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Definition - An elementary family is a collection $\varepsilon$ of subsets of $X$ such that

i.) $\emptyset\in \varepsilon$

ii.) if $E,F\in \varepsilon$ then $E\cap F\in \varepsilon$

iii.) if $E\in \varepsilon$ then $E^c$ is a finite disjoint union of members of $\varepsilon$

1.7 Proposition - If $\varepsilon$ is an elementary family, the collection $\mathcal{A}$ of finite disjoint unions of members of $\varepsilon$ is an algebra.

Attempted proof:

i.) If $A,B\in \varepsilon$, set $B^c = \bigcup_{1}^{n} C_j$ where $\{C_j\}_{1}^{n}\subseteq \varepsilon$. Then $$A\setminus B = A\cap B^c = A\cap \left(\bigcup_{1}^{n}C_j\right) = \bigcup_{1}^{n}C_j\cap A$$ which is a finite disjoint union of members of $\varepsilon$. Thus we have closure under complements.

ii.) Now we can write $A\cup B = (A\setminus B)\cup B$ and we can see that $A\cup B\in \mathcal{A}$ (If more details needs to be provided please let me know)

iii.) Suppose for a fixed $n$ we always have $\bigcup_{1}^{n} A_j\in \mathcal{A}$ whenever $\{A_j\}_{1}^{n}\subset \varepsilon$. Thus, $$\bigcup_{1}^{n+1}A_j = A_{n+1}\cup \bigcup_{1}^{n}A_j$$

I am not sure where to go from here.

Any suggestions on this is greatly appreciated. Please let me know if something is not clear in what I have written so far or if this can be more neatly presented.

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  • $\begingroup$ Step iii.) is not necessary since it follows directly from ii.) via induction. In addition, you might want to show that the entire space belongs to $\mathcal{A}$. $\endgroup$ – Lionel Ricci May 27 '16 at 10:48
  • $\begingroup$ Ok, but I am pretty sure we still need to show by induction that $\bigcup_{1}^{n+1}A_j\in \mathcal{A}$ what do you mean that I should show the entire space belongs to $\mathcal{A}$? $\endgroup$ – Wolfy May 27 '16 at 10:55
  • $\begingroup$ For the induction step, here is a hint: $\bigcup_{1}^{n+1} A_j = \left( \bigcup_{1}^{n} A_j \right) \cup A_{n+1}$. By definition of an algebra, we need $X \in \mathcal{A}$. This is easy to show but you havn't done it in the above? $\endgroup$ – Lionel Ricci May 27 '16 at 10:58
  • $\begingroup$ I just added that step. But I am not sure where to go from there. $\endgroup$ – Wolfy May 27 '16 at 11:00
  • $\begingroup$ Since $B = \bigcup_{i}^{n} A_j \in \mathcal{A}$ and $A_{n+1} \in \mathcal{A}$, what does step ii.) tell you? $\endgroup$ – Lionel Ricci May 27 '16 at 11:02
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Your proof is in the right direction. Here is it in a complete form.

1.7 Proposition - If $\varepsilon$ is an elementary family, the collection $\mathcal{A}$ of finite disjoint unions of members of $\varepsilon$ is an algebra.

Proof:

We begin by proving two auxiliary results:

i.) $\varepsilon \subseteq \mathcal{A}$

Note that if $A\in \varepsilon$, then consider $\{C_j\}_1^1$ such that $C_1=A$. Then $\{C_j\}_1^1$ is a disjoint family of sets in $\varepsilon$, so $A=\bigcup_{j=1}^1C_j\in \mathcal{A}$. So $\varepsilon \subseteq \mathcal{A}$.

ii.) Now let us prove that for any $\{A_j\}_{1}^{n}\subset \varepsilon$, ($\{A_j\}_{1}^{n}$ don't need to be disjoint), we have $\bigcup_{1}^{n}A_j \in \mathcal{A}$.

We will divide the proof in two steps:

ii.a.) First we prove that: if $A \in \mathcal{A}$ and $B\in \varepsilon$, then $A\cup B\in \mathcal{A}$.

