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How do I prove this:

Let $A, B$ and $C$ be sets, $A \subseteq (B \cup C)$ and $B \subseteq (A \cup C)$ then $(A - B) \subseteq C$

How about this:

Let $x \in A$ and $y \in B.$

Since $A \subseteq (B \cup C)$ then $x \in (B \cup C).$

Since $B \subseteq (A \cup C)$ then $y \in (A \cup C)$

$x \in B \cup C$, so if $x \in B$, then $x \notin A - B.$
if $x \notin B$, then $x \in C.$

$y \in A \cup C$, so if $y \in A$, then $y \notin A - B.$ if $y \notin B$, then $y \in C.$

So in both cases, $A - B \subseteq C$

Is it correct? if yes, the converse is false right?

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You don't need both conditions. $A\subset(B\cup C)\Rightarrow (A\setminus B)\subset C$. – Owen Sizemore Aug 16 '13 at 20:18
@OwenSizemore These are actually equivalent. See – Marnix Klooster Nov 19 '13 at 7:09

The question only asks you to prove that $A-B$ is a subset of $C$. This means that we need to take $x\in A-B$ and show that $x\in C$.

Let $x$ be such element, then $x\in A$ and therefore $x\in B\cup C$. However, $x\notin B$ and therefore $x\in C$.

In the other direction, it is true that if $A-B\subseteq C$ then $A\subseteq B\cup C$. See if you can prove it.

(Hint: $x\in A$ then either $x\in B$ or $x\notin B$.)

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SO to recap $A - B \subset C$ if and only if $A \subset B \cup C$. – N. S. Aug 16 '13 at 20:24

You stated more than the necessary assumptions, as pointed out by a comment and an answer that only used $A \subseteq B \cup C$. You can check the converse if you only need to prove the condition $A \subseteq B \cup C$. So for the "full converse" the way you stated it, the only additional question that remains is whether $A-B \subseteq C$ implies $B \subseteq A \cup C$. Hint: Consider $C$ to be the empty set, and when $A \subset B$, where $A \neq B$.

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Well... The converse will be this: Let A,B and C be sets, (A−B)⊆C then A⊆(B∪C) and B⊆(A∪C)

I think this is false. Take the negation and proof:

Negation: Let A,B and C be sets, (A−B)⊆C then A Not⊆ (B∪C) or B Not⊆ (A∪C)

Here is a counter example:

A = {1, 2} B = {1, 4} C = {2, 5}

So A - B = {2} and is a subset of C.

But, A U C = {1, 2, 5} - but B = {1, 4}. So B is not a subset of AUC. Element 4 is missing in AUC.


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Guys please tell me what is wrong with this??? thx... a -1 flag doesn't help much... – Najeeb Aug 16 '13 at 20:50
This is correct. :) Found an example in the book. – Najeeb Aug 16 '13 at 21:07

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