# Σ- algebra

1. Jan 13, 2012

### Fredrik

Staff Emeritus
1. The problem statement, all variables and given/known data

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.

2. Relevant 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 $\Sigma$.

3. 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: Jan 14, 2012
2. Jan 14, 2012

### Staff: Mentor

Re: σ-algebra

couldn't you start with the 1st item where you know ∅ and X are members of Ʃ

and then developer a element generator like: ∅, { ∅ }, { ∅, { ∅ } } .... ε X

I think its set-theoretic numbers -- see:

http://en.wikipedia.org/wiki/Set-theoretic_definition_of_natural_numbers

then do an induction proof where you demonstrate that for ∅ union X = ... is true

and given that prove the 2nd part of the induction.

and the others identities you prove as you said by deriving them from the first.

hope this helps...

I am not a number, I am not a mathematician, I am irish.

3. Jan 14, 2012

### Fredrik

Staff Emeritus
Re: σ-algebra

I don't see a connection between what you're saying and my problem, but thank you for the effort.

Last edited: Jan 14, 2012
4. Jan 14, 2012

### Staff: Mentor

Re: σ-algebra

Yeah I got lost in my thinking too but it seemed like a good idea at the time.

5. Jan 14, 2012

### MathematicalPhysicist

Re: σ-algebra

It's best not to rely on wiki.

Look on Rudin's adult textbook, your third condition is wrong, the E_n need not be mutually disjoint.

6. Jan 14, 2012

### Fredrik

Staff Emeritus
Re: σ-algebra

There are many equivalent definitions. If I drop the words "mutually disjoint", this doesn't change what sort of thing we end up calling a σ-algebra, because we can always rewrite a countable union as a countable disjoint union:
$$\bigcup_{k=1}^\infty E_k=\bigcup_{n=1}^\infty\bigg(E_n-\bigcup_{k=1}^{n-1}E_k\bigg)$$

Last edited: Jan 14, 2012
7. Jan 14, 2012

### micromass

Staff Emeritus
Re: σ-algebra

This is not a $\sigma$-algebra. This is a so-called $\lambda$-system.

We have the implication:

A $\lambda$-system is a $\sigma$-algebra iff it is closed under finite intersections.

But in general, a $\lambda$-system is not a $\sigma$-algebra.

Could you show me where exactly wikipedia uses your definition?

8. Jan 14, 2012

### MathematicalPhysicist

Re: σ-algebra

I think it does matter cause you don't know yet that $$E_n - \bigcup_{k=1}^{n-1} E_k \in \Sigma$$ you said so yourself that you haven't shown yet that a difference of sets is in a sigma algebra, right?

If you take Rudin's definition then it's evident that E-F is in sigma, cause E-F= (E^c U F)^c, and E^c is in sigma, so is E^c U F and thus also E-F by the above.

Unless I missed something, can you prove difference is in sigma with your definition?

9. Jan 14, 2012

### Dick

Re: σ-algebra

Bravo. Here's a λ-system reference from wikipedia that gets it right. http://en.wikipedia.org/wiki/Dynkin_system

10. Jan 14, 2012

### Fredrik

Staff Emeritus
Re: σ-algebra

Yes, I realized this, but at the time I thought it might be possible to prove the "closed under differences" property from my axioms.

That's a good point. I didn't realize this until after I looked inside Rudin and tried to make sure that his definition is equivalent to mine.

That's exactly what I was asking in #1. It looks like the answer is no.

It looks like my definition is broken, and can be fixed in at least two different ways:

Option 1: Replace axiom 2 with the "closed under differences" axiom. Then the stuff I did in #1 show that a σ-algebra is closed under unions and intersections (of two sets), and the trick I included in #6 shows that it's also closed under countable unions (Rudin's version of my axiom 3), and therefore countable intersections (by de Morgan's laws).

Option 2: Drop the words "mutually disjoint" from axiom 3. Now we can prove that a σ-algebra is closed under differences, and then the stuff I did in #1 takes care of everything else.

I can't, because it doesn't. I thought it did, because I thought it doesn't matter if the words "mutually disjoint" are included in axiom 3 or not. I see now that this is only true if I replace my axiom 2 with the stronger "closed under differences" axiom.

In other words, the following would be a good way to state the axioms:
(1) $\emptyset,X\in\Sigma$
(2) $E-F\in\Sigma$, for all $E,F\in\Sigma$.
(3) $\bigcup_{k=1}^\infty E_k\in\Sigma$, for all sequences $\langle E_k\rangle_{k=1}^\infty$ in $\Sigma$.​
Here we can weaken either (2) or (3) by replacing it with the corresponding axiom from my post #1, but we can't do it to both! That would break the definition. We can of course also weaken axiom 1 by dropping either ∅ or X, but that's less interesting.

Thank you both.

Last edited: Jan 14, 2012