Exercise 5.1.1

Proof. It is obvious that

since $K\cup L=L\cup K$ and $K$ and $L$ are disjoint. □

Proof. First we note that clearly

Now suppose that there is an $x\in K\cap (L\cup M)$ so that $x\in K$ and $x\in L\cup M$ If $x\in L$ then $x\in K\cap L$ and if $x\in M$ then $x\in K\cap M$, either of which is a contradiction since all three sets are mutually disjoint. Hence $K$ and $L\cup M$ are disjoint. A similar argument show that $K\cup L$ and $M$ are disjoint. Thus we have the following:

as desired. □

Proof. Define the function $f:K\to K\cup L$ by simply the identity $f(k)=k$ for any $k\in K$. Obviously this is an injective function so that $\kappa =|K|\le |K\cup L|=\kappa +\lambda$. □

Proof. Suppose that

for sets $K_1$, $K_2$, $L_1$, and $L_2$ where $K_1\cap L_1=\varnothing$ and $K_2\cap L_2=\varnothing$. Also suppose that ${\kappa }_1\le {\kappa }_2$ and ${\lambda }_1\le {\lambda }_2$. Thus $|K_1|={\kappa }_1\le {\kappa }_2=|K_2|$ so that there is an injective function $f$ from $K_1$ to $K_2$. Similarly there is an injective function $g:L_1\to L_2$ since $|L_1|={\lambda }_1\le {\lambda }_2=|L_2|$. Now define $h:K_1\cup L_1\to K_2\cup L_2$ by

We show that $h$ is injective so consider $x$ and $y$ in $K_1\cup L_1$ where $x\ne y$.

Case: $x\in K_1$, $y\in K_1$. Then

since $f$ is injective and $x\ne y$.

Case: $x\in L_1$, $y\in L_1$. Then

since $g$ is injective and $x\ne y$.

Case: $x\in K_1$, $y\in L_1$. Then we have $h(x)=f(x)\in K_2$ and $h(y)=g(y)\in L_2$ so that $h(x)\ne h(y)$ since $K_2$ and $L_2$ are disjoint. Note that this is the same as the case in which $x\in L_1$ and $y\in K_1$ since we simply switch $x$ and $y$.

Since these cases are exhaustive and $h(x)\ne h(y)$ in each this shows that $h$ is injective. Hence we have demonstrated that

as desired. □

Proof. First we show that $|A\times B|=|B\times A|$ by constructing a bijection $f:A\times B\to B\times A$. For $(a,b)\in A\times B$ define

which is clearly a function. Then for $(a,b)\in A\times B$ and $(c,d)\in A\times B$ where $f(a,b)=f(c,d)$ we have

so that $b=d$ and $a=c$. Hence $(a,b)=(c,d)$ so that $f$ is injective. Now consider any $(b,a)\in B\times A$ so that clearly $f(a,b)=(b,a)$, noting that $(a,b)\in A\times B$. Clearly this shows that $f$ is surjective.

Hence $f$ is bijective so that

as required. □

Proof. Similar part (e) above, it is trivial to find a bijection from $A\times (B\times C)$ to $(A\times B)\times C$ so that

as desired. □

Proof. Here suppose additionally that $B\cap C=\varnothing$. First we note that since $B$ and $C$ are disjoint that $A\times B$ and $A\times C$ are also disjoint. Suppose that this is not the case so that there is an $(a,b)\in A\times B$ where $(a,b)\in A\times C$ also. Then clearly $b\in B$ and $b\in C$, which is a contradiction since they are disjoint. Now, it is also trivial to show the equality

Hence we have that

as desired. □

Proof. Here suppose that $\lambda >0$ so that $B\ne \varnothing$. Here we construct a bijection $f:A\to A\times B$, from which it follows that

