Proof Of De Morgan`s Laws with indexed set and a universal set.

Question

Let $E$ be a universal set and $\{A_{\alpha}\}_{\alpha \in J},$ for some index set $J$ be a family of subsets of $E.$

Prove that: (a)$E-\bigcup_{\alpha \in J}A_{\alpha} = \bigcap_{\alpha \in J}($R$-A_{\alpha}).$

I do not know what is $R$ or it is a mistake in the question, Could anyone help me ?

(b)$E-\bigcap_{\alpha \in J}A_{\alpha} = \bigcup_{\alpha \in J}($E$-A_{\alpha}).$

Shall I prove it by induction? but what about the index set is it countably infinite or finite or uncountable, and how the proof will differ?

Answer 1

What ever $R$ means should have been identified earlier in your reference book, otherwise it is a mystery.

As to the second, just use the definitions that $$\bigcup_{\alpha\in J} X_\alpha := \{x~:~ \exists \alpha \in J~ (x\in X_\alpha)\} = \{x~:~\bigvee_{\alpha\in J}(x\in X_\alpha)\}\\ \bigcap_{\alpha\in J} X_\alpha := \{x~:~ \forall \alpha\in J~(x\in X_\alpha)\}=\{x~:~ \bigwedge_{\alpha\in J}(x\in X_\alpha)\}\\ B- A = \{x ~:~ x \in B~\wedge~ x\notin A\}$$

Answer 2

‎\begin{eqnarray*}‎ ‎(\bigcap_{i\in\Lambda}A_i)^c &=& \{x|x\notin \bigcap_{i\in\Lambda}A_i\} \\‎ ‎&=& \{x|\exists i\in\Lambda,~~x\notin A_i\}\\‎ ‎&=& \{x|\exists i\in\Lambda,~~x\in A_i^c\} \\‎ ‎&=& \bigcup_{i\in\Lambda}A_i^c‎ ‎\end{eqnarray*}‎