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

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

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}


Suppose $x\in\cup_{n=1}^\infty A_n$. Then $x\in A_m=B_1\backslash B_m$ for some $m>1$. That is $x\in B_1$ and $x\not\in B_m$. This implies $x\in B_1$ and $x\not\in\cap_{n=1}^\infty B_n$. This gives one direction of inclusion.

Conversely, suppose $x\in B_1$ and $x\not\in\cap_{n=1}^\infty B_n$. Then, there is some $m>1$ such that $x\not\in B_m$. And so $x\in A_m$ and we have the reverse inclusion.

So a one-step argument can be written as follows $$ x\in\text{LHS}\iff x\in B_1\;\&\;\exists m>1,\;x\not\in B_m \iff x\in B_1\;\&\;x\not\in\cap_{n=1}^\infty B_n\iff x\in\text{RHS}. $$


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