- #1
B3NR4Y
Gold Member
- 170
- 8
Homework Statement
1. Prove that if [itex] A \cap B = A [/itex] and [itex] A \cup B = A [/itex], then [itex] A = B [/itex]
2. Show that in general [itex] (A-B) \cup B \neq A [/itex]
3. Prove that [itex](A-B) \cap C = (A \cap C) - (B \cap C)[/itex]
4. Prove that [itex] \cup_{\alpha} A_{\alpha} - \cup_{\alpha} B_{\alpha} \subset \cup_{\alpha} (A_{\alpha} - B_{\alpha})[/itex]
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 [itex] \cap [/itex] and [itex] \cup [/itex], because [itex] A \cup B [/itex] is both the sets joined, and if that equals A, B must be the empty set or A. The second part says [itex] A \cap B = A [/itex], 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 [itex] B=A [/itex] or [itex] B = \emptyset [/itex] 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 [itex] A = {1, 2, 3} [/itex] and [itex] B = {5, 6 ,7} [/itex], therefore [itex] A- B = {1 , 2, 3} [/itex] and [itex] (A-B) \cup B = {1,2,3,5,6,7} \neq A [/itex]. 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 [itex] x \in (A-B)\cap C [/itex] then [itex]x\in (A-B)[/itex] AND [itex]x\in C[/itex], since [itex]x \in (A-B), x\notin B[/itex] by the definition of subtraction of sets. Therefore [itex] x\in A\cap C[/itex] and [itex] x\notin B\cap C [/itex]. Therefore x is in the subtraction of [itex]A\cap C[/itex] and [itex]B\cap C[/itex]. Which proves the identity for any arbitrary element [itex]x[/itex]... right? (the best way to end a proof is "... right?")
I am not sure where to start for 4.