Is This Set Theory Proof Correct?

[lus1450]

Homework Statement

For all sets A, B, and C, prove or provide a counterexample the following statements.
(A \setminus B) \cap (C \setminus B) = A \setminus (B \cup C).

Homework Equations

N/A

The Attempt at a Solution

I went ahead and said it was false, and provided a counter example. I'm new to this and just want to make sure my thought process was correct and the statement is indeed false.
Counterexample:
Let A = {1,2,3}, B = {2,3}, and C = {1,2}.
A \setminus B = {1} and C \setminus B = {1}. Then (A \setminus B) \cap (C \setminus B) = {1}.
B \cup C = {1,2,3}. Then A \setminus (B \cup C) = {0}. Since {1} ≠ {0}, the statement is false.

It's more the last part, i.e. "Then A \setminus (B \cup C) = {0}" I want to make sure is correct. Thanks for your help!

 

[jbunniii]

Homework Statement

For all sets A, B, and C, prove or provide a counterexample the following statements.
(A \setminus B) \cap (C \setminus B) = A \setminus (B \cup C).

Homework Equations

N/A

The Attempt at a Solution

I went ahead and said it was false, and provided a counter example. I'm new to this and just want to make sure my thought process was correct and the statement is indeed false.
Counterexample:
Let A = {1,2,3}, B = {2,3}, and C = {1,2}.
A \setminus B = {1} and C \setminus B = {1}. Then (A \setminus B) \cap (C \setminus B) = {1}.
B \cup C = {1,2,3}. Then A \setminus (B \cup C) = {0}. Since {1} ≠ {0}, the statement is false.

This is almost right. Your notation {0} is incorrect. That refers to a set containing one element, the number 0. What you want is \emptyset, the empty set.

Aside from that your counterexample looks fine.

 

[lus1450]

I meant the empty set, but I guess 0 would be element and not empty. Thanks for telling the difference, I won't make that same mistake again.

 

[HallsofIvy]

If you want to use "regular" set notation, rather than ∅, it would be "{}", not "{0}".