Set-Theoretic Probability

Question

Consider $\{B_i | i \in I\}$ be a collection of events where $I$ is an arbitrary index set. I would like to show that $$\left(\bigcup_{i \in I} B_i\right)^c = \bigcap_{i \in I} B_i^c.$$

My friend recommended some form of induction on the size of $I$, i.e. induct on the value $|I| = n.$ I can see how this is trivial for $n = 1$, and I have sketched out a variation of this proof for $n = 2$.

Consider $$x \in (B_1 \cup B_2)^c$$ $$\iff x \notin B_1 \cup B_2$$ $$\iff x \notin B_1 \wedge x \notin B_2$$ $$\iff x \in B_1^c \wedge x \in B_2^c$$ $$\iff x \in B_1^c \cap B_2^c.$$

However, I am having some troubles extending this to the $n > 2$ case. Any recommendation on how to perform this induction? Or would it be better to do this with a direct proof?

Answer 1

Hint: "$x$ is not in all $B_i$" is equivalent to "there is some $i$ such that $x$ is not in $B_i$".

Answer 2

Instead of induction, rather following a similar argument to what you had, we get:

$$x \in \left(\bigcup_{i\in I}B_i\right)^c \iff x \notin \bigcup_{i \in I}B_i\iff \forall i, x \notin B_i \iff \forall i, x \in B_i^c \iff x \in \bigcap_{i \in I}B_i^c $$

Even if you had an induction proof, I believe this is better because it holds for arbitrary set $I$ rather than simply a countable sets.