Index Set Notation Proof I can reason this out intuitively (it seems obvious), but I can't seem to formalize a proof.
Prove $\left(\bigcup_{\alpha\in J} A_{\alpha}\right)^c = \left(\bigcap_{\alpha\in J}  A^c_\alpha\right)$
Any thoughts?
 A: If $A$ and $B$ are two sets and you want to prove that $A=B$, then the standard way to do this is to show that $A\subset B$ and $B\subset A$. In order to show $A\subset B$, you need to show that $x\in A$ implies $x\in B$. Similarly for $B\subset A$ you need to show that $x\in B$ implies $x\in A$. I will help you to get started with the first implication.
Start by choosing $x\in (\bigcup_{\alpha\in J} A_{\alpha})^{c}$ and try to conclude that $x\in\bigcap_{\alpha\in J}A_{\alpha}^{c}$. Since $x\in (\bigcup_{\alpha\in J} A_{\alpha})^{c}$ then $x\notin \bigcup_{\alpha\in J} A_{\alpha}$ by definition of complement, so $x\notin A_{\alpha}$ for every $\alpha\in J$. Because if on the contrary there would exist $\alpha\in J$ so that $x\in A_{\alpha}$, then what would this say about the assumption $x\notin \bigcup_{\alpha\in J}A_{\alpha}$? Can you continue from here?
Remember that in general $x\in \bigcup_{i}A_{i}$ means that there exists $i$ so that $x\in A_{i}$ and $x\in\bigcap_{i}A_{i}$ means that $x\in A_{i}$ for all $i$.
A: Hint: Recall that the definition of $x\in\bigcup_{i\in I} X_i$ is that for some $i \in I$, $x\in X_i$. Use the similar definition for intersection and complement and show two-sided inclusions.
