- #1
- 10,877
- 422
Homework Statement
I want to know if the definition of σ-algebra stated below implies that every σ-algebra is closed under unions, intersections and differences (of only two members). If I assume that one of those three statements is true, I can prove the others, but I don't see how to prove any of them directly from the axioms.
Homework Equations
There are many equivalent ways to state the definition. This is the one I'd like to use:
A set ##\Sigma\subset\mathcal P(X)## is said to be a σ-algebra of subsets of X if
(1) ##\emptyset,X\in\Sigma##
(2) ##E^c\in\Sigma##, for all ##E\in\Sigma##.
(3) ##\bigcup_{k=1}^\infty E_k\in\Sigma##, for all mutually disjoint sequences ##\langle E_k\rangle_{k=1}^\infty## in ##\Sigma##.
The Attempt at a Solution
Suppose that ##E,F\in\Sigma##. We have
\begin{align}
& E\cup F=E\cup(F-E)\\
& E\cap F=(E^c\cup F^c)^c\\
& E-F=E\cap F^c\\
& E\cap F=E-(F-E)\\
& E\cup F=(E^c\cap F^c)^c\\
& E-F=E\cap F^c=(E^c\cup F)^c\\
\end{align}
These equalities tell us respectively that:
If Ʃ is closed under differences, it's also closed under unions.
If Ʃ is closed under unions, it's also closed under intersections.
If Ʃ is closed under intersections, it's also closed under differences.
If Ʃ is closed under differences, it's also closed under intersections.
If Ʃ is closed under intersections, it's also closed under unions.
If Ʃ is closed under unions, it's also closed under differences.
I'm obviously missing something simple, but what?
Edit: I think that my definition of σ-algebra was just wrong, and that I need to replace the "closed under complements" axiom with the stronger "closed under differences".
Edit 2: Hm, Wikipedia defines the term essentially the same way I did (with complements, not differences), but instead of my axiom 1 they require that Ʃ is non-empty. This implies my axiom 1, because if E is in Ʃ, then ##E\cup E^c## and ##(E\cup E^c)^c## are in Ʃ.
Last edited: