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If R is $\sigma -$ring then {$E\in X|E\in R \text { or} \ E^c \in R$ } is a $\sigma$ -algebra

$\sigma -$ring : A family of sets $R\in P(X)$ is called a $\sigma $-ring if it is closed under countable unions and differences of 2 sets.

$\sigma-$ algebra: A family of sets $R\in P(X)$ is called a $\sigma- $algebra if it is closed under countable unions and every element has a complement in it.

I had proved that it is enough to show that if $X\in R$ then $R$ is a $\sigma-$ algebra

Case 1) If $E,E^c\in R$ then by union we are done.

Case 2) If only either $E$ or $E^c \in R,$

then I am not able to proceed.

I wanted to solve this. Please give me only a hint.

Thanks in advance.

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  • $\begingroup$ Can you rephrase the question, what is the meaning of "if...then...then..."? $\endgroup$ Feb 6, 2019 at 12:04
  • $\begingroup$ @CalvinKhor Sir SOrry I had done $\endgroup$ Feb 6, 2019 at 12:08

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Let $\mathcal A:=\{E\in\wp(X)\mid E\in\mathcal R\text{ or }E^{\complement}\in\mathcal R\}$ where $\mathcal R$ denotes a $\sigma$-ring on set $X$.

Then it is immediate that $\mathcal A$ is closed under complements.

Now observe that a $\sigma$-ring is closed under countable intersections.

This because we can write: $$\bigcap_{n=1}^{\infty}R_n=R_1\setminus\bigcup_{n=2}^{\infty}(R_1\setminus R_n)$$

Now let $A_n\in\mathcal A$ for $n=1,2,\dots$ and let it be that $A_n\in\mathcal R$ for $n\in I$ and $A_n^{\complement}\in\mathcal R$ for $n\in J$ where we take $I$ and $J$ to be disjoint and covering subsets of $\mathbb N$.

Then $P:=\bigcap_{n\in I}A_n\in\mathcal R$ and $Q:=\bigcup_{n\in J}A_n^{\complement}\in\mathcal R$.

Consequently $\bigcap_{n=1}^{\infty}A_n=P\setminus Q\in\mathcal R\subseteq\mathcal A$.

Proved is now that $\mathcal A$ is closed under countable intersections, and combined with the fact that it also closed under complements it can be proved now that $\mathcal A$ is closed under countable unions: $\bigcup_{n=1}^{\infty}A_n=\left(\bigcap_{n=1}^{\infty}A_n^{\complement}\right)^{\complement}$.

Proved is now that $\mathcal A$ is a $\sigma$-algebra.

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  • $\begingroup$ Sorry Sir , But I am not understanding WHy A is closed under complement ? From Defination it is not direct I think. $\endgroup$ Feb 6, 2019 at 13:10
  • $\begingroup$ Under what condition do we have $A\in\mathcal A$? Under condition $A\in\mathcal R\vee A^{\complement}\in\mathcal R$. Under what condition do we have $A^{\complement}\in\mathcal A$? Under condition $A^{\complement}\in\mathcal R\vee (A^{\complement})^{\complement}\in\mathcal R$. Now observe that both conditions are exactly the same because $(A^{\complement})^{\complement}=A$. If the condition is satisfied then both $A$ and $A^{\complement}$ are elements of $\mathcal A$. If not then both are not elements of $\mathcal A$. So $A\in\mathcal A\iff A^{\complement}\in\mathcal A$. $\endgroup$
    – drhab
    Feb 6, 2019 at 13:16
  • $\begingroup$ Ohh SO sorry Sir I misunderstood that. Please See my argument : $A\cup A^c=X\in R$ Now $X\A\in R$ this implies closed under complement and cpomtable union is direct . SO we have sigma algebra. THANKS A LOT $\endgroup$ Feb 6, 2019 at 13:21
  • $\begingroup$ Careful: is not proved that $\mathcal R$ is closed under complementation, but that $\mathcal A$ is closed under complementation. Further it is not direct that $\mathcal A$ is closed under countable unions (we need a proof for that, and I gave one). We have $X=A\cup A^{\complement}\in\mathcal A$ but not necessarily $X=A\cup A^{\complement}\in\mathcal R$ as you seem to think. $\endgroup$
    – drhab
    Feb 6, 2019 at 13:25
  • $\begingroup$ Thanks a lot again Sir. Actually I had missed all crux of question . Now cleared...... $\endgroup$ Feb 6, 2019 at 13:33

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