On the Venn diagram, shade the region which represents $(A \cup B') \cap (A' \cup B)$. $\mathcal{E} = \{(x, y): x + y \le 10\}$
$A = \{(0, 1), (1, 2), (1, 4), (1, 5)\}$
$B = \{(1, 0), (1, 2), (1, 3), (2, 1), (3, 7)\}$ Underline the correct statements from the list below. $A \cap B = (1, 2) \qquad \{1, 5\} \subset A \qquad \emptyset \subset B \qquad (2, 6) \in (A \cup B)' \qquad A \subset B$

1. Identify the region for $A \cup B'$

The region $A \cup B'$ includes all of set $A$ and everything outside set $B$. This leaves only the region $B \cap A'$ unshaded.

2. Identify the region for $A' \cup B$

The region $A' \cup B$ includes all of set $B$ and everything outside set $A$. This leaves only the region $A \cap B'$ unshaded.

3. Find the intersection of the two regions

The intersection $(A \cup B') \cap (A' \cup B)$ consists of the regions common to both. This is the intersection $A \cap B$ and the region outside both sets $(A \cup B)'$.

1. Evaluate the first statement $A \cap B = (1, 2)$

The intersection of sets $A$ and $B$ is the set containing the common element $(1, 2)$. Thus, $A \cap B = \{(1, 2)\}$. The statement $A \cap B = (1, 2)$ is false because the right side is a coordinate pair, not a set.

2. Evaluate the second statement $\{1, 5\} \subset A$

The elements of $A$ are coordinate pairs. The pair $(1, 5)$ is in $A$, so $\{(1, 5)\} \subset A$. The set $\{1, 5\}$ contains the numbers 1 and 5, which are not elements of $A$. Thus, the statement is false.

3. Evaluate the third statement $\emptyset \subset B$

The empty set is a subset of every set. Thus, $\emptyset \subset B$ is true.

4. Evaluate the fourth statement $(2, 6) \in (A \cup B)'$

First, check if $(2, 6)$ is in the universal set $\mathcal{E}$: $2 + 6 = 8 \le 10$, so $(2, 6) \in \mathcal{E}$. Next, check if it is in $A \cup B$. Since $(2, 6)$ is neither in $A$ nor in $B$, it is not in $A \cup B$. Therefore, it is in the complement $(A \cup B)'$. This statement is true.

5. Evaluate the fifth statement $A \subset B$

For $A \subset B$ to be true, every element of $A$ must be in $B$. Since $(0, 1) \in A$ but $(0, 1) \notin B$, the statement is false.