1
$\begingroup$

Context: Measure theory.

Reason: Just curious.

Question: Given $\{A_k\}$ with $A_k$ not disjoint, $B_1=A_1$ and $B_n = A_n - \bigcup\limits_{k=1}^{n-1} A_k$ for $n \in \mathbb{N}-\{1\}$ and $k \in \mathbb{N}$, how can I show that $$\bigcup\limits_{n=2}^{\infty}A_n = \bigcup\limits_{n=2}^{\infty}B_n?$$

Attempt: $$\bigcup\limits_{n=2}^{\infty}B_n=\bigcup\limits_{n=2}^{\infty}\left(A_n \cap (\bigcup\limits_{k=1}^{n-1} A_k)^c\right)=\bigcup\limits_{n=2}^{\infty}A_n \bigcap \bigcup\limits_{n=2}^{\infty}\left(\bigcup\limits_{k=1}^{n-1} A_k\right)^c = \cdots$$

Where do I go from here?

$\endgroup$
0
$\begingroup$

(I'm adding this answer since it shows that the statement as written is not completely correct, because it shows a different style of proof, and because I discovered a direct connection with induction.)

$ \newcommand{\calc}{\begin{align} \quad &} \newcommand{\calcop}[2]{\\ #1 \quad & \quad \text{"#2"} \\ \quad & } \newcommand{\endcalc}{\end{align}} \newcommand{\ref}[1]{\text{(#1)}} $Let's first restate the statement in a slightly different notation, with $\;B_n\;$ expanded, and implicitly assuming that $\;k,n\;$ are integers with $\;k \ge 1\;$ and $\;n \ge 2\;$: $$ \tag 0 \langle \cup n :: A_n \rangle \;=\; \langle \cup n :: A_n - \langle \cup k : k < n : A_k \rangle \rangle $$

The most straightforward approach for proofs like this is to start with the most complex side, and calculate the elements $\;x\;$ of that set: $$\calc x \in \langle \cup n :: A_n - \langle \cup k : k < n : A_k \rangle \rangle \calcop\equiv{expand definition of $\;\cup\;$ twice, and of $\;-\;$} \langle \exists n :: x \in A_n \land \lnot \langle \exists k : k < n : x \in A_k \rangle \rangle \calcop\equiv{logic: rewrite using DeMorgan twice, and 'shunting'} \lnot \langle \forall n : \langle \forall k : k < n : x \not\in A_k \rangle : x \not\in A_n \rangle \tag{*} \endcalc$$

The last step was inspired by the shape of the formula, since the last statement reminds us of the principle of complete induction (see Wikipedia and another question): $$ \tag 1 \langle \forall i : i \in V \land \langle \forall j : j \in V \land j < i : P_j \rangle : P_i \rangle \;\equiv\; \langle \forall i : i \in V : P_i \rangle $$ where $\;i,j\;$ range over the same set $\;V\;$ with an ordering $\;<\;$. (Often $\ref 1$ is presented only with the $\;\Rightarrow\;$ direction, but since the other direction is trivial, it is usually simpler to use the stronger more symmetrical version.)

However, there is a difference between $\ref *$ and $\ref 1$: in $\ref *$ the dummies do not range over the same set, since $\;n \in \mathbb N_{\ge 2}\;$ but $\;k \in \mathbb N_{\ge 1}\;$. So we cannot use $\ref 1$ directly.

Therefore, let's try to rewrite $\ref *$ so that we can use $\ref 1$: $$\calc \tag{*} \lnot \langle \forall n : \langle \forall k : k < n : x \not\in A_k \rangle : x \not\in A_n \rangle \calcop\equiv{logic: split off $\;k = 1\;$} \lnot \langle \forall n : x \not\in A_1 \land \langle \forall k : 2 \le k \land k < n : x \not\in A_k \rangle : x \not\in A_n \rangle \calcop\equiv{logic: DeMorgan; move $\;x \not\in A_1\;$ out of $\;\exists n\;$; DeMorgan} x \not\in A_1 \land \lnot \langle \forall n : \langle \forall k : 2 \le k \land k < n : x \not\in A_k \rangle : x \not\in A_n \rangle \calcop{\tag{**} \equiv}{using induction principle $\ref 1$ on $\;\mathbb N_{\ge 2}\;$ with $\;P_n := x \not\in A_n\;$} x \not\in A_1 \land \lnot \langle \forall n :: x \not\in A_n \rangle \calcop\equiv{logic: DeMorgan; definitions of $\;\cup\;$ and $\;-\;$} x \in \langle \cup n :: A_n \rangle - A_1 \endcalc$$

By set extensionality, the above calculation proves $$ \tag 2 \langle \cup n :: A_n \rangle - A_1 \;=\; \langle \cup n :: A_n - \langle \cup k : k < n : A_k \rangle \rangle $$ and the key step was $\ref{**}$ where we applied the principle of complete induction.

Finally, this implies that $\ref 0$ is not true: now that we know $\ref 2$ it is easy to find a counterexample, e.g., if $\;A_1\;$ and $\;A_2\;$ share an element, then that element is in the left hand side of $\ref 0$ but not in its right hand side.

$\endgroup$
2
$\begingroup$

It might be easier to show each side is a subset of the other.

$\bigcup A_n \subset \bigcup B_n$

For $x \in \bigcup A_n$, let $n$ be the smallest integer such that $x \in A_n$. Then $x \in B_n$.

$\bigcup A_n \supset \bigcup B_n$

For $x \in \bigcup B_n$, there exists a [unique] $n$ such that $x \in B_n$. Then $x \in A_n$.

$\endgroup$
0
$\begingroup$

Try showing the double inclusion. One side is easy as $B_i \subseteq A_i$ for all $i$. For the other side, think of $x \in \bigcup A_i$, and let $A_k$ the first $k$ such that $x \in A_k$, what can you say about $x$ and $B_k$?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy