Proving or disproving if sets are equal

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Does this mean we can ignore A because its a subset of B, so now were only dealing with B or C? So you substitute B or C with A. Am I right because that gives me the right answer. (C-C) and empty set = empty set or (B-b) and empty set = empty set. Please help, I made a reasonable attempt on this problem

Also, it's misleading to read $\cup$ as "or" (and $\cap$ as "and"). Read it as "union" (and "intersection"). — Gerry Myerson, Mar 01 '14 at 05:15
if your not gona help id rather have you not comment on this — Roy Kesserwani, Mar 01 '14 at 05:15
I've been here 3 years, you've been here 3 minutes. Please don't tell me what to do here. — Gerry Myerson, Mar 01 '14 at 05:16
OK, that's a much better problem statement. See --- I've helped you already! — Gerry Myerson, Mar 01 '14 at 05:17
I think you are misreading the statement. There is no "or" or "and", like you might find in a logic statement. This is talking about set membership, set unions, and set intersections. The statement says "If set $A$ is contained in the union of sets $B$ and $C$, then the part of $A$ outside of $C$ doesn't meet the part of $A$ outside of $B$." — MPW, Mar 01 '14 at 05:22
@MPW, that's what I wrote, too (I mean the part about "and" and "or"). Maybe Roy will believe it if it comes at him in stereo. Who is Gary? — Gerry Myerson, Mar 01 '14 at 05:24
Sorry I'm not so perfect, I feel like I'm getting slapped in the face, I'm just a student, take it easy :( — Roy Kesserwani, Mar 01 '14 at 05:25
Yea I wasn't so sure to who he was talking to, I was like who's Gary? — Roy Kesserwani, Mar 01 '14 at 05:26
I mean Gerry, I misread the name. Don't you two turn on me now ;) I deleted the comment anyway — MPW, Mar 01 '14 at 05:38
And, at the risk of really catching hell from you, you should have said "to whom he was talking". Hee hee. — MPW, Mar 01 '14 at 05:39

score 1 · Answer 1 · answered Mar 01 '14 at 06:48

Okay you seem to have a very feeble grasp on the fundamentals of Set Theory so I suggest you first try and read these 12 pages before reading my answer.

I will use the following rules and definitions in proving the answer.

$A \subseteq B \iff (x \in A \implies x \in B)$
$A \cup B = \text {the set of all $x$ such that $x \in A$ or $x \in B$ } $
$A \cap B = \text {the set of all $x$ such that $x \in A$ and $x \in B$ } $
$A - B = \text {the set of all $x$ such that $x \in A$ but $x \not \in B$}$

Let there be an element $x$ in $(A-C) \cap(A-B)$.

Then $x \in (A-C) $ and $x \in (A - B)$. Then we have the two statements - "$x$ is in $A$ but not in $C$" and "$x$ is in $A$ but not in $B$" - both of which must be satisfied. Assume the first statement is true. That is $x$ is in $A$ but not in $C$. The only way the second one too is satisfied is if $x$ is not in $C$. That is $x$ must be in $A$ but not in $B$ and not in $C$ i.e. $x$ must be in $A$ but not in $B$ or $C$. Can this occur? NO!! Why?

Because we are given that $A \subseteq B \cup C$. This is true if and only if the statement "If $x$ is in $A$ then $x$ is in $B \cup C$" which contradicts our development in the previous paragraph. Thus we arrive at a contradiction that says our initial assumption was false. What was that?

That there was an element $x$ in $(A-C) \cap(A-B)$.

Thus we assert that $(A-C) \cap(A-B)$ is empty.

Q.E.D.

score 0 · Answer 2 · answered Mar 01 '14 at 05:34

I think you can do it symbolically by just "subtracting" both sides of the hypothesis inclusion from $A$ and reversing the inclusion: $$A\subset B\cup C \implies A-(B\cup C) \subset A-(A)=\varnothing$$ Then use DeMorgan's Law to write $$(A-B)\cap(A-C)\subset\varnothing$$ which clearly implies $$(A-B)\cap(A-C)=\varnothing$$

Proving or disproving if sets are equal

2 Answers2