$ (\cup_{n=1}^{\infty} A_n )^c= \cap _{n=1}^{\infty} (A_n) ^c $

Question

Explain why induction cannot be used to prove that : $$ (\cup_{n=1}^{\infty} A_n )^c= \cap _{n=1}^{\infty} (A_n) ^c $$

I don't think statement is true for infinite sets

example

$ A_1= {1,2,3,4,...} $

$ A_2 ={2,3,4,...}$

$A_3= {3,4,5,...}$

Now Consider the statement as
$ (\cup_{n=1}^{\infty} A_n ^c )^c= \cap _{n=1}^{\infty} (A_n) $ (Just assuming it to be true and replaced $A_n$ by its complement)

Now LHS is just empty set whilst RHS has infinite elements. Is this right ?

How do I do this ?

Answer 1

Someone correct me if I'm wrong, but I believe the issue here is that while induction can prove

$$\left(\bigcup_{n=1}^{N}A_{n}\right)^{c} = \bigcap_{n=1}^{N}(A_{n})^{c}$$

for any positive integer $N$, it cannot be used to prove the infinite union and infinite intersection are equal. The result however is in fact true in general, even for uncountable unions and intersections - this is called DeMorgan's Law. $$\left(\bigcup_{i\in I}A_{i}\right)^{c} = \bigcap_{i\in I}A^{c}_{i}$$ A proof can be seen here https://math.stackexchange.com/a/660687