Simple Set Theory Proofs

1. May 27, 2015

B3NR4Y

1. The problem statement, all variables and given/known data
1. Prove that if $A \cap B = A$ and $A \cup B = A$, then $A = B$
2. Show that in general $(A-B) \cup B \neq A$
3. Prove that $(A-B) \cap C = (A \cap C) - (B \cap C)$
4. Prove that $\cup_{\alpha} A_{\alpha} - \cup_{\alpha} B_{\alpha} \subset \cup_{\alpha} (A_{\alpha} - B_{\alpha})$
2. Relevant equations
None that I can think of.

3. The attempt at a solution
1. For the first one, it seems pretty clear to me going by the logic of the operations $\cap$ and $\cup$, because $A \cup B$ is both the sets joined, and if that equals A, B must be the empty set or A. The second part says $A \cap B = A$, the intersection of two sets is only what the two sets have in common, therefore B must be equivalent to A because in the previous part we found $B=A$ or $B = \emptyset$ and it cannot be the empty set because the empty set intersected with any set is the empty set. I came to the correct conclusion but I don't think my logic is sound.

2. I can show by an example, if $A = {1, 2, 3}$ and $B = {5, 6 ,7}$, therefore $A- B = {1 , 2, 3}$ and $(A-B) \cup B = {1,2,3,5,6,7} \neq A$. This is just one example and I know this doesn't prove it in general. But I'm not sure how to start.

3. If $x \in (A-B)\cap C$ then $x\in (A-B)$ AND $x\in C$, since $x \in (A-B), x\notin B$ by the definition of subtraction of sets. Therefore $x\in A\cap C$ and $x\notin B\cap C$. Therefore x is in the subtraction of $A\cap C$ and $B\cap C$. Which proves the identity for any arbitrary element $x$..... right? (the best way to end a proof is ".... right?")

I am not sure where to start for 4.

2. May 27, 2015

Zondrina

I believe your logic is quite sound for question 1. Near the end to clarify the argument, I would have said $B$ cannot be the empty set because the intersection of the empty set with anything is the empty set. We know $A$ intersect $B$ is equal to $A$ and not the empty set, so it must be the case $A = B$.

2. Your logic for question 2 is good as well. You have found a counter example; One counter example is enough to verify the claim in general.

3. Right.

3. May 27, 2015

B3NR4Y

Okay, cool. I'm not as bad as I thought!

4. May 27, 2015

Zondrina

4. Prove that: $\cup_{\alpha} A_{\alpha} - \cup_{\alpha} B_{\alpha} \subset \cup_{\alpha} (A_{\alpha} - B_{\alpha})$.

To gain some insight, lets write:

$$\cup_{\alpha} A_{\alpha} - \cup_{\alpha} B_{\alpha} \subset \cup_{\alpha} (A_{\alpha} - B_{\alpha})$$
$$(A_1 \cup \cdots \cup A_n) - (B_1 \cup \cdots \cup B_n) \subset (A_1 - B_1) \cup \cdots \cup (A_n - B_n)$$

Can you show every element in the set on the left is contained in the set on the right? That is, $\forall x \in (A_1 \cup \cdots \cup A_n) - (B_1 \cup \cdots \cup B_n)$ can you show $x \in (A_1 - B_1) \cup \cdots \cup (A_n - B_n)$?

5. May 27, 2015

Ray Vickson

6. May 27, 2015

vela

Staff Emeritus
If A = {1, 2, 3, 4} and B = {2, 3}, then $A \cup B = \{1,2,3,4\}=A$ but $B \ne A$ and $B \ne \emptyset$.

7. May 28, 2015

wabbit

Sorry but this is completely wrong. $A\cap B =A$ does not imply in any way that B should be A or the empty set.

@B3NR4Y, try to think again about what $A\cap B=A$ means. First, do you see that this is equivalent to $A\subset A\cap B$ ?

8. May 28, 2015

PeroK

I guess you mean that $A\cap B=A$ is equivalent to $A\subset B$

9. May 28, 2015

wabbit

No, I meant exactly what I said. I was trying to get OP think by himself by only spelling out a minimal step.

10. May 28, 2015

HallsofIvy

I would try to be much more specific. To prove "A= B" you need to prove "A is a subset of B" and "B is a subset of A".

To prove "A is a subset of B" start with "If x is an element of A" and use the properties of both A and B to conclude "x is an element of B".

To prove " if $A \cap B = A$ and $A \cup B = A$, then $A = B$"
start, "if x is an element of A, then, because $A \cap B= A$ x is an element of B. Therefore A is a subset of B.
if y is an element of B then $A\cup B= A$, y is an element of A. Therefore B is a subset of A. Therefore A= B."

11. May 28, 2015

B3NR4Y

Okay so for 3, $x\in (A\cap C) - (B\cap C)$, which implies $x\in A \cap C$ but $x\notin B\cap C$ and from that it follows $x \in A,C$ and $x\notin B$, therefore $\forall x\in (A\cap C) - (B\cap C), \, \, x\in (A-B)\cap C$ which proves $(A\cap C) - (B\cap C)\subset (A-B)\cap C$ and I've already proven that the converse is true, so this completes the proof?

Okay I think I understand it better now, to prove two things are equivalent I need to prove the are subsets of each other, and for problems like 4, I only need to prove that any arbitrary "x" in the first part is also in the second part.

I am working from "Introductory Real Analysis" by Kolmogorov and Fomin.