In an exam, my teacher wrote this problem, I proved it in a similar way to the answer in that post, but he pointed out to me that a step in the proof is wrong.
What my teacher says
Let $A_1 \supset A_2 \supset A_3 \supset \dots$ sequence of decreasing sets
We can construct the following sequence of disjoint sets
$B_1 = A_1 – A_2$
$B_2 = A_2 – A_3$
$B_3 = A_3 – A_4$
$\dots$
let $A = \cap_{k \in \mathbb{N}} A_k$, then $A$ and $\cup_{k = N}^{\infty} B_k$ do not form a partition of $A_N$, are not disjoint, in fact $A$ is contained in the union $\cup_{k = N}^{\infty} B_k$.
But I think he is wrong, and I have this proof to prove that step.
My proof
$A_1 – \cap_{k \in \mathbb{N}} A_k = A_1 \cap (\cap_{k \in \mathbb{N}} A_k)^c = A_1 \cap (\cup_{k \in \mathbb{N}} A_k^c)$
The intersection is distributed over the union, then
$A_1 \cap (\cup_{k \in \mathbb{N}} A_k^c) = \cup_{k \in \mathbb{N}} (A_1 \cap A_k^c) = $
$= \emptyset \cup (A_1 – A_2) \cup (A_1 – A_3) \cup (A_1 – A_4) \cup … = $
since the sequence is decreasing, finally
$= \emptyset \cup (A_1 – A_2) \cup (A_1 – A_3) \cup (A_1 – A_4) \cup … = $
$= \emptyset \cup (A_1 – A_2) \cup (A_2 – A_3) \cup (A_3 – A_4) \cup … =$
$= \cup_{k \in \mathbb{N}} B_k$
This is $A_1 – \cap_{k \in \mathbb{N}} A_k = \cup_{k \in \mathbb{N}} B_k$.
The case $A_N – \cap_{k \in \mathbb{N}} A_k = \cup_{k = N}^{\infty} B_k$ is analogous.
Is my proof correct?
Best Answer
If your teacher says that the sets $A,B_N,B_{N+1},\dots$ do not form a partition of $A_N$ then he is wrong.
It is evident that the sets $B_N, B_{N+1},\dots$ are disjoint.
Further if $x\in A$ then $x\in A_n$ for every $n$ implying, that no $n$ exists with $x\in B_n$.
This because $B_n\subseteq A_{n+1}^c$ by definition.
This makes it evident that the sets $A,B_N,B_{N+1},\dots$ are disjoint, and it is evident as well that their union coincides with $A_N$.
Your proof is correct and confirms that $A_N-A=\bigcup_{k=N}^{\infty}B_k$.
This combined with $A\subseteq A_N$ and disjointness of the $B_i$ (again) proves that $A,B_N,B_{N+1},\dots$ form a partition of $A_N$.