How can I show that these two sets have the same cardinal?

Question

Let ${\{A_i\}}_{i \in I}$ be a family of sets, where $\# A_i < \infty$ for all $i \in I$ and $\# I = \infty$. How can I show that $$ \# \left(\bigcup_{i \in I} A_i\right) = \# I? $$

Answer 1

Claim: Given $I$ well-ordered and non-empty, and cardinals $\{\kappa_i\}_{i\in I}$ such that for any $i\in I,$ we have that either $\kappa_i$ is finite or an aleph, if $I$ or some $\kappa_i$ is infinite, then $$\sum_{i\in I}\kappa_i=\max\left\{|I|,\sup_{i\in I}\kappa_i\right\}.$$

Proof First, note that $\sup_{i\in I}\kappa_i$ is a well-orderable cardinal--in particular, it is the least such cardinal that is $\geq\kappa_i$ for all $i\in I$. Let $f:I\to|I|$ be a bijection, and define $g:\bigcup_{i\in I}\kappa_i\times\{i\}\to|I|\times\sup_{i\in I}\kappa_i$ by $g(\alpha,i)=\langle f(i),\alpha\rangle$. Since $f$ is a bijection, then $g$ is an injection from $\bigcup_{i\in I}\kappa_i\times\{i\}$ into $|I|\times\sup_{i\in I}\kappa_i$, and so $\sum_{i\in I}\kappa_i\leq|I|\cdot\sup_{i\in I}\kappa_i.$

Since it can be shown that, for any well-orderable cardinals $\alpha$ and $\beta$ with at least one of them infinite, we have $\alpha\cdot\beta=\max\{\alpha,\beta\},$ then we have $|I|\cdot\sup_{i\in I}\kappa_i=\max\bigl\{|I|,\sup_{i\in I}\kappa_i\bigr\},$ and so $$\sum_{i\in I}\kappa_i\leq\max\bigl\{|I|,\sup_{i\in I}\kappa_i\bigr\}.$$

Now, clearly, $I\to\bigcup_{i\in I}\kappa_i\times\{i\}$ given by $i\mapsto\langle 0,i\rangle$ is defined (since each $\kappa_i>0$) and injective, so $|I|\leq\sum_{i\in I}\kappa_i$. Define $h:\bigcup_{i\in I}\kappa_i\times\{i\}\to\bigcup_{i\in I}\kappa_i$ by $h(\alpha,i)=\alpha$. Clearly, $h$ is surjective, and since its domain is well-orderable, then it can be shown that there is an injection from $\bigcup_{i\in I}\kappa_i$ into $\bigcup_{i\in I}\kappa_i\times\{i\}$. Since $\sup_{i\in I}\kappa_i=\bigcup_{i\in I}\kappa_i$ by definition, then $\sup_{i\in I}\kappa_i\leq\sum_{i\in I}\kappa_i$. $\Box$

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At this point, I invoke the Axiom of Choice.* As your set $I$ is infinite, but of unspecified cardinality, we can't guarantee that it is well-orderable without assuming that all sets are well-orderable, which is a form of AC. This is also enough to ensure that each of your finite (so well-orderable) sets $A_i$ can have a well-ordering chosen for each of them via choice function, and that $\aleph_0\le|I|.$ Thus, we have the following

Corollary: $$\left|\bigcup_{i\in I}A_i\right|=|I|.$$

Proof: Note that the left-hand side is simply $\sum_{i\in I}|A_i|$, by definition and by the fact that they are pairwise-disjoint. Moreover, by the fact that each $A_i$ is finite, while $I$ is infinite, we have $$\sup_{i\in I}|A_i|\le\aleph_0\le|I|,$$ and hence, applying the Claim, we have $$\left|\bigcup_{i\in I}A_i\right|=\sum_{i\in I}|A_i|\le\max\left\{|I|,\sum_{i\in I}|A_i|\right\}\leq\max\bigl\{|I|,\aleph_0\bigr\}=|I|.\Box$$

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*If you're taking the Axiom of Choice for granted, then we can also adjust the Claim as follows:

Given $I$ non-empty and cardinals $\{\kappa_i\}_{i\in I},$ if $I$ or some $\kappa_i$ is infinite, then $$\sum_{i\in I}\kappa_i=\max\left\{|I|,\sup_{i\in I}\kappa_i\right\}.$$