All of the following sets are subsets of positive integers.
$A = \{x\mid x\ \text{is divisible by 2}\} \\ B = \{x\mid x\ \text{is divisible by 4}\} \\ C = \{x\mid x\ \text{is divisible by 6}\}$
Below you will find what I think are the answers.
Two of the following six statements (A - F) are false. What statements are false?:
A. $A \subset B$ true my reasoning: A is a true subset to B, so A is not equal to B.
B. $C \subset A$ true my reasoning: C is a true subset to B, so C is not equal to B
C. $B \setminus A = \emptyset$ FALSE my reasoning: this almost means "everything in B belongs to A"
D. $A \cap B = B$ true my reasoning: the elements in A and B are the same as the elements in B, in A and B they are all multiples of 2, which they are in B as well.
E. $C \setminus B = \emptyset$ FALSE my reasoning: it means that all the elements in C are in B; in B they are multiples of 4, whilst in C they are multiples of 2 and 3, thus not true.
F. $B \cap C = \{x\mid x\ \text{is divisible by 12}\}$ true
Are my answers correct or not? If not, what are the correct answers, and why?
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$\begingroup$Hint:
Try writing out what each of $A,B,C$ look like explicitly.
$A = \{x~:~x~\text{is a positive integer divisible by 2}\} = \{2,4,6,8,10,12,\dots\}$
$B = \{x~:~x~\text{is a positive integer divisible by 4}\} = \{4,8,12,16,\dots\}$
$C = \{x~:~x~\text{is a positive integer divisible by 6}\} = \{6,12,18,24,\dots\}$
Remember also the definition of each of the symbols:
$A\subset B \Leftrightarrow (x\in A\Rightarrow x\in B)$
$x\in (B\setminus A)\Leftrightarrow (x\in B\wedge x\notin A)$
$x\in (A\cap B)\Leftrightarrow (x\in A \wedge x\in B)$
For the first statement $A\subset B$, this will be true if everything in $A$ is also in $B$. It will not be true if there is something in $A$ that is not in $B$. (I.e. is it true that every multiple of 2 is also a multiple of 4? Is 6 a multiple of 4?)
For the later statement, try proving the following statement:
$$E\subset F \Leftrightarrow E\cap F = E$$
and the similar statement for additional practice:
$$E\subset F \Leftrightarrow E\cup F = F$$
$\endgroup$ $\begingroup$B is a true subset of A: All integers divisible by 4 are also divisible by 2 but not the other way around.
C is a true subset to A as anything that is divisible by 6 must also be divisible by 2
As B is a subset of A, B without A is the empty set.
D: your answer is correct
E: your answer is correct
F: your answer is correct
$\endgroup$ 2 $\begingroup$Let's just look at the first one. Unfortunately you got that one wrong.
$A\subset B$ means that every $x\in A$ must also satisfy $x\in B$. That is, every $x$ that is a multiple of $2$ is also a multiple of $4$. This is false; for example, consider $2$ itself. $2\in A$ but $2\notin B$. Hence $A\not\subset B$.
The general principle in solving such problems is to consider the elements of the sets. They have properties, which are given as the defining properties of the sets.
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