在韦恩图上，阴影表示 $(A \cup B') \cap (A' \cup B)$ 的区域。 $E = \{(x,y): x + y \le 10\}$ $A = \{(0,1), (1,2), (1,4), (1,5)\}$ $B = \{(1,0), (1,2), (1,3), (2,1), (3,7)\}$ 在下方列表中正确的陈述下画横线。 A ∩ B = (1, 2) {1, 5} ⊂ A ∅ ⊂ B (2, 6) ∈ (A ∨ B)' A ⊂ B

1. Identify the set expression as the regions where membership of $A$ and $B$ is the same.

$$(A \cup B') \cap (A' \cup B)$$

For each Venn region:

• $A \cap B$: in $A$ and in $B$, so it is in both $A \cup B'$ and $A' \cup B$.
• $A \cap B'$: in $A$ but not in $B$, so it is not in $A' \cup B$.
• $A' \cap B$: not in $A$ but in $B$, so it is not in $A \cup B'$.
• $A' \cap B'$: not in $A$ and not in $B$, so it is in both $A \cup B'$ and $A' \cup B$.

Hence the required region is

$$(A \cap B) \cup (A' \cap B').$$

1. Find the common elements of $A$ and $B$ and check the first statement.

The only ordered pair that appears in both sets is $(1,2)$.

So
$A \cap B = \{(1,2)\}.$

The statement $A \cap B = (1,2)$ is not correct, since $A \cap B$ should be a set, not an ordered pair.

2. Check the subset statements involving $A$ and $B$.

For $\{1,5\} \subset A$, the elements $1$ and $5$ are not elements of $A$; the elements of $A$ are ordered pairs. So $\{1,5\} \subset A$ is false.

Since $B$ is non-empty, the empty set is a proper subset of $B$.

So $\varnothing \subset B$ is true.

For $A \subset B$, not every element of $A$ is in $B$. For example, $(0,1) \in A$ but $(0,1) \notin B$. So $A \subset B$ is false.

3. Check whether $(2,6)$ is in the complement of $A \cup B$ within $E$.

First check that $(2,6)$ is in the universal set $E$:
$2+6=8 \leq 10.$

So $(2,6) \in E$.

Also, $(2,6)$ is not listed in $A$ or $B$, so $(2,6) \notin A \cup B$.

Therefore
$(2,6) \in (A \cup B)'.$