Set theory equation that "should be easy to see", but is not

Question

I have a set-theory equation in my book that I don't understand how to prove.

I have a disjoint collection of sets $A_1, A_2,\cdots\ , A_n$ and a disjoint collection $A_1\star, A_2\star, ..., A_n\star$ is constructed with the property that $(A_1\star)\cup(A_2\star)\cup(A_3\star)\cdots = (A_1)\cup(A_2)\cup(A_3)\cdots$

$A_i\star$ is defined by $$A_1\star = A_1,\quad A_i\star = A_i \diagdown [(A_1)\cup(A_2)\cup\cdots(A_{i-1})]$$ where $$A_i \diagdown [(A_1)\cup(A_2)\cup\cdots(A_{i-1})] = A_i\cap\overline{[(A_1)\cup(A_2)\cup...(A_{i-1})]}$$.

The book says it "should be easy to see that" $$(A_1\star)\cup(A_2\star)\cup(A_3\star)\cup... = (A_1)\cup(A_2)\cup(A_3)\cup\cdots$$

Should I use induction, and if so how?

Thanks.

Answer 1

If we look at the union step for step, in each step $i$, $A_i^*$ differs from $A_i$ in that it has been taken away the elements from $A_{i-1}, A_{i-2},\dots,A_1$. But these elements are already in your union up to this step, hence it doesnt matter (this is an argument of induction; what I wrote actually proves, that for each $i\in\mathbb{N}$ the two different unions up to this point are the same).

Thus, you know that all of the finite unions are the same. Now, every element that is in an infinite union, already belongs to a finite sub-union. Therefore, if $a$ is in the non-star union, you can find a finite sub-union that contains $a$, but this sub-union is equal to the star sub-union, hence the infinite star-union contains $a$ as well.

Answer 2

You don’t really need induction. It’s obvious that $\bigcup_{k\in\Bbb Z^+}A_k^*\subseteq\bigcup_{k\in\Bbb Z^+}A_k$. Now suppose that $x\in\bigcup_{k\in\Bbb Z^+}A_k$, and let $m=\min\{k\in\Bbb Z^+:x\in A_k\}$; clearly $x\in A_m^*$, and it follows immediately that $\bigcup_{k\in\Bbb Z^+}A_k\subseteq\bigcup_{k\in\Bbb Z^+}A_k^*$ and hence that $\bigcup_{k\in\Bbb Z^+}A_k^*=\bigcup_{k\in\Bbb Z^+}A_k$.

Answer 3

Use induction as below

For $n=1$, the definition tells that the statement $A_1\star=A_1$ holds.

Let this be true for $n=k$.

Now, for $n=k+1$

$$\bigcup_{j=1}^{k+1}A_{j}\star=\left(\bigcup_{j=1}^{k}A_{j}\star\right)\bigcup A_{k+1}\star\\ =\left(\bigcup_{j=1}^{k}A_{j}\right)\bigcup \left(A_{k+1}\bigcap\overline{\bigcup_{j=1}^k A_j}\right)\\ =\left(\bigcup_{j=1}^{k+1}A_{j}\right)\bigcap\left(\bigcup_{j=1}^{k}A_{j}\bigcup\overline{\bigcup_{j=1}^k A_j}\right)\\ =\left(\bigcup_{j=1}^{k+1}A_{j}\right)\bigcap \Omega=\bigcup_{j=1}^{k+1}A_{j}$$ where $\Omega $ is the universal set. Hence the statement is true for all $n\in \mathbb{N}$.