Proving Set Theorems Examples 1

Proving Set Theorems Examples 1

We will now look at some more examples of Proving Set Theorems. Before we look at these examples, please recall the following implications as they are fundamentally important for writing up our proofs.

• If $A \subseteq B$ and $x \in A$ implies $x \in B$.
• $x \in A \cup B$ implies $x \in A$ or $x \in B$.
• $x \not \in A \cup B$ implies $x \not \in A$ and $x \not \in B$
• $x \in A \cap B$ implies $x \in A$ and $x \in B$.
• $x \not \in A \cap B$ implies $x \not \in A$ or $x \not \in B$
• $x \in A \setminus B$ implies $x \in A$ and $x \not \in B$.
• $x \not \in A \setminus B$ implies $x \not \in A$ or $x \in B$.

Example 1

Prove that $A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)$.

• Proof: $\Rightarrow$ If $A \cup (B \cap C) = \emptyset$ then we are done. If not, then let $x \in A \cup (B \cap C)$. Then $x \in A$ or $x \in B \cap C$. If $x \in A$ then $x \in A cup B$ and $x \in A cup C$ so $x \in (A \cup B) \cap (A \cup C)$ and we're done. If $x \in B \cap C$, then $x \in B$ and $x \in C$. So $x \in A$ or $x \in B$ and $x \in A$ or $x \in C$ so $x \in (A \cup B) \cap (A \cup C)$. Therefore $A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)$.

Example 2

Let $S := \{ x \in \mathbb{R} : x^2 > x + 6 \}$ and let $T := \{ x \in \mathbb{R} : x > 3 \: \mathrm{or} \: x < -2 \}$. Prove that $S = T$.

This example is different from our past examples, but we will tackle it on in the same approach.

• Proof: $\Rightarrow$ We first want to show $S \subseteq T$. First let $x \in S$. So $x \in \mathbb{R}$ and $x^2 > x + 6$. Using some algebraic manipulation we see that $x^2 - x - 6 > 0$ or rather $(x + 2)(x - 3) > 0$. There are two cases to consider.