# 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 $|\bigcup\limits_{i\in I}A_i|=|A_i|$?

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.

• @ajotatxe I'm sorry. That's a typo. It should be $|X_i|$. – LE Anh Dung Sep 15 '18 at 7:03
• @ajotatxe It's my mistake a gain. It should be $A_i$. – LE Anh Dung Sep 15 '18 at 7:05
• @bof It's a typo. It should be $A_i$, not $X_i$. – LE Anh Dung Sep 15 '18 at 7:06
• Thanks @bof ! What if $|A_i|>\aleph_0$ for all $i\in I$ and $|I|>\aleph_0$? – LE Anh Dung Sep 15 '18 at 7:13
• Under the assumptions in your header (and assuming the axiom of choice), $$|\bigcup_{i\in I}A_i|=|A_i\times I|=|A_i|\cdot|I|.$$ – bof Sep 15 '18 at 7:21

## 1 Answer

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 $$\left|\bigcup_{i\in I}A_i\right|=|I|\cdot |A|=\max\{|I|,|A|\}$$

• Did you mean $|\bigcup_{i\in I}A_i|$ rather than $\bigcup_{i\in I}|A_i|$? – LE Anh Dung Sep 15 '18 at 7:54