$\epsilon = \{\text{integers } x : 2 \le x \le 13\}$ $A = \{\text{prime numbers}\}$ $B = \{\text{multiples of 4}\}$ $C = \{\text{factors of 12}\}$ 列出以下集合中的元素：

1. List the elements of the universal set and identify the elements of $B$.

$\epsilon=\{2,3,4,5,6,7,8,9,10,11,12,13\}$

$B=\{\text{multiples of }4\}=\{4,8,12\}$

2. Find the complement $B'$ within the universal set.

$B'$ means the elements in $\epsilon$ that are not in $B$.

So remove $4,8,12$ from $\epsilon$:

$B'=\{2,3,5,6,7,9,10,11,13\}$

1. List the elements of $A$ and use $B'$ from part (a).

Within $\epsilon=\{2,3,4,5,6,7,8,9,10,11,12,13\}$,

$A=\{2,3,5,7,11,13\}$.

From part (a),

$B'=\{2,3,5,6,7,9,10,11,13\}$.

2. Find the intersection of $A$ and $B'$.

$A\cap B'$ means the elements that are in both $A$ and $B'$.

So,

$A\cap B'=\{2,3,5,7,11,13\}$.

1. Find the elements in $A \cup B$ within the universal set.

$\epsilon=\{2,3,4,5,6,7,8,9,10,11,12,13\}$

$A=\{2,3,5,7,11,13\}$, since these are the prime numbers in $\epsilon$.

$B=\{4,8,12\}$, since these are the multiples of 4 in $\epsilon$.

So
$A\cup B=\{2,3,4,5,7,8,11,12,13\}$.

2. Find the complement of $A \cup B$ within the universal set.

$(A\cup B)'$ means the elements in $\epsilon$ that are not in $A\cup B$.

$(A\cup B)'=\{6,9,10\}$

1. List the elements of $B$ and $C$ within the universal set.

The universal set is $\epsilon=\{2,3,4,5,6,7,8,9,10,11,12,13\}$.

$B=\{\text{multiples of 4}\}=\{4,8,12\}$

$C=\{\text{factors of 12}\}=\{2,3,4,6,12\}$

2. Find the elements common to both $B$ and $C$.

$B\cap C$ means the elements that are in both $B$ and $C$.

So,
$B\cap C=\{4,12\}.$