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The collection $\{A, A^c, \emptyset, \Omega\}$ is, I think, the smallest field generated by A. Is this also denoted $\sigma$(A) or is that notation reserved for $\sigma$-fields?

Asaf Karagila
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TonyK
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    The collection ${A, A^c, \emptyset, \Omega}$ is a $\sigma$-field. – Kurt G. Apr 12 '22 at 18:39
  • @KurtG. Thanks. I thought it had to be closed under countably infinite unions to be a $\sigma$-field. I guess you could union $A$ with itself infinitely many times ... is that how we get there?? Is it it also a field? – TonyK Apr 12 '22 at 21:36
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    Any finite set that is closed under unions is closed under countable unions, indeed under any sized unions. – Asaf Karagila Apr 12 '22 at 23:25
  • @AsafKaragila. Thank you also. What does countably infinite unions mean for a finite set such as the one above? Can I sa that if $\cup_{i=1}^\infinty A_i$, etc., are in the set then the set is closed under countably infinite unions? – TonyK Apr 13 '22 at 00:01
  • What is an infinite sequence of elements from ${0,1}$? – Asaf Karagila Apr 13 '22 at 00:01
  • @AsafKaragila. Do you mean ${0,1}^\infty$. If yes then I am not getting the point. – TonyK Apr 13 '22 at 00:09
  • You are asking me about a countable union of elements of a finite set. I'm asking you what would an infinite sequence be, if all the elements appearing in it come from a finite set, in this specific example, ${0,1}$. My point is that you're making the mistake of treating unions as somehow "exceptional" and differently than you'd treat other operations. You shouldn't. – Asaf Karagila Apr 13 '22 at 00:11
  • @AsafKaragila. Okay, then what distinguishes a field from a $\sigma$-field if I can monkey around with countably infinite unions of elements of a finite collection,( per my previous comment)? – TonyK Apr 13 '22 at 00:19
  • The fact that it is closed under countable unions and intersections, of course. It's the same thing that differentiates "finite" and "infinite" or "metric space" and "complete metric space", etc. Some definitions in mathematics are just vacuously true in some situations. It's a good idea to get used to that fact. – Asaf Karagila Apr 13 '22 at 00:24
  • I can live with that. – TonyK Apr 13 '22 at 00:27

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To your question: What you describe would be denoted $\sigma(\{A\})$, the smallest $\sigma$-field containing $\{A\}$ as a subset, not $\sigma(A)$.

Jonathan Schilhan
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  • Thanks for posting but now I'm a confused again -- a permanent state with me so I'm getting used to it. In the setting above A is a subset (not an elemental) in a collection. Why should the sigma field it generates be labelled differently? – TonyK Apr 22 '22 at 16:27
  • I probably shouldn't but I will and I'll also apologize in advance for being a nuisance: "What you describe would be denoted σ({A}), the smallest σ-field containing {A} as a subset, not σ(A)" ... the smallest $\sigma$-field containing A as a ....? – TonyK Apr 29 '22 at 03:06