A cardinal number $\alpha$ is a number such that there exists a set $A$ so that $|A| = \alpha.$ The sum of two cardinal numbers $\alpha$ and $\beta$ is given by $\alpha + \beta := |A\cup B|,$ where $A$ and $B$ are sets so that $ A\cap B = \emptyset, |A| = \alpha, |B| = \beta$.
Let $\alpha, \beta,$ and $\gamma $ be cardinal numbers. Prove that
- $\alpha + \beta$ is well defined.
- $\alpha + \beta = \beta + \alpha$ and $(\alpha + \beta) + \gamma = \alpha + (\beta + \gamma).$
To show the first property, it suffices to show that for any sets $A,B,C,D$ with $|A| = |C|, |B| = |D|, A\cap B = C\cap D = \emptyset, |A\cup B| = |C\cup D|.$ Since the sets are disjoint, by the properties of union (for any sets, including infinite ones), $|A\cup B| = |A| + |B| = |C| + |D| = |C\cup D| = \alpha + \beta.$
The property that $\alpha + \beta = \beta + \alpha$ seems to follow for commutativity of $\cup$: for any two sets $A, B, |A\cup B| = |B\cup A|.$ $\alpha + \beta := |A\cup B|,$ where $A\cap B = \emptyset, |A| = \alpha$ and $|B| = \beta.$ And $\beta + \alpha = |B\cup A| = |A\cup B| = \alpha + \beta.$
The second property seems to follow from associativity of union. $(\alpha + \beta) + \gamma = |(A_1\cup A_2)\cup A_3|, A_i \cap A_j = \emptyset$ for $i\neq j, |A_1| = \alpha, |A_2| = \beta, |A_3| = \gamma.$ Since the $A_i$'s are disjoint, $\alpha + (\beta + \gamma) = |A_1 \cup (A_2 \cup A_3)|$ and by associativity of union, $\alpha + (\beta + \gamma) = |A_1 \cup (A_2 \cup A_3)| = |(A_1\cup A_2)\cup A_3| = (\alpha + \beta) + \gamma.$
Is this correct?