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Let $S$ be a set and let $F=\{\{s\}:s\in S\}$. Show that $\sigma(F)=\cap\{A:A \text{ is a sigma-algebra on } S \text{ and }F\subseteq A\}$ is a subset of $\lambda(F)=\{A\subseteq S: A \text{ is countable or }A^c \text{ is countable}\}$.

I'm completely lost in this exercise. Do I need to show that every set in the sigma-algebra generated by $F$ is countable or cocountable? It seems reasonable to assume $S$ is infinite because if it were finite then $\lambda(F)$ would just be the power set of $F$.

If I were to take the direct approach I would start with 'let there be a $A\in\sigma(F)$.' It doesn't seem at all obvious what to do after. So instead maybe contraposition is a better approach? We say 'let $A\not\in\lambda(F)$' s.t. neither $A$ or $A^c$ is countable. Now we could maybe construct a sigma algebra of $F$ that does not contain $A$, which would mean $A\not\in\sigma(F)$.

It seems logical that a sigma-algebra on $F$ can only have countably many elements. Anyhow I can't figure this one out completely. The hints in my book say 'Use the fact that the subset of countable sets are countable and that countable union of countable sets is again countable.'

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  • $\begingroup$ In the definition of $\sigma(F)$, do you mean $A$ is a $\sigma$-algebra on $S$? And in the definition of $\lambda(F)$, do you mean $A\subseteq S$? $\endgroup$ – Eric Wofsey Jan 18 '18 at 21:33
  • $\begingroup$ @EricWofsey I mean $A$ is a sigma-algebra on $F$ and in the other definition I mean $A\subseteq F$. So just like it's there essentialy. $\endgroup$ – J Dijkstra Jan 18 '18 at 21:39
  • $\begingroup$ As written, your definitions don't make any sense...how can $F\subseteq A$ if $A$ is a $\sigma$-algebra on $F$? $\endgroup$ – Eric Wofsey Jan 18 '18 at 21:44
  • $\begingroup$ @EricWofsey I see, you're right. For some reason I messed that up. $\endgroup$ – J Dijkstra Jan 18 '18 at 21:55
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Since $\sigma(F)$ is the minimal sigma-algebra containing $F$, it suffices to show that $\lambda(F)$ is a sigma algebra and that $F\subseteq \lambda(F)$. It is relatively easy to show that $\lambda(F)$ is a sigma-algebra and it is clear that $F\subseteq \lambda(F)$ since $\{s\}\in\lambda(F)$ for all $s\in S$.

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  • $\begingroup$ I already showed $\lambda(F)$ was a sigma-algebra in an earlier exercise. So what did I get the hints for? Your reasoning seems to work. $\endgroup$ – J Dijkstra Jan 18 '18 at 22:19

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