Exercise 1.4.2

Proof. $\text{(→)}$ First assume that $A\subseteq B$.

To show that $A\cap B=A$, first consider $x\in A\cap B$, then clearly $x\in A$ so that $A\cap B\subseteq A$. Now suppose that $x\in A$ so that also $x\in B$ since $A\subseteq B$. Hence, $x\in A$ and $x\in B$ so that $x\in A\cap B$, showing that $A\subseteq A\cap B$.

Next we show that $A\cup B=B$. First consider any $x\in A\cup B$. If $x\in A$ then also $x\in B$ since $A\subseteq B$. Since, in the other case, we have that $x\in B$ directly, it follows that $A\cup B\subseteq B$. Now suppose that $x\in B$ so that clearly also $x\in A\cup B$, and so $B\subseteq A\cup B$ as well.

Finally, we show that $A-B=\varnothing$. To see this, assume first that $A-B\ne \varnothing$ so that there is an $x\in A-B$. Thus, $x\in A$ but $x\notin B$. However, $x\in A$ implies that $x\in B$ since $A\subseteq B$, which is a contradiction.

$\text{(←)}$ First suppose that $A\cap B=A$ and consider any $x\in A$ so that also $x\in A\cap B$, and hence $x\in B$ as well. This shows that $A\subseteq B$.

Next, suppose that $A\cup B=B$ and consider any $x\in A$. Then $x\in A\cup B=B$, which suffices to show that $x\in B$ and thus $A\subseteq B$.

Lastly, suppose that $A-B=\varnothing$ and consider any $x\in A$. Were it the case that $x\notin B$ then it would follow that $x\in A-B=\varnothing$, an impossibility. Hence, it has to be that $x\in B$ as well so that $A\subseteq B$ again.

This all suffices to show that

as desired. □

Proof. $\text{(→)}$ Suppose first that $A\subseteq B\cap C$ and consider any $x\in A$. Then also $x\in B\cap C$ so that both $x\in B$ and $x\in C$. As $x$ was arbitrary, this shows that both $A\subseteq B$ and $A\subseteq C$.

$\text{(←)}$ Now suppose that $A\subseteq B$ and $A\subseteq C$ and consider any $x\in A$. Then both $x\in B$ since $A\subseteq B$, and $x\in C$ since $A\subseteq C$. Therefore, $x\in B\cap C$, showing that $A\subseteq B\cap C$. □

Proof. $\text{(→)}$ First suppose that $B\cup C\subseteq A$ and consider any $x\in B$. Then $x\in B\cup C$ so that also $x\in A$, showing that $B\subseteq A$. Similarly, for any $x\in C$ we have that again $x\in B\cup C$ so that $x\in A$. This shows that $C\subseteq A$ as well.

$\text{(←)}$ Now suppose that both $B\subseteq A$ and $C\subseteq A$ and consider any $x\in B\cup C$. In the case in which $x\in B$, we have that $x\in A$ since $B\subseteq A$. In the other case in which $x\in C$ we also have $x\in A$ since $C\subseteq A$ as well. This shows that $B\cup C\subseteq A$ since $x$ was arbitrary. □

Proof. $\text{(⊆)}$ Suppose that $x\in A-B$ so that $x\in A$ and $x\notin B$. Then of course also $x\in A\cup B$ since $x\in A$. Since also $x\notin B$ it follows that $x\in (A\cup B)-B$, showing that $A-B\subseteq (A\cup B)-B$. We also have that $x\notin A\cap B$ since $x\notin B$. Hence, also $x\in A-(A\cap B)$ so that $A-B\subseteq A-(A\cap B)$ as well.

$\text{(⊇)}$ Suppose that $x\in (A\cup B)-B$. Then $x\in A\cup B$ and $x\notin B$. Since it must be that $x\in A$ or $x\in B$, but $x\notin B$, it must be that $x\in A$. Hence, $x\in A-B$, which proves that $(A\cup B)-B\subseteq A-B$.

Now suppose that $x\in A-(A\cap B)$ so that $x\in A$ and $x\notin A\cap B$. Since $x\in A$ it has to be that $x\notin B$ since otherwise it would be that $x\in A\cap B$. Therefore, $x\in A$ and $x\notin B$ so that $x\in A-B$, showing that also $A-(A\cap B)\subseteq A-B$.

This all is sufficient to show that

as desired. □

Proof. This is shown with some basic rules of logic. For any $x$, we have

and it is as simple as that. □

Proof. Consider any $x$. Let $F$ be the proposition that $x\in A$ and $x\notin A$, which is of course always false. We then have

showing the desired result. □

Proof. Here we have, for any $x$,

which proves the result. □

Proof. $\text{(→)}$ Suppose that $A=B$. Then $A△B=A△A=\varnothing$ by a property of the symmetric difference that was listed in the text and proven in Exercise 1.4.1 above.

$\text{(←)}$ Now suppose that $A△B=\varnothing$. Then it follows that both $A-B=\varnothing$ and $B-A=\varnothing$ since otherwise we would have that $A△B=(A-B)\cup (B-A)\ne \varnothing$.

Now consider any $x\in A$ so that of course $x\notin A-B=\varnothing$. Then, by Lemma 1, we have that $x\notin A$ or $x\in B$. Hence, $x\in B$ since we have already established that $x\in A$. This shows that $A\subseteq B$.

Now consider any $x\in B$ so that, similarly, $x\notin B-A=\varnothing$. Then, again by Lemma 1, $x\notin B$ or $x\in A$ so that $x\in A$ since $x\in B$. This shows that $B\subseteq A$ as well, and therefore that $A=B$ as desired. □