Logan Wyatt

Answered

2022-06-30

Let $X=Y$ be uncountable and define $\mathcal{A},\mathcal{B}$ to be the countable-cocountable $\sigma $-algebras on $X,Y$ respectively. Let $C=\sigma (A\times B)$. Prove that for each subset $E\in C$, there exist $A,B$ in $\mathcal{A},\mathcal{B}$ countable such that either $E\subseteq (A\times Y)\cup (X\times B)$ or ${E}^{c}\subseteq (A\times Y)\cup (X\times B)$.

My idea is to define $D$ as the collection of all such subsets of $C$, prove that $D$ contains ${A}_{1}\times B1$ for each ${A}_{1}\in \mathcal{A},{B}_{1}\in \mathcal{B}$ and then show that $D$ is a $\sigma $-algebra.

I am able to prove that $D$ is a $\sigma $ algebra, and I can show that in the case ${A}_{1},{B}_{1}$ are either both countable or cocountable, ${A}_{1}\times {B}_{1}$ is in $D$. How do I do it for the case that exactly one of them is cocountable?

Something that may be useful is

$({A}_{1}\times {B}_{1}{)}^{c}=({A}_{1}^{c}\times Y)\cup (X\times {B}_{1}^{c})\cup ({A}_{1}^{c}\times {B}_{1}^{c})$

My idea is to define $D$ as the collection of all such subsets of $C$, prove that $D$ contains ${A}_{1}\times B1$ for each ${A}_{1}\in \mathcal{A},{B}_{1}\in \mathcal{B}$ and then show that $D$ is a $\sigma $-algebra.

I am able to prove that $D$ is a $\sigma $ algebra, and I can show that in the case ${A}_{1},{B}_{1}$ are either both countable or cocountable, ${A}_{1}\times {B}_{1}$ is in $D$. How do I do it for the case that exactly one of them is cocountable?

Something that may be useful is

$({A}_{1}\times {B}_{1}{)}^{c}=({A}_{1}^{c}\times Y)\cup (X\times {B}_{1}^{c})\cup ({A}_{1}^{c}\times {B}_{1}^{c})$

Answer & Explanation

Bruno Dixon

Expert

2022-07-01Added 14 answers

Suppose ${A}_{1}$ is countable and ${B}_{1}$ is cocountable. Then it follows that

${A}_{1}\times {B}_{1}\subseteq {A}_{1}\times Y\subseteq ({A}_{1}\times Y)\cup (X\times B)$

where $B\in \mathcal{B}$ is an arbitrary countable set (e.g., $B=\mathrm{\varnothing}$). This shows that ${A}_{1}\times {B}_{1}\in D$.

${A}_{1}\times {B}_{1}\subseteq {A}_{1}\times Y\subseteq ({A}_{1}\times Y)\cup (X\times B)$

where $B\in \mathcal{B}$ is an arbitrary countable set (e.g., $B=\mathrm{\varnothing}$). This shows that ${A}_{1}\times {B}_{1}\in D$.

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