3

"For if $ A=\bigcup A^{'}_{n}$ with $A^{'}_{n} \in M_F(\mu)$, write $A_1=A^{'}_{1} $, and $$ A_n=(A^{'}_1\cup ...\cup A^{'}_n)-(A^{'}_n \cup ... \cup A^{'}_{n-1})$$ $(n=2,3,4,...)$.

Then $$ A=\bigcup_{n=1}^{\infty}A_n$$ I can't understand why $A_n$ is expressed like the above? Should the correct one be $$ A_n=A^{'}_n-(A^{'}_1 \cup ... \cup A^{'}_{n-1})$$

JamesWang
  • 49
  • The two formulas are equivalent. Rudin's choice might make more apparent what $A_1\cup\cdots\cup A_n$ is and consequently why $A=\bigcup\limits_nA_n$. – Did Sep 01 '14 at 06:16
  • You mean that the following 2 are equivalent ?$$ A_n=(A^{'}1\cup ...\cup A^{'}_n)-(A^{'}_n \cup ... \cup A^{'}{n-1})$$

    $$ A_n=A^{'}n-(A^{'}_1 \cup ... \cup A^{'}{n-1})$$ why?

     – JamesWang Sep 01 '14 at 06:24
  • Yes, this is what I mean (starting from $A_1=A_1'$). Why: double inclusion, if you wish, or drawing the case n=3. – Did Sep 01 '14 at 06:29
  • for n=3 the 1st one is $$ A_3=(A^{'}1\cup A^{'}_2\cup A^{'}_3)-(A^{'}_3\cup A^{'}{2})$$ the 2nd one is

    $$ A_3=A^{'}3-(A^{'}_1 \cup A^{'}{2})$$ are both equivalent ?

     – JamesWang Sep 01 '14 at 06:40
  • 1
    Aaahh... typo! One should define $(A_n)$ by $A_1=A_1'$ and $$ A_n=(A^{'}1\cup ...\cup A^{'}_n)-(A^{'}_1 \cup ... \cup A^{'}{n-1}).$$ Then my previous comments apply. Sorry to have missed that. Are you sure Rudin uses your formula? – Did Sep 01 '14 at 06:47
  • Yes. I am so confused. Page 307 of Rudin Principles of mathematical analysis. thanks a lot – JamesWang Sep 01 '14 at 06:51

1 Answers1

3

At least in some editions of the book (say, the 3rd edition) there seems to be a typo. The defining identity $$ A_n=(A^{'}_1\cup ...\cup A^{'}_n)-(A^{'}_n \cup ... \cup A^{'}_{n-1})$$ should be replaced by $$ A_n=(A^{'}_1\cup ...\cup A^{'}_n)-(A^{'}_1 \cup ... \cup A^{'}_{n-1}).$$ Since $A_1=A^{'}_1$, this is equivalent to $$ A_n=A^{'}_n-(A^{'}_1 \cup ... \cup A^{'}_{n-1}).$$ Each definition leads to $$A_1\cup ...\cup A_n=A^{'}_1\cup ...\cup A^{'}_n,$$ for every $n$, and the rest follows.

Did
  • 279,727