is it true that $(\bigcap_{i=1}^{n} U_i)^c = \bigcup_{i=1}^{n} U_{i}^{c}$? Is it true that $(\bigcap_{i=1}^{n} U_i)^c  = \bigcup_{i=1}^{n} U_{i}^{c}$? Where $c$ denotes the compliment. I drew a picture for two sets and tried to convinced myself that it was true, but a more formal proof/explenation would be nice.
 A: $$x\in\left(\bigcap_{i=1}^nU_i\right)^c\iff x\notin\bigcap_{i=1}^nU_i\iff\cdots\iff x\in\bigcup_{i=1}^nU_i^c$$
Try to work this out yourself.
A: De Morgan's laws give you that $(A \bigcap B)^c = A^c \bigcup B^c$ ($\star$). It can easily be verified as follows:
\begin{align*}
x \in (A \bigcap B)^c \iff x \not\in A\bigcap B \iff x \not\in A \text{ or } x \not\in B \iff x \in A^c \bigcup B^c
\end{align*}
Now let us prove by induction on $n$ that  $ \mathcal{P}(n) : (\bigcap_{i=1}^{n} U_i)^c  = \bigcup_{i=1}^{n} U_{i}^{c}$ is true for every $n \in \mathbb{N}$. Suppose $\mathcal{P}(n-1)$ is true. Then use ($\star$): 
$(\bigcap_{i=1}^{n} U_i)^c = ((\bigcap_{i=1}^{n-1} U_i) \bigcap U_n)^c = (\bigcap_{i=1}^{n-1} U_i)^c \bigcup U_n^c = \bigcup_{i=1}^{n-1} U_{i}^{c}\bigcup U_n^c = \bigcup_{i=1}^{n} U_{i}^{c}$
Thus $\mathcal{P}(n)$ is true, yielding that $\mathcal{P}(n)$ is true for every $n$ by induction.
A: You can prove this by doing the inclusions from left to right and right to left.


*

*Let $x \in \left(\bigcap\limits_{i=1}^n U_i\right)^c$. This means there exists some index $j$ such that $x \notin U_j$. Hence $x \in U_j^c$ and therefore $x \in \bigcup\limits_{i=1}^n U_i^c$. Therefore $\left(\bigcap\limits_{i=1}^n U_i\right)^c \subseteq \bigcup\limits_{i=1}^n U_i^c$.

*Let $x \in \bigcup\limits_{i=1}^n U_i^c$. Then there is some $j$ such that $x \in U_j^c$. Hence $x \notin U_j$ and thus $x \notin \bigcap\limits_{i=1}^n U_i$. Thus $x \in \left(\bigcap\limits_{i=1}^n U_i\right)^c$. Therefore $\bigcup\limits_{i=1}^n U_i^c \subseteq \left(\bigcap\limits_{i=1}^n U_i\right)^c$.

*Since both inclusions are true, we must have that
$$\bigcup\limits_{i=1}^n U_i^c = \left(\bigcap\limits_{i=1}^n U_i\right)^c$$
