# $\bigcap\limits_{n=1}^{\infty} ( \bigcup\limits_{i=1}^{n} G_i )^c = ( \bigcup\limits_{n=1}^{\infty} (\bigcup\limits_{i=1}^{n} G_i ))^c$

Why is $\bigcap\limits_{n=1}^{\infty} \left( \bigcup\limits_{i=1}^{n} G_i \right)^c = \left( \bigcup\limits_{n=1}^{\infty} \left( \bigcup\limits_{i=1}^{n} G_i \right) \right)^c$? What set properties are being applied here? (The $^c$ is set complement)

$$A^c \cap B^c = \left(A \cup B\right)^c$$
For each $n\in\Bbb Z^+$ let $A_n=\bigcup_{i=1}^nG_i$, and let $C_n=A_n^c$; then $A_1\subseteq A_2\subseteq A_3\subseteq\ldots$, and therefore $C_1\supseteq C_2\supseteq C_3\supseteq\ldots\;$, a fact that may help you visualize the situation. You’re being asked to show that
$$\bigcap_{n\ge 1}C_n=\left(\bigcup_{n\ge 1}A_n\right)^c\;.$$
You can do this directly by element-chasing: assume that $x\in\bigcap_{n\ge 1}C_n$ and show that $x\notin\bigcup_{n\ge 1}A_n$, and then assume that $x\notin\bigcup_{n\ge 1}A_n$ and show that $x\in\bigcap_{n\ge 1}C_n$.