I have two problems that are giving me a bit of a difficulty. I've already completed one of them, but I have a feeling I'm doing something wrong, and would greatly appreciate feedback. For the other one, I'm lost entirely.
For each of the following sets, determine whether 4 is an element of that set.
a) {x ∈ R | x is an integer greater than 4}
b) {x ∈ R | x is the square of an integer|
c) {4,{4}}
d) {{4},{{4}}}
e) {{4},{4,{4}}}
f) {{{4}}}
I answered as the following:
a. As x is greater than 4, 4 is not in the set.
b. If the integer is 2, the square of 2 is 4, meaning that 4 is an element of the set. (Truthfully, I'm on the fence about this. How do I know that the square is 2?)
c. The set contains the value 4, and the subset {4}. 4 is in the set.
d. The set contains the subsets {4} and {{4}}. 4 is not an element in the set. (I assume that {4} isn't the same as 4)
e. The set contains the subset {4} and {4,{4}}. 4 is an element of the subset, but not the set.
f. The set consists wholly of the subset {{{4}}}. 4 is not an element.
The second problem I'm lost on completely is
Prove or disprove:
A - (B n C) = (A-B) u (A-C)
I know that A is equivalent to (A u A), so I can go from
A- (B n C) = (A-B) u (A-C)
(A u A) – (B n C) = (A-B) u (A-C)
But from there, I'm lost. I don't even know how A-B or A-C would translate into anything else, so any assistance would be greatly, greatly appreciated.
Best Answer
Your logic in 1 is fine. For part 2, to prove two sets are equal, show they are subsets of each other.
So, assume $x\in A-(B\cap C)$ Xo, $x\in A$ and $x\notin (B\cap C)$ Thus, we have $x\notin B$ or $x\notin C$ along with $x\in A$. thus, we have $x\in A-B$ or $x\in A-C$ Hence, $x\in (A-B)\cup(A-C)$
To argue the reverse, just start with an element of the right hand side and follow the same logic backward to the left hand side