Let $⟨A_i\mid i\in I⟩$ be a system of mutually disjoint sets, and let $|A_i|={\kappa }_i$ for all $i\in I$. We define the sum of $⟨{\kappa }_i\mid i\in I⟩$ by

If $⟨A_i\mid i\in I⟩$ and $⟨A_i^{\prime }\mid i\in I⟩$ are systems of mutually disjoint sets such that $|A_i|=|A_i^{\prime }|$ for all $i\in I$, then $\left|\underset{i\in I}{⨆}{A}_i\right|=\left|\underset{i\in I}{⨆}{A}_i^{\prime }\right|$

This lemma makes the definition of $\sum _{i\in I}{\kappa }_i$ legitimate.

If ${\kappa }_i\le {\lambda }_i$ for all $i\in I$, then $\sum _{i\in I}{\kappa }_i\le \sum _{i\in I}{\lambda }_i$.

We can see that $$\left|\bigcup_{i \in I} A_i\right| \le \sum_{i \in I} |A_i|$$

If the summands are all equal, then the following holds, as the finite case: If $\kappa ={\kappa }_i$ for all $i\in \lambda$, then

Th: Let $\lambda$ be an infinite ordinal, let ${\kappa }_{\alpha }$ $(\alpha <\lambda )$ be a nonzero cardinal numbers, and let $\kappa =sup\{{\kappa }_{\alpha }\mid \alpha <\lambda \}$

If $J_i$ $(i\in I)$ are mutually disjoint sets and $J=\bigcup _{i\in I}{J}_i$, and ${\kappa }_j$ $(j\in J)$ are cardinals, then

We also get the distributive law, let $\lambda$ be a cardinal, and ${\kappa }_i$ be a cardinal for every $i\in I$, then $$\lambda \cdot \left(\sum_{i \in I}\kappa_i\right) = \sum_{i \in I} (\lambda \cdot \kappa_i)$$

Let $⟨A_i\mid i\in I⟩$ be a family of sets such that $|A_i|={\kappa }_i$ for all $i\in I$. We define the product $⟨{\kappa }_i\mid i\in I⟩$ by

Let $⟨A_i\mid i\in I⟩$ and $⟨A_i^{\prime }\mid i\in I⟩$ are such that $|A_i|=|A_i^{\prime }|$ for all $i\in I$, then $|\prod _{i\in I}{A}_i|=|\prod _{i\in I}{A}_i^{\prime }|$

The infinite products have many properties of finite products of natural numbers. For instance, if at least on ${\kappa }_i$ is $0$, then $\prod _{i\in I}{\kappa }_i=0$.

If $J_i$ $(i\in I)$ are mutually disjoint sets and $J=\bigcup _{i\in I}{J}_i$, and ${\kappa }_j$ $(j\in J)$ are cardinals, then

If ${\kappa }_i\le {\lambda }_i$ for all $i\in I$, then $\prod _{i\in I}{\kappa }_i\le \prod _{i\in I}{\lambda }_i$.

If the factors are all equal, then the following holds, as the finite case: If $\kappa ={\kappa }_i$ for all $i\in \lambda$, then

The following rules, involving exponentiation, also generalize from the finite to the infinite case

König's Theorem

If ${\kappa }_i$ and ${\lambda }_i$ $(i\in I)$ are cardinal numbers, and if ${\kappa }_i<{\lambda }_i$, for all $i\in I$, then

