Tao Analysis I - 3.1.10

Exercise 3.1.10

Let $A$ and $B$ be sets. Show that the three sets $A \setminus B$, $A \cap B$, and $B \setminus A$ are disjoint, and that their union is $A \cup B$.

To show the three sets $A \setminus B$, $A \cap B$, and $B \setminus A$ are disjoint, we need to show that a member of one is not a member of the other two.

Let's use the definitions in the textbook to write out what these set descriptions mean.

• $(A \setminus B)$ means a member is in $A$ but not in $B$:

$$x \in (A \setminus B) \iff (x \in A) \land (x \not \in B) $$

• $(A \cap B)$ means a member is in both $A$ and $B$:

$$x \in (A \cap B) \iff (x \in A) \land (x  \in B) $$

• $(B \setminus A)$ means a member is in $B$ but not $A$:

$$x \in (B \setminus A) \iff (x \in B) \land (x \not \in A) $$

Now if $x$ is a member of $(A \setminus B)$ then it is not a member of $B$. The descriptions above tell us $x$ is not in $(A \cap B)$, and it is not in (B \setminus A), because both would require it to be in $B$.

If $x$ is a member of $(A \cap B)$ then it is a member of both $A$ and $B$. The descriptions above tell us $x$ is not in $(A \setminus B)$, because that would require $x$ to be in $B$. Also, the descriptions tell us $x$ is not in $(B \setminus A)$, because that would require it to be in $A$.

Finally, if $x$ is a member of $(B \setminus A)$ then it is not in $(A \cap B)$, and it is not in $(A \cap B)$., because both would require it to be in $A$.

We have shown that the three sets $A \setminus B$, $A \cap B$, and $B \setminus A$ are disjoint. $\square$

We now want to show the union of all three $A \setminus B$, $A \cap B$, and $B \setminus A$ is $A \cup B$. To do this we need to show that a member of any of the three sets is in $A \cup B$, and also that a member of $A \cup B$ is in one of the three sets.

First let's remind ourselves of the definition of union.

$$x \in (A \cup B) \iff (x \in A) \lor (x \in B)$$

Now, 

• If $x \in (A \setminus B)$ then because by definition $x \in A$ and so $x \in (A \cup B)$.
• If $x \in (A \cup B)$ then because by definition $x \in A$  and $x \in B$, and so $x \in (A \cup B)$.
• If $x \in (B \setminus A)$ then because by definition $x \in B$ and so $x \in (A \cup B)$.

So we have that if $x$ is a member of any of the three sets, then it is a member of $A \cup B$.

Finally, if $x \in (A \cup B)$ then we have two cases, $x \in A$ or $x \in B$:

• $x \in A$. Using the description earlier, this means $x$ is in $(A \setminus B)$ or $x \in (A \cup B)$
• $x \in B$. Using the description earlier, this means $x$ is in $(B \setminus A)$ or $x \in (A \cup B)$

We have shown that a member of any of the three sets is in $A \cup B$, and also that a member of $A \cup B$ is in one of the three sets. Thus we have proved $A \setminus B$, $A \cap B$, and $B \setminus A$ is $A \cup B$. $\square$