If $A \in \mathcal{A}$ and $B\in \varepsilon$, then we have $A=\bigcup_{k=1}^m D_k$ where $\{D_k\}_{1}^{n}$ is a family of disjoint sets in $\varepsilon$ and $B^c = \bigcup_{j=1}^{n} C_j$ where $\{C_j\}_{1}^{n}\subseteq \varepsilon$ and $\{C_j\}_{1}^{n}$ is a disjoint family of sets. Then $$A\setminus B = A\cap B^c =\left( \bigcup_{k=1}^m D_i\right)\cap \left(\bigcup_{j=1}^{n}C_j\right) = \bigcup_{k=1}^m\bigcup_{j=1}^{n}C_j\cap D_k$$ which is a finite disjoint union of members of $\varepsilon$. So $A\setminus B \in \mathcal{A}$.

Now we can write $A\cup B = (A\setminus B)\cup B$. Since $A\setminus B$ is finite disjoint unions of members of $\varepsilon$, $B\in \varepsilon$ and $(A\setminus B)\cap B =\emptyset$, we can see that $A\cup B\in \mathcal{A}$.

ii.b.) It follows by induction that, for all $n\geqslant 2$, if $\{A_j\}_{1}^{n}\subset \varepsilon$, , then $\bigcup_{1}^{n}A_j \in \mathcal{A}$.

In fact, for $n=2$ the result is follows immediately from items i.) and ii.a.).

Assume $n>2$ and that we know by induction hypothesis that $\bigcup_{1}^{n-1}A_j \in \mathcal{A}$, then just apply item ii.a.) to $\bigcup_{1}^{n-1}A_j$ and $A_n$ and we get that $\bigcup_{1}^{n}A_j= \left(\bigcup_{1}^{n-1}A_j\right)\cup A_n\in \mathcal{A}$.

So $\mathcal{A}$ is the the collection of finite unions of members of $\varepsilon$.

Now let us prove that $\mathcal{A}$ is an algebra.

  1. $\emptyset \in \varepsilon \subseteq \mathcal{A}$. So $\emptyset \in \mathcal{A}$.

  2. Since $\mathcal{A}$ is the the collection of finite unions of members of $\varepsilon$, it is immediate that $\mathcal{A}$ is closed under finite unions. (finite union of finite unions are still just a finite union).

  3. If $A \in \mathcal{A}$, then we have $A=\bigcup_{k=1}^m D_k$ where $\{D_k\}_{1}^{n}$ is a family of disjoint sets in $\varepsilon$, then we have, for each $k\in \{1,\dots,n\}$,
    $$D_k^c= \bigcup_{j=1}^{n_k}B_{k,j}$$ where $B_{k,j}\in \varepsilon$. So we have $$A^c=\bigcap_{k=1}^m D_k^c=\bigcap_{k=1}^m\left( \bigcup_{j=1}^{n_k}B_{k,j} \right)= \bigcup\{B_{1,j_1}\cap \cdots \cap B_{m,j_m} : 1\leqslant j_k \leqslant n_k, 1 \leqslant k \leqslant m \}$$ So $A^c$ is a finite union of of members of $\varepsilon$. Thus $A^c \in \mathcal{A}$.

So we have proved that $\mathcal{A}$ is an algebra.

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  • $\begingroup$ Was there anything wrong in part (i) and (ii) of my proof? $\endgroup$ – Wolfy May 28 '16 at 18:19
  • $\begingroup$ Also, when showing $A\cup B\in\mathcal{A}$ once we show $A\setminus B\in \mathcal{A}$ then we can write $$A\cup B = (A\setminus B)\cup B = \cup_{1}^{n}\cup_{1}^{m}C_i\cap D_j \cup B = \cup_{1}^{n}\cup_{1}^{m}\emptyset \cup B = B$$ does this imply since $B\in \varepsilon$ we have $A\cup B\in\mathcal{A}$? $\endgroup$ – Wolfy May 28 '16 at 19:24
  • $\begingroup$ When proving $\mathcal{A}$ is an algebra. For number 2, isn't $\mathcal{A}$ a collection of finite disjoint unions? not just finite unions. $\endgroup$ – Wolfy May 30 '16 at 15:49
  • $\begingroup$ @Wolfi By definition $\mathcal{A}$ is the collection of finite disjoint unions of sets in $\varepsilon$. HOWEVER, before proving that $ \mathcal{A}$ is an algebra, we proved that: for any $\{A_j\}_{1}^{n}\subset \varepsilon$, ($\{A_j\}_{1}^{n}$ don't need to be disjoint), we have $\bigcup_{1}^{n}A_j \in \mathcal{A}$. It measn that we proved that $\mathcal{A}$ is the collection of finite unions of sets in $\varepsilon$. I have edited the answer to better highlight this point. $\endgroup$ – Ramiro May 30 '16 at 20:19
  • $\begingroup$ I see that now thanks, also I am working on exercise 16 here: math.stackexchange.com/questions/1806842/… $\endgroup$ – Wolfy May 31 '16 at 12:29
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The proposition can be reconstructioned to: Let $\mathcal{E}$ be a elementary family. Let $$ \mathcal{A}_1:=\bigcup_{n=1}^{\infty}\left\{\bigcup_{j=1}^n E_j\in 2^X; (E_1,\ldots,E_n)\in \mathcal{E}^n\right\}, $$ $$ \mathcal{A}_2:=\bigcup_{n=1}^{\infty}\left\{\bigcup_{j=1}^n E_j\in 2^X; (E_1,\ldots,E_n)\in \mathcal{E}^n,\ (i\ne j)\Rightarrow(E_i\cap E_j)=\emptyset\right\}. $$ Then $\mathcal{A}_1=\mathcal{A}_2=\mathcal{A}$, and $\mathcal{A}$ is an algebra.

