Exercise 1.2

Proof. Suppose that $A\subset B$ and $A\subset C$ and consider any $x\in A$. Then clearly also $x\in B$ since $A\subset B$ so that $x\in B\cup C$. Since $x$ was arbitrary, this shows that $A\subset (B\cup C)$ as desired.

To show that the converse is not true, suppose that $A=\left\{1,2,3\right\}$, $B=\left\{1,2\right\}$, and $C=\left\{3,4\right\}$. Then clearly $A\subset \left\{1,2,3,4\right\}=B\cup C$ but it neither true that $A\subset B$ (since $3\in A$ but $3\notin B$) nor $A\subset C$ (since $1\in A$ but $1\notin C$). □

Proof. Suppose that $A\subset B$ or $A\subset C$ and consider any $x\in A$. If $A\subset B$ then clearly $x\in B$ so that $x\in B\cup C$. If $A\subset C$ then clearly $x\in C$ so that again $x\in B\cup C$. Since $x$ was arbitrary, this shows that $A\subset (B\cup C)$ as desired.

The counterexample that disproves the converse of part (a), also serves as a counterexample to the converse here. Again this is because $A\subset B\cup C$ but neither $A\subset B$ nor $A\subset C$, which is to say that $A\not\subset B$ and $A\not\subset C$. Hence it is not true that $A\subset B$ or $A\subset C$. □

Proof. $\text{(⇒)}$ Suppose that $A\subset B$ and $A\subset C$ and consider any $x\in A$. Then clearly also $x\in B$ and $x\in C$ since both $A\subset B$ and $A\subset C$. Hence $x\in B\cap C$, which proves that $A\subset B\cap C$ since $x$ was arbitrary.

$\text{(⇐)}$ Now suppose that $A\subset B\cap C$ and consider any $x\in A$. Then $x\in B\cap C$ as well so that $x\in B$ and $x\in C$. Since $x$ was an arbitrary element of $A$, this of course shows that both $A\subset B$ and $A\subset C$ as desired. □

Proof. To show the converse, suppose that $A\subset B\cap C$. It was shown in part (c) that this implies that both $A\subset B$ and $A\subset C$. Thus it is clearly true that $A\subset B$ or $A\subset C$.

As a counterexample to the forward implication, let $A=\left\{1\right\}$, $B=\left\{1,2\right\}$, and $C=\left\{3,4\right\}$ so that clearly $A\subset B$ and hence $A\subset B$ or $A\subset C$ is true. However we have that $B$ and $C$ are disjoint so that $B\cap C=\varnothing$, therefore $A\not\subset \varnothing =B\cap C$ since $A\ne \varnothing$. □

Proof. First consider any $x\in A-(A-B)$ so that $x\in A$ but $x\notin A-B$. Hence it is not true that $x\in A$ and $x\notin B$. So it must be that $x\notin A$ or $x\in B$. However, since we know that $x\in A$, it has to be that $x\in B$. Thus $A-(A-B)\subset B$ since $x$ was arbitrary.

Now let $A=\left\{1,2\right\}$ and $B=\left\{2,3\right\}$. Then we clearly have $A-B=\left\{1\right\}$, and thus $A-(A-B)=\left\{2\right\}$. So clearly $B$ is not a subset of $A-(A-B)$ since $3\in B$ but $3\notin A-(A-B)$. □

Proof. First suppose that $x\in A-B$ so that $x\in A$ but $x\notin B$. Then it is certainly true that $x\notin B$ or $x\in A$ so that, by logical equivalence, it is not true that $x\in B$ and $x\notin A$. That is, it is not true that $x\in B-A$, which is to say that $x\notin B-A$. Since also $x\in A$, it follows that $x\in A-(B-A)$, which shows the desired result since $x$ was arbitrary.

To show that the other direction does not hold consider the counterexample $A=\left\{1,2\right\}$ and $B=\left\{2,3\right\}$. Then $B-A=\left\{3\right\}$ so that $A-(B-A)=\left\{1,2\right\}=A$. We also have that $A-B=\left\{1\right\}$ so that $2\in A-(B-A)$ but $2\notin A-B$. This suffices to show that $A-(B-A)\not\subset A-B$. □

Proof. $\text{(⊂)}$ Suppose that $x\in A\cap (B-C)$ so that $x\in A$ and $x\in B-C$. Thus $x\in B$ but $x\notin C$. Since both $x\in A$ and $x\in B$ we have that $x\in A\cap B$. Also since $x\notin C$ it clearly must be that $x\notin A\cap C$. Hence $x\in (A\cap B)-(A\cap C)$, which shows the forward direction since $x$ was arbitrary.