Since $B\ne \varnothing$ there exists a $b\in B$. So for any $a\in A$ define

which is clearly a function. So for $a_1,a_2\in A$ where $a_1\ne a_2$ we have that

so that $f$ is injective. □

Proof. Suppose that

for sets $A_1$, $A_2$, $B_1$, and $B_2$ where ${\kappa }_1=|A_1|\le |A_2|={\kappa }_2$ and ${\lambda }_1=|B_1|\le |B_2|={\lambda }_2$. Hence there is an injective function $f:A_1\to A_2$ and injective function $g:B_1\to B_2$. We shall construct an injective function $h:A_1\times B_1\to A_2\times B_2$ so that it immediately follows that

as required. So for $(a,b)\in A_1\times B_1$ define

Suppose then $(a,b)\in A_1\times B_1$ and $(c,d)\in A_1\times B_1$ where $(a,b)\ne (c,d)$. If $a\ne c$ then $f(a)\ne f(c)$ since $f$ is injective so that

Similarly if $b\ne d$ then $g(b)\ne g(d)$ since $g$ is injective. Hence again

Thus in all cases $h(a,b)\ne h(c,d)$ so that $h$ is injective. □

Proof. Suppose here that $\kappa \ge 2$. Then $2\le \kappa$ and $\kappa \le \kappa$ so that by property (i) we have

Then by property (j) we have

as desired. □

Proof. Here suppose that $\lambda =|B|>0$ so that $B\ne \varnothing$. Hence there exists a $b\in B$. We shall construct an injective $f:A\to A^B$, from which it follows that

So for any $a\in A$ define $f(a)=g$ where $g:B\to A$ is a function defined by $g(b)=a$ for all $b\in B$, noting that $g\ne \varnothing$ since $B\ne \varnothing$.

Now consider any $a_1,a_2\in A$ where $a_1\ne a_2$ so that for any $b\in B$ we have

From this it follows that $f(a_1)\ne f(a_2)$ so that $f$ is injective. □

Proof. Here suppose that $\kappa =|A|>1$ so that there are $a_1,a_2\in A$ where $a_1\ne a_2$. We shall construct an injective function $f:B\to A^B$ so that

So for any $b\in B$ define $f(b)=g$ where $g:B\to A$ is a function defined by

for $c\in B$. Now suppose that $b_1,b_2\in B$ where $b_1\ne b_2$. We then have

since $b_1\ne b_2$. From this it follows that $f(b_1)\ne f(b_2)$ so that $f$ is injective. □

Proof. Suppose that

for sets $A_1$, $A_2$, $B_1$, and $B_2$ where ${\kappa }_1=|A_1|\le |A_2|={\kappa }_2$ and ${\lambda }_1=|B_1|\le |B_2|={\lambda }_2$.

The theorem as presented in the text is actually not true in full generality. As a counterexample suppose that $A_1=A_2=B_1=\varnothing$ so that ${\kappa }_1={\kappa }_2={\lambda }_1=0$ and $B_2=1$ so that ${\lambda }_2=1$. Then certainly the hypotheses above are true but we also have

where we have used the results of Exercises 5.1.2 and 5.1.3.

However, if we add the restriction that ${\kappa }_2>0$ then it becomes true. To prove this first note that this implies that $A_2\ne \varnothing$ so that there is an $a_2\in A_2$. Also there is an injective function $f:A_1\to A_2$ and an injective function $g:B_1\to B_2$. We shall construct an injective $F:A_1^{B_1}\to A_2^{B_2}$, from which it follows that

So for any $h_1\in A_1^{B_1}$ define $F(h_1)=h_2$ where $h_2\in A_2^{B_2}$ is defined by

for any $b\in B_2$, noting that $g^{-1}$ is a function on $\mathrm{ran}(h_1)$ since $g$ is injective. Clearly $F$ is a function but now we show that it is injective.

So consider any $h_1,h_2\in A_1^{B_1}$ where $h_1\ne h_2$. Then there is a $b_1\in B_1$ such that $h_1(b_1)\ne h_2(b_1)$. So let $b_2=g(b_1)$ so that clearly $b_2\in \mathrm{ran}(g)$ and $b_1=g^{-1}(b_2)$. Hence we have

since $h_1(b_1)\ne h_2(b_1)$ and $f$ is injective. It thus follows that $F(h_1)\ne F(h_2)$ so that we have shown that $F$ is injective. □