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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.

Faraam
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2 Answers2

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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

Alan
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For your set equality, draw a Venn diagram with circles for A, B, and C. You can then visually see what the two sides are and why they are equal.

marty cohen
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