$\text{(⊃)}$ Now suppose that $x\in (A\cap B)-(A\cap C)$. Hence $x\in A\cap B$ but $x\notin A\cap C$. From the former of these we have that $x\in A$ and $x\in B$, and from the latter it follows that either $x\notin A$ or $x\notin C$. Since we know that $x\in A$, it must therefore be that $x\notin C$. Hence $x\in B-C$ since $x\in B$ but $x\notin C$. Since also $x\in A$ we have that $x\in A\cap (B-C)$, which shows the desired result since $x$ was arbitrary. □

Proof. First consider any $x\in (A\cup B)-(A\cup C)$ so that $x\in A\cup B$ and $x\notin A\cup C$. From the latter, it follows that $x\notin A$ and $x\notin C$ since otherwise we would have $x\in A\cup C$. From the former, we have that $x\in A$ or $x\in B$ so that it must be that $x\in B$ since $x\notin A$. Therefore we have that $x\in B$ and $x\notin C$ so that $x\in B-C$. From this it obviously follows that $x\in A\cup (B-C)$, which shows that $A\cup (B-C)\supset (A\cup B)-(A\cup C)$ since $x$ was arbitrary.

To show that the forward direction does not always hold, consider the sets $A=\left\{1,2\right\}$, $B=\left\{2,3\right\}$, and $C=\left\{2\right\}$. Then we clearly have that $B-C=\left\{3\right\}$, and hence $A\cup (B-C)=\left\{1,2,3\right\}$. On the other hand, we have $A\cup B=\left\{1,2,3\right\}$ and $A\cup C=\left\{1,2\right\}$ so that $(A\cup B)-(A\cup C)=\left\{3\right\}$. Hence, for example, $1\in A\cup (B-C)$ but $1\notin (A\cup B)-(A\cup C)$, which suffices to show that $A\cup (B-C)\not\subset (A\cup B)-(A\cup C)$ as desired. □

Proof. We show this with a chain of logical equivalences:

where we note that “True” denotes the fact that $x\in B\vee x\notin B$ is always true by the excluded middle property of logic. □

Proof. Suppose that $A\subset C$ and $B\subset D$. Consider any $(x,y)\in A\times B$ so that $x\in A$ and $y\in B$ by the definition of the cartesian product. Then also clearly $x\in C$ and $y\in D$ since $A\subset C$ and $B\subset D$. Hence $(x,y)\in C\times D$, which shows the result since the ordered pair $(x,y)$ was arbitrary. □

Proof. Consider the following sets:

Then we have that $A\times B=\varnothing$ since there are no ordered pairs $(x,y)$ such that $x\in A$ (since $A=\varnothing$). Hence it is vacuously true that $(A\times B)\subset (C\times D)$. However, clearly it is not the case that $B\subset D$, and so, even though $A\subset C$, it is not true that $A\subset C$ and $B\subset D$. □

Proof. Suppose that $(A\times B)\subset (C\times D)$. First consider any $x\in A$. Then, since $B\ne \varnothing$, there is a $b\in B$. Then $(x,b)\in A\times B$ so that clearly also $(x,b)\in C\times D$. Hence $x\in C$ so that $A\subset C$ since $x$ was arbitrary. An analogous argument shows that $B\subset D$ since $A$ is nonempty. Hence it is true that $A\subset C$ and $B\subset D$ as desired. □

Proof. First consider any $(x,y)\in (A\times B)\cup (C\times D)$ so that either $(x,y)\in A\times B$ or $(x,y)\in C\times D$. In the first case $x\in A$ and $y\in B$ so that clearly $x\in A\cup C$ and $y\in B\cup D$. Hence $(x,y)\in (A\cup C)\times (B\cup D)$. In the second case we have $x\in C$ and $y\in D$ so that again $x\in A\cup C$ and $y\in B\cup D$ are still both true. Hence of course $(x,y)\in (A\cup C)\times (B\cup D)$ here also. This shows the result in either case since $(x,y)$ was an arbitrary ordered pair.

