Difference between revisions of "Notes:Hereditary sigma-ring"

Revision as of 03:14, 8 April 2016

I'm writing down some "facts" so I don't keep redoing them on paper.

What I want to show

• $\mathcal{H}(\sigma_R(S))=\sigma_R(\mathcal{H}(S))$ for a system of sets, $S$.

Both are hereditary, and both are $\sigma$-rings.

The result is true!

TODO: Make a theorem out of this!

Showing $\mathcal{H}(\sigma_R(S))=\sigma_R(\mathcal{H}(S))$

Which way first?

$\mathcal{H}(\sigma_R(S))\subseteq\sigma_R(\mathcal{H}(S))$

• Let $A\in\mathcal{H}(\sigma_R(S))$ be given. Then:
  • $\exists B\in\sigma_R(S)[A\in\mathcal{P}(B)]$
    • Notice that $S\subseteq\mathcal{H}(S)$ thus:
      • $\sigma_R(S)\subseteq\sigma_R(\mathcal{H}(S))$
    • So $B\in\sigma_R(\mathcal{H}(S))$
  • As $\sigma_R(\mathcal{H}(S))$ is hereditary $\forall C\in\mathcal{P}(B)[C\in\sigma_R(\mathcal{H}(S))$
    • So $A\in\sigma_R(\mathcal{H}(S))$
• This shows $\mathcal{H}(\sigma_R(S))\subseteq\sigma_R(\mathcal{H}(S))$

$\sigma_R(\mathcal{H}(S))\subseteq\mathcal{H}(\sigma_R(S))$

• Let $A\in\sigma_R(\mathcal{H}(S))$, then, either (by fact 3):
  1. $A\in\mathcal{H}(S)$
     • This means: $\exists B\in S[A\in\mathcal{P}(B)]$ (also said as: $\exists B\in S[A\subseteq B]$)
     • But $S\subseteq\sigma_R(S)$, thus:
       • $B\in\sigma_R(S)$
     • But $\sigma_R(S)\subseteq\mathcal{H}(\sigma_R(S))$ thus:
       • $B\in\mathcal{H}(\sigma_R(S))$
     • As $\mathcal{H}(\sigma_R(S))$ is hereditary, all subsets of $B$ are in it.
     • As $A\in\mathcal{P}(B)$ we see $A\in\mathcal{H}(\sigma_R(S))$ - this completes this half of the proof.
  2. $\exists ({ A_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{H}(S) [\bigcup_{n=1}^\infty A_n=A]$
     • Using part 1 we see that: $\forall i\in\mathbb{N}[A_i\in\mathcal{H}(\sigma_R(S))]$
       • But we know $\mathcal{H}(\sigma_R(S))$ is a $\sigma$-ring, thus closed under countable union
     • Thus $\bigcup_{n=1}^\infty A_n\in\mathcal{H}(\sigma_R(S))$
       • But $A:=\bigcup_{n=1}^\infty A_n$ so
     • $A\in\mathcal{H}(\sigma_R(S))$
• We have shown that in either case, $A\in\mathcal{H}(\sigma_R(S))$

Facts

1. An hereditary system is a sigma-ring $\iff$ it is closed under countable unions.
   • Thus $\sigma_R(\mathcal{H}(S))$ is just $\mathcal{H}(S)$ with the additional property:
     • $\forall(A_n)_{n=1}^\infty\subseteq\mathcal{H}(S)\left[\bigcup_{n=1}^\infty A_n\in\sigma_R(\mathcal{H}(S))\right]$
2. $\mathcal{H}(\mathcal{R})$ is a $\sigma$-ring (for any $\sigma$-ring, $\mathcal{R}$)
   • This means $\sigma_R(\mathcal{H}(\mathcal{R}))=\mathcal{H}(\mathcal{R})$
   • It also means $\mathcal{H}(\sigma_R(S))$ is a $\sigma$-ring
3. $\sigma_R(\mathcal{H}(S))$ is just $\mathcal{H}(S)$ closed under countable union.
4. $\sigma_R(\mathcal{H}(S))$ is hereditary

