If $A_1,A_2,…,A_n$ are independent then $A_1^c,A_2^c,…,A_n^c$ are also independent. Let $A_1,A_2,…,A_n$ be events in a probability space $(\Omega,\Sigma,P)$.
If $A_1,A_2,…,A_n$ are independent then  $A_1^c,A_2^c,…,A_n^c$ are also independent, (where $A^c = \Omega \setminus A$).
I have found a proof by induction for this exercise, however, I have not been able to understand the conclusion of the proof, which I have marked in red. That is, why can it be immediately concluded that $A_1^c , A_2^c ,..., A_{k+1}^c$ are independent? I would really appreciate if someone can give me a clear explanation of what happens in that conclusion.
Proof by induction.
Basis for the Induction.
If $A_1$ and $A_2$ are independent then  $A_1^c$ and $A_2^c$ are independent.
Assume $A_1$ and $A_2$ are independent. Then
\begin{align*}
P(A_1^c \cap A_2^c) 
&= 1 - P(A_1 \cup A_2) \\
&= 1 - P(A_1) - P(A_2) + P(A_1 \cap A_2) \\
&= 1 - P(A_1) - P(A_2) + P(A_1)P(A_2) \\
&= (1-P(A_1))(1-P(A_2)) \\
&= P(A_1^c)P(A_2^c).
\end{align*}
Induction Hypothesis.
This is our induction hypothesis:
If $A_1,A_2,…,A_k$ are independent then $A_1^c,A_2^c,…,A_k^c$ are independent.
Then we need to show:
If $A_1,A_2,…,A_{k+1}$ are independent then $A_1^c,A_2^c,…,A_{k+1}^c$ are independent.
Induction Step.
This is our induction step.
Suppose $A_1,A_2,…,A_{k+1}$ are independent.
Then:
\begin{align}
P\left( {\bigcap_{i = 1}^{k + 1} A_i}\right) &= P\left( \bigcap_{i=1}^{k}A_i \cap A_{k+1} \right) \\
&= \prod_{i=1}^{k}P(A_i) \cdot P(A_{k+1})\\
&= P\left(\bigcap_{i=1}^{k}A_i\right) \cdot P(A_{k+1})
\end{align}
So we see that $\bigcap_{i=1}^{k}A_i$ and $A_{k+1}$ are independent.
So $\bigcap_{i=1}^{k}A_i$ and $A_{k+1}^c$ are independent.
$\color{red}{\text{So, from the above results, we can see that} A_1^c,A_2^c,…,A_{k+1}^c \text{are independent}}.$
 A: Partial attempt:
Let $B_1 := \bigcap_{i=1}^k A_i^c$ and $B_2 := A_{k+1}^c$.
\begin{align}
P\left(\bigcap_{i=1}^{k+1} A_i^c\right)
&= P(B_1 \cap B_2)
\\
&= 1 - P(B_1^c \cup B_2^c)
\\
&= 1 - P(B_1^c) - P(B_2^c) + P(B_1^c \cap B_2^c)
\\
&\overset{*}{=} 1 - P(B_1^c) - P(B_2^c) + P(B_1^c) P(B_2^c)
\\
&= P(B_1) P(B_2)
\\
&= P\left(\bigcap_{i=1}^{k} A_i^c\right) P(A_{k+1}^c)
\\
&= \prod_{i=1}^{k+1} P(A_i^c).
\end{align}
It remains to verify the starred equality $P(B_1^c \cap B_2^c) = P(B_1^c) P(B_2^c)$.
\begin{align}
P(B_1^c \cap B_2^c)
&= P\left(
\left(\bigcap_{i=1}^k A_i^c\right)^c
\cap A_{k+1}
\right)
\\
&= P\left(
\left(\bigcup_{i=1}^k A_i\right)
\cap A_{k+1}
\right)
\\
&\overset{?}{=} P\left(\bigcup_{i=1}^k A_i \right) P(A_{k+1})
\\
&= P(B_1^c) P(B_2^c).
\end{align}
For the "?" equality, I think you can show this by writing $\bigcup_{i=1}^k A_i$ as the disjoint union of intersections of $A_1, \ldots, A_k, A_1^c, \ldots, A_k^c$.
A: Hi Inquirer: My apologies for the delay. I've had a lot going on so I just got back to this today.
I figured I'd put it in an answer since you get more space that way.
So, you agree that you proved the statement for $i = 2$ case, right. You showed that if $A_1$ and $A_2$ are independent, then $A^{c}_{1}$ and $A^{c}_{2}$ are independent. So, what induction says is that, if we have proven the statement for $i = 2$ and then we can show that, it being true for $i = 2$, implies that it is also true for the $i = 3$ case, then we are done with the proof by an induction argument.
So, let us assume that $A_{1}, A_{2}$ and $A_3$ are independent. We want to show that, given that the statement is true for $A_{1}$ and $A_{2}$, then this implies that $A^{c}_{1}, A^{c}_{2}$ and $A^{c}_{3}$ are independent.
So, what we can do for clarity is let $k = 2$ and then use the same proof that you used in your question.
\begin{align}
P\left( {\bigcap_{i = 1}^{2 + 1} A_i}\right) &= P\left( \bigcap_{i=1}^{2}A_i \cap A_{2+1} \right) \\
&= \prod_{i=1}^{2}P(A_i) \cdot P(A_{2+1})\\
&= P\left(\bigcap_{i=1}^{2}A_i\right) \cdot P(A_{2+1}) \\
&= P\left(\bigcap_{i=1}^{2}A_i\right) \cdot P(A_{3})
\end{align}
Next we can use the trick of letting $A_{1}^{*} = \bigcap_{i=1}^{2}A_i$ and letting $A_{2}^{*} = A_{3}$.
So, what we have shown above is that that $A^{*}_1$ and $A^{*}_2$ are independent. But we then know that (because it's already been proven for $i = 2$) that $A^{c*}_{1}$ and $A^{c*}_{2}$ are independent. But $A^{c*}_{1} = \bigcap_{i=1}^{2}A^{c}_i$  and $A^{*c}_{2} = A^{c}_{3}$ so this
means that $A^{c}_{1}, A^{c}_{2}$ and $A^{c}_{3}$ are independent. This is because $\left(\bigcap_{i=1}^{2}A^{c}_i\right) \bigcap A^{c}_{3} = \bigcap_{i=1}^{3}A^{c}_i$.
So, what we have shown is that the $i = 2$ case being true implies that the $i = 3$ case is also true so, since this argument holds for $k = 2$, it implies that it also holds for any value of $k$. So we have proved the statement using induction. Does that make sense ? If not, let me know what part doesn't and I'll try to explain it more clearly.
