Let S={{a}, {b}, {c}} contain three elements. Find out the sigma-field generated by {a,b}.

My question concerns the set notation. {a}, {b}, and {c} are the elements of S, but they are also sets, correct? (indicated by the brackets)

What is the difference between the given S and the following S'={a,b,c }?

In finding the sigma-algebra, would {{a},{b}} be equivalent to {a,b}?


1 Answer 1

  1. Yes, but notice that $S$ is a set of sets, i.e. every element of the set $S$ is a set. For example, $\{a\}$ is a set containing one element $a$, but the set itself is an element of the set $S$.
  2. S' is a set that contains three elements $a,b,c$; while $S$ is a set that contains three elements, and each element is a set of one element, namely $\{a\},\{b\},\{c\}$.
  3. No, they are not equivalent. Sigma-algebra is always a set of sets.
  • $\begingroup$ I understand your explanation for 1 and 2. But am still confused for #3. I am trying to find the sigma-algebra of {a,b}, which by your explanation would be a set that contains two elements a,b; but they aren't sets. So in this case, it appears to me that I am trying to find the sigma-algebra of a set of two elements that aren't sets, which by definition, is not a sigma algebra (always a set of sets). $\endgroup$
    – Amaziah
    Aug 14, 2017 at 22:52
  • $\begingroup$ When we talk about sigma algebra generated by family of sets, we are referring to the smallest sigma algebra which contains all the sets. And if we just say generated by one subset, it should mean that implicitly we have a family of sets that contains this one element/set. In this case, we specify the family of sets to be of one set $\{a,b\}$, and thus the sigma algebra should be $\{\emptyset, X, \{a,b\} \{a,b\}^c\}$ - but wired, since $\{a,b\} \notin S$ $\endgroup$
    – Jay Zha
    Aug 14, 2017 at 23:02
  • $\begingroup$ But I do feel your question setting is a bit wired, as $\{a,b\} \notin S$, but I assume $S$ should be the space that want to have the sigmal-algebra over, right? - a bit wired.. If $S=\{a,b,c\}$, it will be more natural. $\endgroup$
    – Jay Zha
    Aug 14, 2017 at 23:05
  • $\begingroup$ Yes, over S. That leads to my confusion on the complement of {a,b}. I believe the complement of {a,b} is the element c. But then wouldn't the union of {a,b} with S be necessary? I know that a union of a set and its subset would just be the set itself. But the fact that {a,b} is not specifically contained in S is giving me problems. $\endgroup$
    – Amaziah
    Aug 14, 2017 at 23:10
  • $\begingroup$ @Amaziah Right, I think it could be a typo for $S$, and it makes most of the sense for $S=\{a,b,c\}$. But it is certain that $\{a,b\}\ne \{\{a\},\{b\}\}$. $\endgroup$
    – Jay Zha
    Aug 14, 2017 at 23:18

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