We have that if $1<{\kappa }_i\le {\lambda }_i$ for all $i\in I$, then

Cardinal Exponentiation

By Cantor's Theorem, we get that $2^{{\mathrm{\aleph }}_{\alpha }}>{\mathrm{\aleph }}_{\alpha }$, in other words, $$\aleph_{\alpha +1} \le 2^{\aleph_\alpha}$$We can look at the Generalised Continuum Hypothesis, and we get that $$ \aleph_{\alpha +1} = 2^{\aleph_\alpha}$$Without assuming the Generalised Continuum Hypothesis, there is not much one can prove about $2^{{\mathrm{\aleph }}_{\alpha }}$ except, the first property and the trivial property $$2^{\aleph_\alpha} \le 2^{\aleph_\beta} \quad \text{whenever} \quad \alpha\le \beta$$
Lemma: For every $\alpha$, $$\aleph_\alpha < \text{cf}(2^{\aleph_\alpha})$$Thus $2^{{\mathrm{\aleph }}_0}$ cannot be ${\mathrm{\aleph }}_{\omega }$, since $\text{cf}(2^{{\mathrm{\aleph }}_{\omega }})={\mathrm{\aleph }}_0$, but the lemma doesn't prevent $2^{{\mathrm{\aleph }}_0}$ from being ${\mathrm{\aleph }}_{{\omega }_1}$. Similarly, $2^{{\mathrm{\aleph }}_1}$ cannot be either ${\mathrm{\aleph }}_{{\omega }_1}$ or ${\mathrm{\aleph }}_{\omega }$ or ${\mathrm{\aleph }}_{\omega +\omega }$, etc.

If there's a $\gamma <\alpha$ such that ${\mathrm{\aleph }}_{\gamma }^{{\mathrm{\aleph }}_{\beta }}\ge {\mathrm{\aleph }}_{\alpha }$, then ${\mathrm{\aleph }}_{\alpha }^{{\mathrm{\aleph }}_{\beta }}={\mathrm{\aleph }}_{\gamma }^{{\mathrm{\aleph }}_{\beta }}$

Th: Let ${\mathrm{\aleph }}_{\alpha }$ be a singular cardinal. Let us assume that the value of $2^{{\mathrm{\aleph }}_{\xi }}$ for all $\xi <\alpha$, say $2^{{\mathrm{\aleph }}_{\xi }}={\mathrm{\aleph }}_{\beta }$. Then $2^{{\mathrm{\aleph }}_{\alpha }}={\mathrm{\aleph }}_{\beta }$.

Lemma: If $\alpha \le \beta$, then ${\mathrm{\aleph }}_{\alpha }^{{\mathrm{\aleph }}_{\beta }}=2^{{\mathrm{\aleph }}_{\beta }}$

Lemma: Let $\alpha \ge \beta$, and let $S$ be the set of all subsets $X\subseteq {\omega }_{\alpha }$ such that $|X|={\mathrm{\aleph }}_{\beta }$. Then $|S|={\mathrm{\aleph }}_{\alpha }^{{\mathrm{\aleph }}_{\beta }}$.

Th: Let us assume the Generalised Continuum Hypothesis. If ${\mathrm{\aleph }}_{\alpha }$ is a regular cardinal, then $$ \aleph_\alpha^{\aleph_\beta} = \begin{cases} \aleph_\alpha & \beta < \alpha\ \aleph_{\beta +1} & \beta \ge \alpha\end{cases}$$
Lemma: For every cardinal $\kappa >1$, and every $\alpha$, $\text{cf}({\kappa }^{{\mathrm{\aleph }}_{\alpha }})>{\mathrm{\aleph }}_{\alpha }$

Th: Let us assume the Generalised Continuum Hypothesis. If ${\mathrm{\aleph }}_{\alpha }$ is a singular cardinal, then $$ \aleph_\alpha^{\aleph_\beta} = \begin{cases}
\aleph_\alpha & \aleph_\beta < \text{cf}(\aleph_\alpha)\
\aleph_{\alpha +1} & \text{cf}(\aleph_\alpha) \le \aleph_\beta \le \aleph_\alpha\
\aleph_{\beta+1} & \aleph_\beta \ge \aleph_\alpha
\end{cases}$$
Hausdorff's Formula:
For every $\alpha$ and every $\beta$ $$\aleph_{\alpha+1}^{\aleph_\beta} = \aleph_\alpha^{\aleph_\beta}\cdot \aleph_{\alpha +1}$$
We get the result that for $n<\omega$, then $$ \aleph_n ^{\aleph_\beta} = \aleph_n \cdot 2^{\aleph_\beta}$$ We can also get the result $$\prod_{n <\omega} \aleph_n = \aleph_\omega ^{\aleph_0} $$We get a small corollary from this let $\beta >0$, then $$\aleph_\omega^{\aleph_\beta} = \aleph_\omega^{\aleph_0} \cdot 2^{\aleph_\beta}$$