Suppose I wanted to prove $\Big(\bigcup_{n=1}^{\infty} B_n \Big)^c = \Big(\bigcap_{n=1}^{\infty} B_n^c\Big)$. I've written out a proof of this, and was hoping someone could tell me whether it appears to be on the right track. I've proved several types of theorems of this kind, but only with a finite number of sets.
Take the set $X \in \Big(\bigcup_{n=1}^{\infty} B_n \Big)^c$. Then, $X \not \in \bigcup_{n=1}^{\infty} B_n$, and so for $n \in [1, \infty)$, $X \in B_k^c$, equivalent to $\Big(\bigcap_{n=1}^{\infty} B_n^c\Big)$. So $\Big(\bigcup_{n=1}^{\infty} B_n \Big)^c \subset \Big(\bigcap_{n=1}^{\infty} B_n^c\Big)$.
(The last step seems like a jump to me, but I don't quite know another way other than to say $X \not \in B_1, X \not \in B_2, \ldots$ and thus $X \in B_1^c$, $X \in B_2^c$, $X \in B_c^x, \ldots$, but that doesn't seem convincing without some sort of inductive argument.)
Now take $X \in\Big(\bigcap_{n=1}^{\infty} B_i^c\Big)$. This requires that $X \not \in B_i$ for $i \in [1, \infty)$, and $x \not \in \bigcup_{n=1}^{\infty} B_n$, which implies $X \in \Big(\bigcup_{n=1}^{\infty} B_n\Big)^c$, and thus $\Big(\bigcap_{n=1}^{\infty} B_n^c\Big) \subset \Big(\bigcup_{n=1}^{\infty} B_n\Big)^c$.
From $\Big(\bigcup_{n=1}^{\infty} B_n \Big)^c \subset \Big(\bigcap_{n=1}^{\infty} B_n^c\Big)$ and $\Big(\bigcap_{n=1}^{\infty} B_n^c\Big) \subset \Big(\bigcup_{n=1}^{\infty} B_n\Big)^c$, we conclude $\Big(\bigcup_{n=1}^{\infty} B_n \Big)^c = \Big(\bigcap_{n=1}^{\infty} B_n^c\Big)$.
How does this proof look? Are there jumps in my logic?
Thanks in advance.
Best Answer
Your logic is fine. There is really no distinction here between a finite number of sets and an infinite number of sets. Induction isn't necessary.
If $\{B_i\}_{i \in I}$ is an indexed family of sets (one set $B_i$ for each element $i$ of the set $I$) the union is defined by $$x \in \bigcup_{i \in I} B_i \iff (\exists i \in I) x \in B_i$$ and the intersection is defined by $$x \in \bigcap_{i \in I} B_i \iff (\forall i \in I) x \in B_i.$$
The whole argument comes down to properly negating quantified statements. Thus you could write \begin{align*} x \in \left( \bigcup_{i \in I} B_i \right)^c &\iff x \notin \bigcup_{i \in I} B_i \\ &\iff \neg (\exists i \in I) x \in B_i \\ &\iff (\forall i \in I) x \notin B_i \\ &\iff (\forall i \in I) x \in B_i^c \\ &\iff x \in \bigcap_{i \in I} B_i^c. \end{align*}