Let $(A_i\mid i\in I)$ be a family of disjoint sets where $|A_i|=|A_j|\ge\aleph_0\space\forall i\in I$. Is $|\bigcup\limits_{i\in I}A_i|=|A_i|$?

Question

Let $(A_i\mid i\in I)$ be a family of sets where $|A_i|=|A_j|\ge\aleph_0$ and $A_i\cap A_j=\emptyset$ for all $i\neq j$.

I found that we can prove $|\bigcup\limits_{i\in I}A_i|=|A_i|$ for all $i\in I$ in case $I$ is countablle by induction. I have no idea in case $I$ is uncountable.

Answer 1

Assuming Choice, just consider infinite sets $A$ and $I$. Let $A_i=\{i\}\times A$, for every $i\in I$. Then $$\bigcup_{i\in I}A_i=I\times A$$ and then $$\left|\bigcup_{i\in I}A_i\right|=|I|\cdot |A|=\max\{|I|,|A|\}$$