I am trying to solve the following,


Let $\mathcal{A}$ be an algebra,

Given any sequence $\{A_n\}_{n \in \mathbb{N}} \subseteq \mathcal{A}$, with $\cup_{n = 1}^{\infty} A_n = A \in \mathcal{A}$, it is possible to construct a pairwise disjoint sequence ${B_n}_{n \in \mathbb{N}} \subseteq \mathcal{A}$ with $B_n \subseteq A_n$ and $A = \sqcup_{n = 1}^{\infty} B_n$, where $\sqcup$ denotes a disjoint union.

My Solution

I use the following construction,

$B_1 = A_1$, $B_n = A_n - (\cup_{j=1}^{n-1}A_j) \ \forall n \geq 2$.

Since $\mathcal{A}$ is an algebra we have that $B_n \in \mathcal{A} \ \forall n$.

By construction $B_n \subseteq A_n$ for each $n$.

If we let $i > j$, $$\begin{align*} B_i \cap B_j & = (A_i - (\cup_{n = 1}^{i - 1} A_n)) \cap (A_j - \cup_{n = 1}^{j-1} A_n) \\ & = (A_i \cap (\cap_{n = 1}^{i - 1} A_n^c)) \cap (A_j \cap (\cap_{n = 1}^{j-1} A_n^c)) \\ & = (A_i \cap (\cap_{n = 1}^{i - 1} A_n^c)) \cap (A_j) \\ & = A_i \cap A_j \cap A_j^c \cap (\cap_{n = 1}^{j - 1} A_n) \cap (\cap_{n=j+1}^{i-1} A_n) = \emptyset \end{align*}$$

Now I want to show that $\sqcup_{n = 1}^{\infty} B_n = \cup_{n = 1}^{ \infty} A_n$.

I can show $$\begin{align*}B_1 \sqcup B_2 & = A_1 \sqcup (A_2 - A_1) \\ & = A_1 \sqcup (A_2 \cap A_1^c) \\ & = (A_1 \sqcup A_2) \cap (A_1 \sqcup A_1^c) \\ &= A_1 \cup A_2 \end{align*}$$

From here I could go on by induction to show that it holds for all $n \in \mathbb{N}$.

However if I understand induction correctly this argument is not valid when I want to proof something for a countable union.


How can I show directly (without induction) that $\sqcup_{n = 1}^{\infty} B_n = \cup_{n = 1}^{ \infty} A_n$ in this case ?


1 Answer 1


Based on your construction, we get $B_1\subset A_1$ and for $n\geq 2$, $B_n\subset A_n$. Hence,

$$\bigcup_{n=1}^{+\infty}B_n\subset \bigcup_{n=1}^{+\infty}A_n.$$

Let $$x\in\bigcup_{n=1}^{+\infty}A_n.$$ Then there exists $p\in\Bbb N$ such that $x\in A_p$. Define $$S=\{m\in\Bbb N:x\in A_m\}.$$

Then $p\in S$. Thus, $\emptyset\neq S\subset \Bbb N$. By the Well Ordering Principle, $S$ has a least element, say $k$. Then $k\geq 1$. If $k=1$, then $x\in A_1$, and so $x\in B_1\subset \bigcup_{n=1}^{+\infty}B_n.$ So, suppose that $k\geq 2$. We have $$x\notin A_1,\quad x\notin A_2,\quad \dots\quad,x\notin A_{k-1},\quad \text{and}\quad x\in A_k.$$ Thus, $$x\notin \bigcup_{i=1}^{k-1}A_i.$$ Hence, $$x\in A_k\smallsetminus\bigcup_{i=1}^{k-1}A_i=B_k\subset\bigcup_{n=1}^{+\infty}B_n.$$ Thus, $$x\in\bigcup_{n=1}^{+\infty}B_n.$$ Hence, $$\bigcup_{n=1}^{+\infty}A_n\subset\bigcup_{n=1}^{+\infty}B_n.$$ Equality follows.


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