjggh Oct 30 '12 at 19:18
    
The comment I made was when you had $B\cup C$ at the beginning, not $B\cap C$. Since it is not relevant to the changed question, I am deleting it. –  André Nicolas Oct 30 '12 at 19:25
    
Just so you're clear and for future reference: $B \cap C$ reads: "B intersect C"; i.e., the word to use is "intersect", not "intercept"...Also, to clear up any confusion: $(C - A)$ is often/usually denoted: "$(C \setminus A$)". (In Tex, that's (C \setminus A), enclosed in "$" signs here at MathSE.) –  amWhy Oct 30 '12 at 19:54
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5 Answers

With the intersection, think about what the equation means. Suppose you have an element of B which is not in A, can it be in $B\cap C$? Can it be in $C$? Can it be in $C-A$?

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Revised to match the corrected problem statement:

If $B\cap C\subseteq A$, then $(B\cap C)\setminus A=\varnothing$. But $$(B\cap C)\setminus A=(B\setminus A)\cap(C\setminus A)\;,\tag{1}$$

so if $(B\cap C)\setminus A=\varnothing$, then $(B\setminus A)\cap(C\setminus A)=\varnothing$.

Proving $(1)$ is a good exercise, and not hard.

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And the premise is about neither intercept nor intersect, but union. –  Hagen von Eitzen Oct 30 '12 at 19:07
    
i used the wrong icon, i wanted to use this:⊆ –  hjggh Oct 30 '12 at 19:15
    
i used the wrong one again ∪, i wanted to use the cap –  hjggh Oct 30 '12 at 19:16
    
@amWhy: I went a bit overboard and wrote nothing instead of \varnothing! Thanks. –  Brian M. Scott Oct 30 '12 at 19:33
    
np...I knew what you meant...just didn't want to leave any room for confusion! :) –  amWhy Oct 30 '12 at 19:36
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The statement is true.

We prove the contrapositive.

Suppose that $(B\setminus A)\cap (C\setminus A) \neq \emptyset$. Then there is some $x \in (B\setminus A) \cap (C\setminus A)$. By definition $x\notin A$, however $(B\setminus A) \cap (C\setminus A) \subset B \cap C$. So we have $x \in B\cap C$ and $x \notin A$,

So $B\cap C$ is not a subset of $A$.

EDIT: To expand this explaining every step more carefully, hopefully this helps.

Proving the contrapositive means we show that $\neg (C\setminus A) \cap (B\setminus A) = \emptyset \rightarrow \neg B\cap C \subset A$, which is equivalent to the original statement.

If we assume $(B\setminus A) \cap (C\setminus A) \neq \emptyset$ then this means that this set has an element, so we take some $x\in (B\setminus A) \cap (C\setminus A)$. Then since $(B\setminus A)\cap (C\setminus A)$ is the set of elements which are in $B$ and $C$ and not in $A$, this means that $x\in B\cap C$ and $x\notin A$. Therefore we have produced an element, $x$, which is in $B\cap C \setminus A$ which means that $B\cap C$ can't be a subset of $A$.

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i can't understand anything you wrote. –  hjggh Oct 30 '12 at 20:25
    
@MathildaPitt I have edited hopefully it is more clear. –  Deven Ware Oct 30 '12 at 20:31
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$$B \cup C \subset A \implies B \subset A \text{ and } C \subset A$$ Hence, $$C - A = \emptyset = B-A$$

EDIT

Since the OP changed the question to $B \cap C \subset A$, instead of $B \cup C \subset A$, below is the revised answer. $$(C-A) \cap (B-A) = (C \cap A^c) \cap (B \cap A^c) = (B \cap C) \cap A^c \subseteq A \cap A^c = \emptyset$$

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Where did $B \cup C \subseteq A$ come from? –  Daniel Littlewood Oct 30 '12 at 21:15
    
@DanielLittlewood THE OP had $B \cup C$ initially instead of $B \cap C$. Will change it now. –  user17762 Oct 30 '12 at 21:16
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$(C \setminus A) \cap (B \setminus A) = \emptyset$ means that all elements are in $((C \setminus A) \cap (B \setminus A) )'=(C-A)'\cup(B-A)'$ by De Morgan,
$=(C'\cup A) \cup (B' \cup A)$ combining an alternate definition of $A \setminus B$ and De Morgan again,
$=(C'\cup B')\cup(A\cup A)=(C\cap B)' \cup A$
Since $B \cap C \subseteq A$, $A' \subseteq (B\cap C)'$, so our union is indeed the whole set.

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