What result is $\left | \bigcup_{i \in I}A_i \right | =\sum_{i \in I} |A_i|$?

Question

I'm reading a text that uses the following equality for disjoint sets $(A_i)_{i \in I}$:

$$\left | \bigcup_{i \in I}A_i \right |=\sum_{i \in I} |A_i|$$

This has to do with disjoint unions, but I'd like to understand, how does the equivalence hold?

Answer 1

This is a generalization of the usual rule that $|A|+|B|=|A\cup B|$, granted that $A\cap B=\varnothing$. Of course the additivity would be false if $A\cap B\neq\varnothing$. If $A=B=\{0\}$, then $|A|+|B|=2$ but $|A\cup B|=|A|=1$.

So in general, if you want to guarantee that a union's cardinality is the sum of cardinals, you would like that union to be a disjoint union.

Now, since it is generally the definition of the generalized cardinal summation, I can't quite give you an answer explaining why this is "true" without being circular (it's true because this is how we define it). But we can prove why this definition works in general.

In order to prove that, we need to show that it does not depend on the choice of each $A_i$. Namely, if $\{B_i\mid i\in I\}$ is a family of pairwise disjoint sets such that $|A_i|=|B_i|$, then $\bigcup A_i$ and $\bigcup B_i$ have the same cardinality.

So for each $i\in I$ there is a bijection between $A_i$ and $B_i$, say $f_i\colon A_i\to B_i$. We define a function $f$ whose domain is $\bigcup A_i$ by defining: $$f(a)=f_i(a)\iff a\in A_i$$ We need to show that $f$ is well-defined, and that it is a bijection. I will leave it to you to verify these details, but essentially the key point is that the $A_i$'s are disjoint (for it to be well-defined) and the $B_i$'s are disjoint (for it to remain injective) and surjectivity comes from the assumption that each $f_i$ is surjective.

Let me finish with a note that we had to use the axiom of choice to choose each $f_i$. Without the axiom of choice it is possible for $I=\Bbb N$, and for each $i$, $|A_i|=|B_i|=2$, but $\bigcup A_i$ is countable whereas $\bigcup B_i$ is not countable.