# Proof

Let $$X, Y, U$$ and $$W$$ be finite sets.

### (1)

We want to show that from $$|X|=|Y|, |U|=|W|$$ and $$X\cap U=\emptyset, Y\cap W=\emptyset$$ it follows for set unions that $$|X\cup U|=|Y\cup W|=|X|+|U|=|Y|+|W|$$.

• From the hypothesis $$|X|=|Y|, |U|=|W|$$ it follows from rules of comparing cardinal numbers that there are bijective maps $$f:X\mapsto Y$$ and $$g:U\mapsto W$$.
• The hypothesis $$X\cap U=\emptyset$$ means that from $$a\in X\cup U$$ it follows that either $$a\in X$$ or $$a\in U$$ (but not both).
• Thus, we can define a map $$h:$$. $h:=\begin{cases} X\cup U & \mapsto Y\cup W\\ a & \mapsto f(a) \Leftrightarrow a\in X\\ a & \mapsto g(a) \Leftrightarrow a\in U\\ \end{cases}$
• The hypothesis $$Y\cap W=\emptyset$$ means that from $$h(a)\in Y\cup W$$ it follows that either $$h(a)\in Y$$ or $$h(a)\in W$$ (but not both), so $$|X\cup U|=|X|+|U|$$ and $$|Y\cup W|=|Y|+|W|$$.
• Moreover, $$h$$ is bijective, because $$f$$ and $$g$$ are bijective.
• Thus, $$|X\cup U|=|Y\cup W|$$.

### (2)

We want to show that from $$|X|=|Y|$$ and $$|U|=|W|$$ it follows for the Cartesian products that $$|X\times U|=|Y\times W|=|X|\cdot|U|=|Y|\cdot|W|$$ * As in (1) from $$|X|=|Y|, |U|=|W|$$ it follows that there are bijective maps $$f:X\mapsto Y$$ and $$g:U\mapsto W$$. * We define a new map $h:\begin{cases} X\times U&\mapsto Y\times W\\ (a,b)&\mapsto (f(a),g(b)) \end{cases}$ * For a fixed cardinal number $$|X|$$ we prove by induction on the cardinal number $$|U|=n\in\mathbb N$$. * For $$n=0$$ we have $$U=\emptyset$$ and $$W=g(U)=g(\emptyset)=\emptyset$$. Therefore $$h(X\times \emptyset)=h(\emptyset)=f(\emptyset)=\emptyset$$. Thus it follows trivially that $$h$$ is bijective as $$f$$. * Thus $$|X\times\emptyset|=|Y\times\emptyset|=|X|\cdot|\emptyset|=|Y|\cdot|\emptyset|=0$$, as required. * Let the claim $(|X|=|Y|\wedge |U_0|=|W_0|)\Rightarrow|X\times U_0|=|Y\times W_0|=|X|\cdot|U_0|=|Y|\cdot|W_0|$ be proven for all $$U_0,W_0$$ with $$|U_0|=|W_0|=n_0\ge 0$$. * We observe that $$|X|=|X \times \{U_0\}|$$, since the set $$\{U_0\}$$ has only one element1.
* For the set $$U=U_0 \cup \{U_0\}$$ we have then $X\times U=X\times(U_0 \cup \{U_0\})=(X\times U_0) \cup (X\times \{U_0\}).$ * Taking the cardinal numbers on both sides we find $|X\times U|=|(X\times U_0) \cup (X\times \{U_0\})|=|(X\times U_0)| + |(X\times \{U_0\})|,$ where the last equation follows from (1) and the fact that $$(X\times U_0) \cap (X\times \{U_0\})=\emptyset$$.2 * Therefore we have $|X\times U|=|X|\cdot|U_0|+|X|\cdot|\{U_0\}|=|X|\cdot n_0+|X|\cdot 1=|X|\cdot(n_0+1)=|X|\cdot(|U_0|+1)=|Y|\cdot(|W_0|+1)=|Y\times W|$ * The claim can analogously be proven by induction for a fixed cardinal number $$|U|$$ on the cardinal number $$|X|=n\in\mathbb N$$.

Github: ### References

#### Bibliography

1. Ebbinghaus, H.-D.: "Einführung in die Mengenlehre", BI Wisschenschaftsverlag, 1994, 3th Edition

#### Footnotes

1. We also could have used $$\{\emptyset\}$$ or any other set with only one element.

2. Note: There is no pair $$(a,b)$$ for which both $$(a,b)\in (X\times \{U_0\})$$ and $$(a,b)\in (X\times U_0)$$ can be true simultaneously, otherwise $$b$$ would equal a set $$U_0$$ which contains itself $$\{U_0\}$$, which is forbidden by the axiom of foundation