I would like to know if my reasoning is right, or I have errors.

Question

Let $\left\{B_{i}\right\}_{i \in I}$ and $\left\{C_{i}\right\}_{i \in I}$ each be a family of mutually disjoint sets. If $B_{i}\approx C_{i}$for each $i \in I $, prove that $$\bigcup_{i\in I}B_{i} \approx \bigcup_{i \in I}C_{i}$$ Proof. suppose that $B_{i}\approx C_{i}$for each $i \in I $, then by defintion of equipotence there is a function biyective: $f:B_{i}\rightarrow C_{i}$, for each $i\in I$.

By hypothesis $$\bigcap_{i\in I}B_{i}=\emptyset \wedge \bigcap_{i\in I}C_{i}=\emptyset $$ hence,$$g:\bigcup_{i\in I}B_{i} \rightarrow \bigcup_{i \in I}C_{i}$$ is a function biyective. Hence $$ \bigcup_{i\in I}B_{i}\approx \bigcup_{i \in I}C_{i}$$

Note: Theorem.Let $A,B,C$ and $D$ be sets where $A\cap C=\emptyset$ and $B \cap D=\emptyset$. If $f: A\rightarrow B $ and $g:C \rightarrow D$ are bijective functions, then $(f\cup g):A\cup C \rightarrow B \cup C$ is a bijective function.

my question is, if I can use this theorem to say that $$g:\bigcup_{i\in I}B_{i} \rightarrow \bigcup_{i \in I}C_{i}$$ is a function biyective.

Answer 1

To your question, yes. You can say that. In its general form the theorem states that:

Suppose that $\{A_i\mid i\in I\}$ and $\{B_i\mid i\in I\}$ are families of pairwise disjoint sets, if for all $i\in I$, $f_i\colon A_i\to B_i$ is a function, then $f\colon\bigcup A_i\to\bigcup B_i$ defined by $f(a)=f_i(a)$ for the unique $i$ such that $a\in A_i$ is a function. Moreover, if each $f_i$ is surjective, then $f$ is surjective, and if each $f_i$ is injective, then $f$ is injective.

You can prove this with exactly the same argument as the case where $I$ is a set of two elements.

The problem, as pointed out by TonyK in the comments, is that you need to move from "There is a bijection $f$ for each $B_i$ and $C_i$" to "There is a chosen bijection, $f_i\colon B_i\to C_i$". If $I$ is infinite, this requires the use of the axiom of choice. Which doesn't mean it's bad, it's just something to note.

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Also, note that $\bigcap B_i=\varnothing$ is a weaker condition than being pairwise disjoint. For example, $\{\{0,1\},\{1,2\},\{2,0\}\}$ has an empty intersection, but no two sets are disjoint.