Arithmetic of cardinal numbers

Suppose $S=\{1,2\}$ and $T=\{1,2,3\}$. Relabelling the elements of $T$ yields a set $T'=\{a,b,c\}$ of the same cardinality, but which is disjoint from $T$. Then $2+3 = \char93 S + \char93 T = \char93 (S \cup T') = \char93 \{1,2,a,b,c\} = 5$. This observation lets one add cardinal numbers in general: if $S$ and $T$ are arbitrary sets then the sum of the cardinal numbers $\char93 S$ and $\char93 T$ is defined to be the cardinality of $S \cup T'$ where $T'$ is obtained from $T$ by relabeling elements so that $S \cap T' = \emptyset$. For example, in the disjoint union

$$\{0,2,4,6,\dots\} \cup \{1,3,5,7,\dots\} = {\mathbb{N}},$$

all three sets are of cardinality $\aleph_0$, so

$$\aleph_0 + \aleph_0 = \aleph_0.$$

The Cartesian product $S \times T$ of two sets $S$ and $T$ is the set of all ordered pairs $(s,t)$ where $s \in S$ and $t \in T$. For example, if $S=\{1,2\}$ and $T=\{a,b,c\}$, then

$$S \times T = \{(1,a),(1,b),(1,c),(2,a),(2,b),(2,c)\}$$

so $\char93 (S \times T) = 6 = 2 \cdot 3$. If $S$ and $T$ are arbitrary sets, then the product of the cardinal numbers $\char93 S$ and $\char93 T$ is defined to be $\char93 (S \times T)$. For instance, the elements of ${\mathbb{N}}\times {\mathbb{N}}$ can be described by strings of typewriter symbols like $\verb*+(5,7)+$, so the Typewriter Principle shows that $\char93 ({\mathbb{N}}\times {\mathbb{N}}) = \aleph_0$. Therefore

$$\aleph_0 \cdot \aleph_0 = \aleph_0.$$

These equalities may look strange, but in fact, such behavior is typical: one can prove that if $\aleph$ and $\aleph'$ are cardinal numbers such that $\aleph \le \aleph'$ and $\aleph'$ is infinite, then $\aleph + \aleph' = \aleph \cdot \aleph' = \aleph'$.

There is no nice way of subtracting or dividing cardinal numbers. But one can exponentiate. If $S$ and $T$ are arbitrary sets, let $S^T$ denote the set of functions from $T$ to $S$. Note the reversal of order! Then $(\char93 S)^{(\char93 T)}$ is defined to be the cardinality of $S^T$. For example, if $S={\mathbb{N}}$ and $T=\{1,2,3\}$, then a sample element of $S^T$ might be described by

the function {1,2,3}->N sending 1 to 12, 2 to 753, and 3 to 489.

The Typewriter Principle shows that $\char93 (S^T) = \aleph_0$. Hence $(\aleph_0)^3 = \aleph_0$.