To show that the other direction does not always hold, consider $A=B=\left\{1\right\}$ and $C=D=\left\{2\right\}$. Then we clearly have $A\times B=\left\{(1,1)\right\}$ and $C\times D=\left\{(2,2)\right\}$ so that $(A\times B)\cup (C\times D)=\left\{(1,1),(2,2)\right\}$. We also have $A\cup C=B\cup D=\left\{1,2\right\}$ so that $(A\cup C)\times (B\cup D)=\left\{(1,1),(1,2),(2,1),(2,2)\right\}$. This clearly shows that $(A\times B)\cup (C\times D)\not\supset (A\cup C)\times (B\cup D)$ as desired. □

Proof. We can show this by a series of logical equivalences:

as desired. □

Proof. $\text{(⊂)}$ First consider any $(x,y)\in A\times (B-C)$ so that $x\in A$ and $y\in B-C$. From the latter of these we have that $y\in B$ but $y\notin C$. We clearly then have that $(x,y)\in A\times B$ since $x\in A$ and $y\in B$. It also has to be that $(x,y)\notin A\times C$ since $y\notin C$ even though it is true that $x\mathit{inA}$. Therefore $(x,y)\in (A\times B)-(A\times C)$ as desired.

$\text{(⊃)}$ Now suppose that $(x,y)\in (A\times B)-(A\times C)$ so that $(x,y)\in A\times B$ but $(x,y)\notin (A\times C)$. From the former we have that $x\in A$ and $y\in B$. It then must be that $y\notin C$ since $(x,y)\notin (A\times C)$ but we know that $x\in A$. Then we have $y\in B-C$ since $y\in B$ but $y\notin C$. Since also $x\in A$, it follows that $(x,y)\in A\times (B-C)$ as desired. □

Proof. $\text{(⊂)}$ Suppose that $(x,y)\in (A-B)\times (C-D)$ so that $x\in A-B$ and $y\in C-D$. Then we have that $x\in A$, $x\notin B$, $y\in C$, and $y\notin D$. So first, clearly $(x,y)\in A\times C$. Then, since $x\notin B$, we have that $(x,y)\notin B\times C$, and hence $(x,y)\in A\times C-B\times C$. Since $y\notin D$, we also have that $(x,y)\notin A\times D$, and thus $(x,y)\in (A\times C-B\times C)-A\times D$. This clearly shows the desired result since $(x,y)$ was arbitrary.

$\text{(⊃)}$ Now suppose that $(x,y)\in (A\times C-B\times C)-A\times D$ so that $(x,y)\in A\times C-B\times C$ but $(x,y)\notin A\times D$. From the former we have that $(x,y)\in A\times C$ and $(x,y)\notin B\times C$. Thus $x\in A$ and $y\in C$ so that it has to be that $x\notin B$ since $(x,y)\notin B\times C$ but we know that $y\in C$. It also must be that $y\notin D$ since $(x,y)\notin A\times D$ but $x\in A$. Therefore we have that $x\in A$, $x\notin B$, $y\in C$, and $y\notin D$, from which it readily follows that $x\in A-B$ and $y\in C-D$. Thus clearly $(x,y)\in (A-B)\times (C-D)$, which shows the desired result since $(x,y)$ was arbitrary. □

Proof. First consider any $(x,y)\in (A-C)\times (B-D)$ so that $x\in A-C$ and $y\in B-D$. Thus we have $x\in A$, $x\notin C$, $y\in B$, and $y\notin D$. From this clearly $(x,y)\in A\times B$ but $(x,y)\notin C\times D$. Hence $(x,y)\in (A\times B)-(C\times D)$, which clearly shows the desired result since $(x,y)$ was arbitrary.

To show that the forward direction does not hold, consider $A=\left\{1,2\right\}$, $B=\left\{a,b\right\}$, $C=\left\{2,3\right\}$, and $D=\left\{b,c\right\}$. We then clearly have the following sets:

This clearly shows that $(A\times B)-(C\times D)$ is not a subset of $(A-C)\times (B-D)$. □