# Countability of arbitrary classes $\mathcal{A}$ of subsets of $\Omega$ and the partitions of $\Omega$ induced by $\mathcal{A}$

Partitions of $$\Omega$$

I denote a partiton of $$\Omega$$ to be $$\Omega_{\tau}$$ which satisfies $$\Omega=\cup_{\beta} B_{\beta}$$ for $$B_{\beta}\in\Omega_{\tau}$$ and $$B_{\beta}\cap B_{\beta'}=\emptyset$$ for all $$\beta\not=\beta'$$ where I use the $$\beta$$ notation to denote the index set $$\beta\in I_{\beta}$$ is possible uncountable. I also assume $$\Omega$$ could be infinite in that it contains uncountably many elements $$\omega\in\Omega$$

Edit: To give an example, for $$\Omega=[0,1]$$, $$\Omega_{\tau}=\{[0,1/2),[1/2,1]\}$$ and $$\Omega_{\tau}=\{[0,1/10),[1/10,2/10),[2/10,1]\}$$ would be finite partitions, $$\Omega_{\tau}=\{\{x\}:x\in[0,1]\}$$ would be an uncountable partition (the finest possible), and for $$\Omega=[1,\infty)$$, $$\Omega_{\tau}=\{[a_{i},a_{i}+1):a_{i}\in\mathbb{N}\}$$ would be a countable partition

An arbitrary class $$\mathcal{A}$$ of subsets of $$\Omega$$

I denote a collection of subsets of $$\Omega$$ as $$\mathcal{A}$$ so that $$A\subseteq\Omega$$ for all $$A\in\mathcal{A}$$. The class $$\mathcal{A}$$ could contain uncountably many sets, each set of which could itself be possibly infinite in size. This class is not assumed to have any structure - for example it is not assumed to be a ring, semi-ring, field or $$\sigma$$-field

Edit: A $$\sigma$$-field $$\mathscr{F}$$ on $$\Omega$$ satisfies i) $$\Omega\in\mathscr{F}$$, ii) $$F\in\mathscr{F}$$ implies $$F^{c}\in\mathscr{F}$$, iii) $$F_{1},F_{2},...\in\mathscr{F}$$ implies $$\cup_{i}F_{i}\in\mathscr{F}$$, and these imply (iv) $$F_{1},F_{2},...\in\mathscr{F}$$ implies $$\cap_{i}F_{i}\in\mathscr{F}$$ and v) $$\emptyset\in\mathscr{F}$$. The notation $$\mathscr{F}=\sigma(\mathcal{A})$$ means $$\mathscr{F}$$ is formed from the intersection of all $$\sigma$$-fields containing $$\mathcal{A}$$ and so $$\mathscr{F}$$ is the smallest $$\sigma$$-field on $$\Omega$$ containing $$\mathcal{A}$$

Partitions of $$\Omega$$ induced by $$\mathcal{A}$$

I denote $$\mathcal{A}_{\tau}$$ to be the partition of $$\Omega$$ induced by the following equivalance relation where $$1_{A}(\omega)=1$$ if $$\omega\in A$$ and $$0$$ if not

$$\forall \omega,\omega'\in\Omega:\hspace{10pt}\omega\sim_{\mathcal{A}}\omega'\Longleftrightarrow 1_{A_{\alpha}}(\omega)=1_{A_{\alpha}}(\omega')\hspace{10pt}\forall A_{\alpha}\in\mathcal{A}$$

Edit (correction of definition): The equivalance class of any $$\omega\in\Omega$$ will be denoted $$[\omega]_{\mathcal{A}}$$ and is explicitly given by

$$[\omega]_{\mathcal{A}}:=\bigcap\{A:\omega\in A\in\mathcal{A}\}\cap\bigcap\{\Omega\backslash A:\omega\not\in A\in\mathcal{A}\}$$

Question regarding the relationship between $$\Omega_{\tau}$$, $$\mathcal{A}$$ and $$\mathcal{A_{\tau}}$$

I found myself naturally gravitating towards assuming $$\Omega_{\tau}$$ and $$\mathcal{A}$$ both contained countably many sets (implying $$\mathcal{A_{\tau}}$$ contains countably many sets) and trying to prove the following, which seemed intuitively obvious;

$$\mathcal{A}_{\tau}=\Omega_{\tau}\Longleftrightarrow \mathcal{A}=\Omega_{\tau}$$

