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Let $X$ be a set and $E$ and $F$ are classes of subsets on X. then $E \cup F = \{ D \subseteq X : D \in E \text{ or } D \in F\}$.

  1. Show that by example $\sigma [E \cup F]$ is not necessarily equal to $\sigma[E] \cup \sigma[F]$
  2. Prove that $\sigma [E \cup F] = \sigma[\sigma[E] \cup \sigma [F]]$
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this is my home work and i could not find the solution can any one help me – math Oct 24 '12 at 15:12
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. – Did Oct 24 '12 at 15:18
@did: I have the impression OP posted a literal quote from his problem source. Of course, then, the statement should still be indicated as such properly by angle brackets. – Lord_Farin Oct 24 '12 at 15:20
can u help me i m running late for this problem – math Oct 24 '12 at 15:22
2  
madhav: Are you going to mention to your tutor the help you received here? – Did Oct 24 '12 at 15:50

1 Answer

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We know that

  • $A\subseteq B$ implies $\sigma(A)\subseteq\sigma(B)$
  • $\sigma(\sigma(A))=\sigma(A)$.

(These two properties are easy to prove - if you have not seen them in your lecture, you should try to find a proof.)

What do you get if you apply this to $E\cup F\subseteq \sigma(E)\cup\sigma(F)$? What do you get if you apply this to $\sigma(E)\cup\sigma(F) \subseteq \sigma(E\cup F)$? Can you show $\sigma(E)\cup\sigma(F) \subseteq \sigma(E\cup F)$ using these two properties?

For the first part (counterexample) try to think what happens if you divide $X$ into two disjoint parts and you choose sets in $E$ from one of them and sets in $F$ from the other one.

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@Lord_Farin Thanks for correcting the typo. When I saw your username, I wondered whether you're in some relation to Farin Urlaub. – Martin Sleziak Oct 24 '12 at 15:30
A fresh pair of eyes spots those more easily :). I'm sure you'll return the favour. – Lord_Farin Oct 24 '12 at 15:33
Nope, not in the least. It's an inheritance from more innocent years; I've simply stuck with it as my preferred internet pseudonym. – Lord_Farin Oct 24 '12 at 15:37

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