Cardinality of Infinite Union of Infinite Sets

Theorem
Let $X_i$ be sets of infinite cardinality $\kappa$ indexed by a set $I$ of cardinality at most $\kappa$.

Then $\displaystyle{\bigcup_{i\in I} X_i}$ has cardinality $\kappa$.

Proof
We can assume that $I$ has cardinality $\kappa$ and that the sets $X_i$ are disjoint. The more general case follows since these other possibilities can only decrease the cardinality of the union, but the cardinality of the union cannot be decreased below $\kappa$, since it contains copies of cardinality $\kappa$ sets.

The idea is to prove by transfinite induction that for limit ordinals $\omega_\alpha$, the order type of $\omega_\alpha \times \omega_\alpha$ is at most $\omega_\alpha$. From this we can conclude that the cardinality of $\aleph_\alpha \times \aleph_\alpha$ is still $\aleph_\alpha$. We can then identify $\displaystyle{\bigcup_{i\in I} X_i}$ with some $\aleph_\alpha \times \aleph_\alpha$ to yield the theorem.

We begin by defining a well-ordering $\prec$ for each $\omega_\alpha \times \omega_\alpha$.

Let $(\beta_1, \beta_2) \prec (\gamma_1, \gamma_2)$ if and only if one of the following holds:


 * 1) $\max{\beta_i} < \max{\gamma_i}$, or
 * 2) $\max{\beta_i} = \max{\gamma_i}$ and $\beta_1 < \gamma_1$, or
 * 3) $\max{\beta_i} = \max{\gamma_i}$ and $\beta_1 = \gamma_1$ and $\beta_2 < \gamma_2$.

Informally rephrased, this says that one pair is less than another pair if its max is smaller or if their max is the same but it is lexicographically smaller.


 * Base case: $(\omega_0 \times \omega_0, \prec)$ has order type at most $(\omega_0, <)$

The elements of $\omega_0 \times \omega_0$ can be arranged in an array as follows, with arrows drawn so that $(a,b)\rightarrow (c,d)$ iff $(a,b)$ is the predecessor of $(c,d)$ in the ordering $\prec$. This diagram may be familiar from the proofs of theorems such as this and this. We include it here to suggest motivation for the definition of $\prec$ by illuminating its behavior.

$\begin{array} {*{4}c} (0,0) & \rightarrow & (0,1) &            & (0,2) & \cdots \\ & \swarrow   &       & \nearrow    & \downarrow     \\ (1,0) & \rightarrow & (1,1) &            & (1,2) & \cdots \\ &            &       & \swarrow    &                \\ &            &\leftarrow                           \\ & \swarrow   &                                     \\ (2,0) & \rightarrow & (2,1) & \rightarrow & (2,2) & \cdots \\ \vdots &           & \vdots & & \vdots & \ddots \\ \end{array}$

It is not hard to see the order-isomorphism suggested by this diagram between $(\omega_0 \times \omega_0, \prec)$ and $(\omega_0, <)$. The existence of such an order-isomorphism verifies the base case of our inductive proof.


 * Inductive step: Assuming $(\omega_\beta \times \omega_\beta, \prec)$ has order type at most $(\omega_\beta, <)$ for all $\beta < \alpha$, we prove that $(\omega_\alpha \times \omega_\alpha, \prec)$ has order type at most $(\omega_\alpha, <)$

First, note that if the order type of $\omega_\alpha \times \omega_\alpha$ exceeds that of $\omega_\alpha$, then $(\omega_\alpha + 1,<)$ embeds into $\omega_\alpha \times \omega_\alpha$, and the image of $\omega_\alpha$ (viewed as an element of $\omega_\alpha + 1$) under this embedding must be a pair $(\beta_1, \beta_2)$ in $\omega_\alpha \times \omega_\alpha$ such that the set $\downarrow (\beta_1, \beta_2) = \{(\gamma_1, \gamma_2)\in \omega_\alpha \times \omega_\alpha \mid (\gamma_1, \gamma_2)\prec(\beta_1, \beta_2)\}$ has cardinality $|\omega_\alpha| = \aleph_\alpha$.

Thus, it will be sufficient to show that for every $(\beta_1,\beta_2)\in \omega_\alpha \times \omega_\alpha$, the set $\downarrow (\beta_1, \beta_2)$ has cardinality less than $\aleph_\alpha$. We show this by containing $\downarrow (\beta_1, \beta_2)$ in a set whose cardinality we can bound using the inductive hypothesis.

Let $\beta = \max{\beta_i} + 1$. We have $\beta \in \omega_\alpha$ since $\omega_\alpha$ is a limit ordinal. Moreover, by the definition of $\prec$ and the equivalence of $<$ and $\in$ for ordinals, for each $(\gamma_1, \gamma_2)\in\ \downarrow (\beta_1, \beta_2)$ we have $\gamma_1\in \beta$ and $\gamma_2 \in \beta$. Thus, $\downarrow (\beta_1, \beta_2)$ is a subset of $\beta \times \beta$.

However, since $\beta \in \omega_\alpha$ and $\omega_\alpha$ is a limit ordinal, we must have $|\beta|= \aleph_\gamma$ for some $\gamma < \alpha$ (unless it is finite, in which case the proof is easy). It follows that the cardinality of $\downarrow (\beta_1, \beta_2)$ is at most $|\aleph_\gamma \times \aleph_\gamma|=|\omega_\gamma \times \omega_\gamma|$. But, by the inductive hypothesis, the order-type of $(\omega_\gamma\times\omega_\gamma, \prec)$ is at most $(\omega_\gamma, <)$, so $|\omega_\gamma \times \omega_\gamma|$ is at most $|\omega_\gamma|=\aleph_\gamma < \aleph_\alpha$.


 * Statement in terms of cardinalities

The inductive proof demonstrates that $(\omega_\alpha \times \omega_\alpha, \prec)$ has order type at most $(\omega_\alpha, <)$ for all limit ordinals $\omega_\alpha$. But since order-isomorphisms are bijections and hence preserve cardinality, and since $\omega_\alpha$ has cardinality $\aleph_\alpha$, this means that the cardinality of $\aleph_\alpha \times \aleph_\alpha$ is at most $\aleph_\alpha$. Since $\aleph_\alpha$ clearly injects into $\aleph_\alpha \times \aleph_\alpha$, the Cantor-Bernstein-Schroeder Theorem allows us to conclude that $|\aleph_\alpha \times \aleph_\alpha|=\aleph_\alpha$.


 * Identification of $\displaystyle{\bigcup_{i\in I} X_i}$ with some $\aleph_\alpha \times \aleph_\alpha$

As a corollary of the Well-Ordering Theorem, the cardinal $\kappa$ is the cardinality of some initial ordinal $\omega_\alpha$ and hence $\kappa$ is $\aleph_\alpha$ for some $\alpha$. Thus, since $\displaystyle{\bigcup_{i\in I} X_i}$ embeds in $I\times \aleph_\alpha$, which in turn embeds in $\aleph_\alpha \times \aleph_\alpha$, the above result yields the theorem.