Generalized Union and Intersection by Induction

Question

Our teacher, told us to prove,

$$\left( \bigcap_{i=1}^n A_i\right)^c = \bigcup_{i=1}^n (A_i^c) $$

By induction. He told us that it has something to do with DeMorgan.

So my question is on knowing what's on the sets. I think that the left one has all the numbers to n except for the number 1, but in the right I get lost.

Can you explain how I could know the elements on the right-handed set? Also, if you could tell me another hint I would really thank you.

Answer 1

\begin{align} \left( \bigcap_{i=1}^{n+1} A_i\right)^c & = \left(\left( \bigcap_{i=1}^n A_i \right) \cap A_{n+1} \right)^c \\[8pt] & = (B\cap A_{n+1})^c \\[8pt] & = B^c \cup A_{n+1}^c & & \text{de Morgan} \\[8pt] & = \left( \bigcap_{i=1}^n A_i \right)^c \cup A_{n+1}^c \\[8pt] & = \left( \bigcup_{i=1}^n (A_i^c) \right) \cup A_{n+1}^c & & \text{by the induction hypothesis} \\[8pt] & = \bigcup_{i=1}^{n+1} (A_i^c) \end{align}

Answer 2

$$P(n)=\left( \bigcap_{i=1}^n A_i\right)^c = \bigcup_{i=1}^n (A_i^c) $$

Now for P(1) $A_1^c=A_1^c$

Let P(k) be true i.e. $\left( \bigcap_{i=1}^k A_i\right)^c = \bigcup_{i=1}^k (A_i^c)$

Let $\left( \bigcap_{i=1}^{k} A_i \right)=B$

Now for P(k+1):

$\left( \bigcap_{i=1}^{k+1} A_i\right)^c=\left( \bigcap_{i=1}^{k} A_i\cap A_{k+1}\right)^c=(B \cap A_{k+1})^c=(B^c \cup A_{k+1}^c)=\left( \bigcap_{i=1}^{k} A_i \right)\cup A_{k+1}^c$

$=\left( \bigcup_{i=1}^{k} A_i^c \right)\cup A_{k+1}^c=\left( \bigcup_{i=1}^{k+1} A_i^c \right)$