Tao Analysis I - 3.1.12

Exercise 3.1.12

Suppose that $A$, $B$, $A'$, $B'$ are sets such that $A' \subseteq A$ and $B' \subseteq B$.

(i) Show that $A' \cup B' \subseteq A \cup B$ and $A' \cap B' \subseteq A \cap B$.

(ii) Give a counterexample to show that the statement $A' \setminus B' \subseteq A \setminus B$ is false. Can you find a modification of this statement involving the set difference operation $\setminus$ which is true given the stated hypotheses? Justify your answer.

Show that $A' \cup B' \subseteq A \cup B$

Let's start by writing down the meaning of $A' \cup B'$ using Axiom 3.5 which defines union.

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

The RHS is true in two cases, $x \in A'$ and $x \in B'$. Let's consider each case.

• If $x \in A'$ then, because we're given $A' \subseteq A$, we have $x \in A$.
• If $x \in B'$ then, because we're given $B' \subseteq B$, we have $x \in B$.

Either, or both, of these cases is true, which gives us a disjunctive implication,

$$x \in (A' \cup B') \implies (x \in A) \lor (x \in B)$$

This is equivalent to the statement $A' \cup B' \subseteq A \cup B$. $\square$

Show that $A' \cap B' \subseteq A \cap B$

Again, let's start with the definition of intersection to write down the meaning of $A' \cap B'$.

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

The RHS is only true if both $(x \in A')$ and $(x \in B')$ are true. Let's consider each statement.

• If $x \in A'$ then, because we're given $A' \subseteq A$, we have $x \in A$.
• If $x \in B'$ then, because we're given $B' \subseteq B$, we have $x \in B$.

Because both have to be true, we have a conjunctive implication.

$$x \in (A' \cap B') \implies (x \in A) \land (x \in B)$$

This is equivalent to the statement $A' \cap B' \subseteq A \cap B$. $\square$

Show $A' \setminus B' \subseteq A \setminus B$ is false

We'll show the statement is false with a counter-example. Consider the following sets:

• $A = \{1,2,3,4\}$, and $A' = \{2,3\}$, noting that $A' \subseteq A$ as required
• $B = \{2,3,4,5\}$, and $B' = \{2\}$, noting that $B' \subseteq B$ as required

Then $A' \setminus B' = \{3\}$, and $A \setminus B = \{1\}$.

Here $A' \setminus B' \subseteq A \setminus B$ means $ \{3\} \subseteq \{1\}$, which is false.

We have shown $A' \setminus B' \subseteq A \setminus B$ is false with the above counterexample. $\square$

Modification

We're asked to consider modifying the statement to make it true. Visualising the Venn diagram can help our thinking.

The following shows the sets $A$, $A'$, $B$ and $B'$ and every combination of intersection and union.

The visualising makes it easier to see that for $A' \setminus B' \subseteq A \setminus B$ to be true, the set $(A' \setminus B') \cap B$ must be empty.

$$\boxed{(A' \setminus B') \cap B = \emptyset}$$

Let's check this with an example. Consider the following sets:

• $A = \{1,2,3,4\}$, and $A' = \{2,3\}$, noting that $A' \subseteq A$ as required
• $B = \{2,4,6,8\}$, and $B' = \{2,4\}$, noting that $B' \subseteq B$ as required

Then $A' \setminus B' = \{3\}$, and $A \setminus B = \{1,3\}$. 

Here $A' \setminus B' \subseteq A \setminus B$ means $ \{3\} \subseteq \{1,3\}$, which is true.

The condition $(A' \setminus B') \cap B = \emptyset$ is  $\{3\} \cap \{2,4,6,8\} = \emptyset$ is true.