Tao Analysis I - 3.1.7

Exercise 3.1.7

Let $A$, $B$, $C$ be sets. Show that $A \cap B \subseteq A$ and $A \cap B \subseteq B$. Furthermore, show that $C \subseteq A$ and $C \subseteq B$ if and only if $C \subseteq A \cap B$. In a similar spirit, show that $A \subseteq A \cup B$ and $B \subseteq A \cup B$, and furthermore that $A \subseteq C$ and $B \subseteq C$ if and only if $A \cup B \subseteq C$.

Show $A \cap B \subseteq A$ and $A \cap B \subseteq B$

Let's start with Definition 3.1.22 of intersection,

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

Both $(x \in A)$ and $(x \in B)$ are true. This gives us

$$\begin{align}x \in (A \cap B) &\implies (x \in A) \\ \\ x \in (A \cap B) &\implies (x \in B) \end{align}$$

This is equivalent to the two statements we want to prove $A \cap B \subseteq A$ and $A \cap B \subseteq B$. $\square$

Show $C \subseteq A$ and $C \subseteq B$ if and only if $C \subseteq A \cap B$

Let's start by writing down the meanings of $C \subseteq A$ and $C \subseteq B$,

$$\begin{align}(x \in C) &\implies (x \in A) \\ \\ (x \in C) &\implies (x \in B)  \end{align}$$

Both of these statements are true, so we have

$$(x \in C) \implies (x \in A) \land (x \in B)$$

This gives us $(C \subseteq A) \land (C \subseteq B) \implies C \subseteq (A \cap B)$.

We now need to show $C \subseteq (A \cap B) \implies (C \subseteq A) \land (C \subseteq B)$. Let's apply the definition of intersection to $C \subseteq (A \cap B)$

$$\begin{align}x \in C &\implies x \in (A \cap B) \\ \\ & \implies (x \in A) \land (x \in B)\end{align}$$

This gives us two statements which are true,

$$\begin{align}(x \in C) &\implies (x \in A) \\ \\ (x \in C) &\implies (x \in B)  \end{align}$$

This is equivalent to $C \subseteq (A \cap B) \implies (C \subseteq A) \land (C \subseteq B)$.

Having shown both

$$(C \subseteq A) \land (C \subseteq B) \implies C \subseteq (A \cap B)$$

and

$$C \subseteq (A \cap B) \implies (C \subseteq A) \land (C \subseteq B)$$

we can conclude 

$$(C \subseteq A) \land (C \subseteq B) \iff C \subseteq (A \cap B) \; \square$$

Show $A \subseteq A \cup B$ and $B \subseteq A \cup B$

The definition of union, Axiom 3.5, tells us that $x \in X \implies x \in (X \cup Y)$ for any $Y$. 

So we have

$$\begin{align}x \in A &\implies x \in (A \cup B) \\ \\ x \in B &\implies x \in (A \cup B) \end{align}$$

This immediately gives us $A \subseteq A \cup B$ and $B \subseteq A \cup B$. $\square$

Show $A \subseteq C$ and $B \subseteq C$ if and only if $A \cup B \subseteq C$

Let's start by writing down the meanings of $A \subseteq C$ and $B \subseteq C$,

$$\begin{align}(x \in A) &\implies (x \in C) \\ \\ (x \in B) &\implies (x \in C)  \end{align}$$

Either, or both, statements $(x \in A)$ and $(x \in B)$ implies $(x \in C)$, which we write as

$$(x \in A) \lor (x \in B) \implies (x \in C)$$

This is equivalent to $(A \subseteq C) \lor (B \subseteq C) \implies (A \cup B \subseteq C)$.

We need to show also $(A \cup B \subseteq C) \implies (A \subseteq C) \lor(B \subseteq C)$. Let's apply the definition of union,

$$(x \in A) \lor (x \in B) \implies x \in C$$

This tells us that either, or both, $(x \in A) \implies (x \in C) $ and $(x \in B) \implies (x \in C)$ are true. That is, $(A \subseteq C) \lor(B \subseteq C)$.

Having shown both

$$(A \subseteq C) \lor (B \subseteq C) \implies (A \cup B \subseteq C)$$

and

$$(A \cup B \subseteq C) \implies (A \subseteq C) \lor(B \subseteq C)$$

we can conclude

$$A \subseteq C \land B \subseteq C \iff A \cup B \subseteq C \; \square$$