(new answer)

a concise proof from Alexander Grigoryan's lecture notes (Theorem 2.1 https://www.math.uni-bielefeld.de/~grigor/pt2010.htm)

Denote $a(\mathcal{E})$ to be the minimal algebra that contain $\mathcal{E}$. Clearly, $\mathcal{E}\subset \mathcal{A}_2\subset \mathcal{A}_1 \subset a(\mathcal{E})$. So we only need to prove than $\mathcal{A}_2$ is a algebra, that is,

(i) $\emptyset,X\in \mathcal{A}_2$.

(ii) $A,B\in \mathcal{A}_2\Rightarrow A\cap B\in \mathcal{A}_2$.

(iii) $A\in \mathcal{A}_2\Rightarrow A^c\in \mathcal{A}_2$.

For (i), $\emptyset \in \mathcal{E}$ and $X=\emptyset^c \in \mathcal{A}_2$.

For (ii), $(\sqcup_{i=1}^m A_i)\cap (\sqcup_{j=1}^n B_j)=\sqcup_{1\le i\le m\atop 1\le j\le n}(A_i\cap B_j)\in\mathcal{A}_2$.

For (iii), $(\sqcup_{i=1}^m A_i)^c=\cap_{i=1}^m A_i^c$, but $A_i^c\in \mathcal{A}_2$, so by (ii), $\cap_{i=1}^m A_i^c\in \mathcal{A}_2$

(old answer)

sketch of proof:

It's obvious that $\mathcal{A}_2\subset \mathcal{A}_1$. So we have to prove that $\mathcal{A}_1\subset \mathcal{A}_2$. This follow by three step: (to prove (1) and (2), use $A\cup B=(A\backslash B)\sqcup B$ and the properties of elementary family.)

  1. $A,B\in \mathcal{E}\Rightarrow A\cup B\in \mathcal{A}_2$.
  2. $A\in \mathcal{A}_2,\ B\in \mathcal{E}\Rightarrow A\cup B\in \mathcal{A}_2$.
  3. By (1),(2) and induction, we get $n\in \mathbb{N};A_1,\ldots,A_n\in \mathcal{E}\Rightarrow \bigcup_{j=1}^n A_j\in \mathcal{A}_2$.

Finally, check that $\mathcal{A}$ is an algebra. It may need (one of) the general distributive law: $$ \bigcap_{i\in I}\left(\bigcup_{j\in J_i}A_j^i\right)=\bigcup_{(j_i)_{i\in I}\in \underset{i\in I}{\prod}J_i} \left(\bigcap_{i\in I} A_{j_i}^{i}\right).$$ When $I=\{1,\ldots,n\},n\in \mathbb{N}$ and $\underset{i\in I}{\forall}(|J_i|<\infty)$, the union and intersection on right hand side of above eqaution is finite. Indeed, the finite case can just be proved by induction: $$A\cap (\cup_{j=1}^n B_j)=\cup_{j=1}^n (A\cap B_j)\\ (\cup_{i=1}^m A_i)\cap (\cup_{j=1}^n B_j)=\cup_{j=1}^n (\cup_{i=1}^m (A_i\cap B_j))\\ (\cup_{i=1}^m A_i)\cap (\cup_{j=1}^n B_j)\cap (\cup_{k=1}^{\ell}C_k)=\cup_{k=1}^{\ell}(\cup_{j=1}^n (\cup_{i=1}^m (A_i\cap B_j))\cap C_k)...$$

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