Proof that $\left(\bigcup_{i}^{}A_i\right)^{\mathsf{c}}=\bigcap_{i}^{}A_i^{\mathsf{c}}$

Question

Here is my attempt. Do I have a correct approach?

Answer 1

To extend your approach to the case where there are infinitely many $A_i$, we'll use quantifiers:$$x\in\left(\bigcup_i A_i\right)^c\iff x\not\in\bigcup_i A_i\iff\forall i (x\not\in A_i)\\\iff\forall i (x\in A_i^c)\iff x\in\bigcap_i A_i^c.$$

Answer 2

$\Rightarrow$ Suppose $x\in \left(\bigcup_{i}A_i\right)^c$ then by definition $x\notin \bigcup_{i}A_i$ and $x\notin A_i$ for all $i$. Thus we have that $x\in A_i^{c}$ for all $i$ and must then be in intersection of the complements i.e. $x\in\bigcap_{i}A_i^c$.

$\Leftarrow$ Suppose $x\in\bigcap_{i}A_i^c$ then $x\in A_i^{c}$ for all $i$. By defintion of the complement $x\notin A_i$ and $x\notin \bigcup_{i}{A_i}$ then we must have $x\in \left(\bigcup_{i}A_i\right)^c$.