Show that $\bigcup_{n=1}^\infty A_n= B_1 \backslash \bigcap_{n=1}^\infty B_n$ Let $\{B_n\}$ be a decreasing set
$B_1 \supseteq B_2 \supseteq B_3 \supseteq ....$
Define $A_n = B_1 \backslash B_n$ i.e. $A_1 = \varnothing, A_2 = B_1 \backslash B_2$
If we imagine $\{B_n\}$ as a donut then it is clear that $\{A_n\}$ is increasing

Show: 

$\bigcup_{n=1}^\infty A_n = B_1 \backslash \bigcap_{n=1}^\infty B_n$

Seems like a proof by exhaustion?
$A_1 \cup A_2 = (B_1 \backslash B_1) \cup (B_1\backslash B_2) = B_1\backslash B_2$
$A_1 \cup A_2 \cup A_3 = (B_1\backslash B_2) \cup (B_1 \backslash B_3) = \text{ ...hoping... } = B_1 \backslash (B_2 \cap B_3)$
It seems the derivation is a little bit heavy:
$(B_1\backslash B_2) \cup (B_1 \backslash B_3)  = (B_1 \cap B_2^c) \cup (B_1 \cap B_3^c) = (B_1 \cup (B_1 \cap B_3^c)) \cap (B_2^c \cup (B_1 \cap B_3^c)) = (B_1 \cup B_1) \cap (B_1 \cup B_3^c) \cap (B_2^c \cup B_1) \cap (B_2^c \cup B_3^c) = B_1 \cap (B_2 \cap B_3)^c  = B_1 \backslash (B_2 \cap B_3) $
Continue this way, we can see that the claim is true. 
Is there any easier way to see relation? The proof in my book did it in one step...
I am thinking something along the line where we can use the property of $\backslash$ to directly show $(B_1\backslash B_2) \cup (B_1 \backslash B_3) = B_1 \backslash (B_2 \cap B_3)$

This holds even without nested condition on $B_n$.
\begin{align}
\bigcup_{n=1}^{\infty}A_n&=\bigcup_{n=1}^{\infty}(B_1-B_n)
\\
&=\bigcup_{n=1}^{\infty}(B_1\cap B_n^c)
\\
&=B_1\cap\bigcup_{n=1}^{\infty}B_n^c
\\
&=B_1\cap\left(\bigcap_{n=1}^{\infty}B_n\right)^c
\\
&=B_1-\bigcap_{n=1}^{\infty}B_n
\end{align}
