Question

Prove that if are subsets of a universal set U, then

$\overline{\underset{k=1}{\overset{n}{\cup }} A_k}=\underset{k=1}{\overset{n}{\cap }} \overline{A_k}$

Short answer

$\overline{\underset{k=1}{\overset{n}{\cup }} A_k}=\underset{k=1}{\overset{n}{\cap }} \overline{A_k}$

Step-by-step solution

Step: 1

$\begin{array}{r}\overline{\underset{k=1}{\overset{n}{\cup }} A_k}=\overline{\underset{k=1}{\overset{n}{\cup }} }{A}_k\cup A_n\\ \overline{\underset{k=1}{\overset{n}{\cup }} A_k\cup A_n}=\overline{\left(\underset{k=1}{\overset{n}{\cup }} A_k\right)}\cap \overline{A_n}\end{array}$

Step: 2

Using hypothesis,

$\begin{array}{r}\underset{k=1}{\overset{n-1}{\cup }} A_k\cap \overline{A_n}=\left(\underset{k=1}{\overset{n-1}{\cap }} \overline{A_k}\right)\cap \overline{A_n}\\ \underset{k=1}{\overset{n-1}{\cup }} A_k\cap \overline{A_n}=\left(\underset{k=1}{\overset{n}{\cap }} \overline{A_k}\right)\\ \underset{k=1}{\overset{n}{\cup }} A_k=\underset{k=1}{\overset{n}{\cap }} \overline{A_k}\\ \overline{\underset{k=1}{\overset{n}{\cup }} A_k}=\underset{k=1}{\overset{n}{\cap }} \overline{A_k}\\ \end{array}$