Simple Set Theory Proofs to Proving Set Identities - Homework Help

[B3NR4Y]

Homework Statement

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 $\cup_{\alpha} A_{\alpha} - \cup_{\alpha} B_{\alpha} \subset \cup_{\alpha} (A_{\alpha} - B_{\alpha})$

Homework Equations

None that I can think of.

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.

 

[STEMucator]

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.

4. I will think about this more after dinner.

 

[B3NR4Y]

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.

4. I will think about this more after dinner.

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

 

[STEMucator]

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

To gain some insight, let's 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 $x \in (A_1 - B_1) \cup \cdots \cup (A_n - B_n)$?

 

[Ray Vickson]

Homework Statement

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 $\cup_{\alpha} A_{\alpha} - \cup_{\alpha} B_{\alpha} \subset \cup_{\alpha} (A_{\alpha} - B_{\alpha})$

Homework Equations

None that I can think of.

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.

*********************
I think your first assertion (that if A and B joined equals A, then B = A or B = empty) is true, but unproved; just saying it does not constitute a proof.

Better, I think is an argument like the following: (i) $A \cap B$ is a subset of $B$, and we are saying the former $=A$, so $A \subset B$. (ii) $B$ is a subset of $A \cup B$ and we are saying the latter $=A$, so $B \subset A$. Since $A \subset B$ and $B \subset A$ we have $A = B$.

***********************

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.

**********************
I think giving a counterexample illustrates falsity of the equality and so proves very clearly that the equality is not always true. I'm not sure what the person posing the problem regards as an "in general" proof.

***********************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?")

************************

You have proved that $(A-B) \cap C \subset A \cap C - B \cap C$. In order to establish equality, you need to prove as well that the right-hand set is a subset of the left-hand set.

************************

I am not sure where to start for 4.

 

[vela]

Homework Statement

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 $\cup_{\alpha} A_{\alpha} - \cup_{\alpha} B_{\alpha} \subset \cup_{\alpha} (A_{\alpha} - B_{\alpha})$

Homework Equations

None that I can think of.

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.

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

 

[wabbit]

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 $A = B$.

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 $A\subset A\cap B$ ?

 

[PeroK]

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 $A\subset A\cap B$ ?

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

 

[wabbit]

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

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

 

[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."

 

[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?

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."

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.