To solve for the intersection of set $A$ and the complement of set $B$ within the universal set $U$, we follow these steps:

Step 1: Define the universal set $U$

The universal set $U$ is the set of all real numbers $x$ such that $3 < x < 14$. Therefore: $U = \{4, 5, 6, 7, 8, 9, 10, 11, 12, 13\}$

Step 2: Define sets $A$ and $B$

• $A = \{6, 7, 9, 10, 13\}$
• $B = \{4, 5, 11\}$

Step 3: Find the complement of set $B$ in $U$

The complement of $B$, denoted $B^c$, is the set of all elements in $U$ that are not in $B$: $B^c = U - B = \{6, 7, 8, 9, 10, 12, 13\}$

Step 4: Find the intersection of $A$ and $B^c$

Now, we compute the intersection $A \cap B^c$, which is the set of elements that are both in $A$ and in $B^c$: $A \cap B^c = \{6, 7, 9, 10, 13\} \cap \{6, 7, 8, 9, 10, 12, 13\}$ $A \cap B^c = \{6, 7, 9, 10, 13\}$

Final Answer:

The intersection $A \cap B^c = \{6, 7, 9, 10, 13\}$.

Would you like more details on this solution, or have any other questions?

Here are five related questions for further practice:

1. What is $A \cup B$?
2. What is $B \cap A^c$?
3. What is $A^c \cup B^c$?
4. What is $(A \cup B)^c$?
5. What is $(A \cap B)^c$?

Tip: When dealing with set complements, always remember to reference the universal set to ensure you include all elements not in the subset.