Proof of facts

1. An hereditary system is a sigma-ring $\iff$ it is closed under countable unions.
   1. Hereditary system is a sigma-ring $\implies$ closed under countable unions
      • It is a $\sigma$-ring which means it is closed under countable unions. Done
   2. A hereditary system closed under countable union $\implies$ it is a $\sigma$-ring
      1. closed under set-subtraction
         • Let $A,B\in\mathcal{H}$ for some hereditary system $\mathcal{H}$. Then:
           • $A-B\subseteq A$, but $\mathcal{H}$ contains $A$ and therefore all subsets of $A$
         • Thus $\mathcal{H}$ is closed under set subtraction.
      2. Closed under countable union is given.
2. $\mathcal{H}(\mathcal{R})$ is a $\sigma$-ring (for any $\sigma$-ring, $\mathcal{R}$)
   1. It is already shown that a hereditary system is closed under set subtraction, only remains to be shown closed under countable union
   2. Closed under countable union
      • Let $(A_n)_{n=1}^\infty\subseteq\mathcal{H}(\mathcal{R})$ (we need to show $\implies\bigcup_{n=1}^\infty A_n\in\mathcal{H}(\mathcal{R})$)
        • This means, for each $A_n\in\mathcal{H}(\mathcal{R})$ there is a $B_n\in\mathcal{R}$ with $A_n\subseteq B_n$ thus:
          • $\forall(A_n)_{n=1}^\infty\subseteq\mathcal{H}(\mathcal{R})\exists(B_n)_{n=1}^\infty\subseteq\mathcal{R}\forall i\in\mathbb{N}[A_i\subseteq B_i]$
        • However $\mathcal{R}$ is a $\sigma$-ring, thus:
          • Define $B:=\bigcup_{n=1}^\infty B_n$, notice $B\in\mathcal{R}$
        • But a union of subsets is a subset of the union, thus:
          • $\bigcup_{n=1}^\infty A_n\subseteq\bigcup_{n=1}^\infty B_n:=B$, thus
            • $\bigcup_{n=1}^\infty A_n\subseteq B$
          • BUT $\mathcal{H}(\mathcal{R})$ contains all subsets of all things in $\mathcal{R}$, thus contains all subsets of $B$.
        • Thus $\bigcup_{n=1}^\infty A_n\in\mathcal{H}(\mathcal{R})$
      • Thus $\mathcal{H}(\mathcal{R})$ is closed under countable union.
3. $\sigma_R(\mathcal{H}(S))$ is just $\mathcal{H}(S)$ closed under countable union.
   • Follows from fact 1. As $\mathcal{H}(S)$ is an hereditary system, the sigma-ring generated by it (the smallest sigma ring containing $\mathcal{H}(S)$ is just the set with whatever is needed to close it under the operators)
4. $\sigma_R(\mathcal{H}(S))$ is hereditary
   • Let $A\in\sigma_R(\mathcal{H}(S))$ be given. We want to show that $\forall B\in\mathcal{P}(A)$ that $B\in\sigma_R(\mathcal{H}(S))$.
     1. If $A\in\mathcal{H}(S)$, then $\forall B\in\mathcal{P}(A)[B\in\mathcal{H}(S)$ but $B\in\mathcal{H}(S)\implies B\in\sigma_R(\mathcal{H}(S))$
        • We're done in this case.
     2. OTHERWISE: $\exists(A_n)_{n=1}^\infty\subseteq\mathcal{H}(S)\left[\bigcup_{n=1}^\infty A_n=A\right]$ (by fact 3)
        • Let $B\in\mathcal{P}(A)$ be given.
          • Define a new sequence, $({ B_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{H}(S)$, where $B_i:=A_i\cap B$
            • $A_i\cap B$ is a subset of $A_i$ and $A_i\in\mathcal{H}(S)$, as "hereditary" means "contains all subsets of" $A_i\cap B\subseteq A_i$ thus $A_i\cap B:=B_i\in\mathcal{H}(S)$
          • Clearly $B=\bigcup_{n=1}^\infty B_n$ (as $B\subseteq A$ and $A=\bigcup_{n=1}^\infty A_n$)
          • As $\sigma_R(\mathcal{H}(S)$ contains all countable unions of things in $\mathcal{H}(S)$ we know:
            • $\bigcup_{n=1}^\infty B_n=B\in\sigma_R(\mathcal{H}(S))$
        • We have shown $B\in\sigma_R(\mathcal{H}(S))$
   • We have completed the proof

TODO: It seems, "hereditary sigma-ring" is the same as "sigma-ideal".