Edit: $$\mathcal{A}_{\tau}=\Omega_{\tau}$$ means equality of sets: for every $$A_{\tau}\in\mathcal{A}_{\tau}$$ we have $$A_{\tau}\in \Omega_{\tau}$$ and for every $$B\in \Omega_{\tau}$$ we have $$B\in A_{\tau}$$. As an example if $$\mathcal{A}=\{A_{1},A_{2}\}$$ for $$A_{1}\subset\Omega$$, $$A_{2}\subset\Omega$$, $$A_{1}\cap A_{2}\not=\emptyset$$ then $$\mathcal{A}_{\tau}=\{A_{1}\backslash A_{2},A_{2}\backslash A_{1},A_{1}\cap A_{2},\Omega\backslash\{A_{1}\cup A_{2}\}\}$$ so that $$\mathcal{A}\not\subset\mathcal{A}_{\tau}$$. However if $$A_{1}\cap A_{2}=\emptyset$$ then $$\mathcal{A}_{\tau}=\{A_{1},A_{2},\Omega\backslash\{A_{1}\cup A_{2}\}\}$$ so that $$\mathcal{A}\subset\mathcal{A}_{\tau}$$

I am almost convinced of this and I spare readers the proof at the moment since I am still not convinced I have the right approach. Indeed this is my first question - does the above claim appear correct? In particular is countablilty of both $$\Omega_{\tau}$$ and $$\mathcal{A}$$ required for this result or does one or the other suffice? I have yet to try and construct counter-examples that might arise due to uncountability in either $$\Omega_{\tau}$$ or $$\mathcal{A}$$ since my understanding of the uncountable infinite is very poor. So any comments and thoughts would be appreciated.

Finally my interest is in taking this result further into some measure theory for probability. In this respect if $$\mathscr{F}$$ is a $$\sigma$$-field generated by $$\mathcal{A}$$, that is $$\mathscr{F}=\sigma(\mathcal{A})$$, then I am more sure uncountability is a real issue here. This is because of the structure of $$[\omega]_{\mathcal{A}}=\bigcap\{A:\omega\in A\in\mathcal{A}\}\cap\bigcap\{\Omega\backslash A:\omega\not\in A\in\mathcal{A}\}$$, in being a possibly uncountable intersection of $$\mathcal{A}$$-sets, is not guaranteed to be contained within $$\mathscr{F}$$? This means if I wish to work with both $$\mathscr{F}_{\tau}$$ and $$\mathcal{A}_{\tau}$$ and make statments such as $$[\omega]_{\mathcal{A}}\in\mathscr{F}$$ I run into trouble?

Again any comments and thoughts are most welcome.

• Undefined term in the partitions induced by A. – William Elliot Feb 9 at 23:36
• very helpful thank you - I am much clearer now – dandar Feb 10 at 9:48
• What is $1_A$?? – William Elliot Feb 10 at 13:41
• Thank you - I have amended – dandar Feb 10 at 16:14
• That intuitive obvious is false because it equates a partition of the set to a subset of the set. An approach of notational overload is not good. – William Elliot Feb 10 at 23:35

It seems that sometimes your notation is overcomplicated. So I tried to simplify it in my answer to more standard.

A partition of a set $$\Omega$$ is its cover by its mutually disjoint non-empty subsets. A family $$\mathcal A$$ of subsets of $$\Omega$$ generates the partition $$\mathcal A_\tau$$ as you described.

$$\mathcal{A}_{\tau}=\Omega_{\tau}\Longleftrightarrow \mathcal{A}=\Omega_{\tau}$$

$$\mathcal A$$ is not always a partition of $$\Omega$$, but $$\mathcal A_\tau$$ always is, so it can be denoted as $$\Omega_{\tau}$$. Thus the implication $$\Rightarrow$$ should not hold in general. On the other hand, it is easy to see that $$\mathcal{A}_\tau=\mathcal{A}$$ if and only if $$\mathcal{A}$$ is a partition of $$\Omega$$ and probably you meant this fact.

I guess that your definition of a $$\sigma$$-field means that it is closed with respect to countable unions and intersections. Thus if $$\mathcal A$$ is countable then $$\mathcal{A}_\tau \subset\sigma(\mathcal{A})$$. But, in general, as you guessed, this inclusion does not hold. For instance, let $$\Omega$$ be any uncountable set. Fix any element $$\omega\in \Omega$$ and let $$\mathcal A$$ consists of all subsets $$A\ni\omega$$ of $$\Omega$$ such that $$\Omega\setminus A$$ is countable. It is easy to check that $$\sigma\mathcal(A)$$ consists of all subsets $$A$$ of $$\Omega$$ such that either $$A\not\ni\omega$$ and $$A$$ is countable or $$A\ni\omega$$ and $$\Omega\setminus A$$ is countable. Then

$$[\omega]_{\mathcal{A}}:=\bigcap\{A:\omega\in A\in\mathcal{A}\}\cap\bigcap\{\Omega\backslash A:\omega\not\in A\in\mathcal{A}\}= \{\omega\}\in \mathcal A_\tau\setminus\sigma(\mathcal A).$$

• Many thanks Alex for your helpful response - I have I think amended my notation sufficiently to answer your questions. – dandar Feb 12 at 16:53
• Thank you very much Alex for your answer here! – dandar Feb 12 at 23:30