# Thread: Sigma algebra, open and closed set.

1. ## Sigma algebra, open and closed set.

Hi,
I have a lot of problems with proving that M is algebra.
$M=\{ F \cap G: G,F \subset R$ and F is open set and G is closed set $\}$

Any clues?

2. Eh...I assume with R you mean the set of real numbers.

You need to check the following:

1. $\emptyset\in M ?$
2. $X\in M \Rightarrow X^c\in M$?
3. $X_1,\cdots ,X_n\in M\Rightarrow \bigcup_{k=1}^n X_n\in M?$

These are all pretty straightforward to check:

So, I don't really see your problem. Any specific point you're having trouble with?

3. Could you explain for example 2nd point? Because i cannot see it ;/

4. Originally Posted by lohengrin
Could you explain for example 2nd point? Because i cannot see it ;/
Do you know the requirements for a collection of subsets of a set to be a sigma algebra?
What are they?

Lets take $X \in M$ so we present X as for example $X=F_1 \cap G_1$ and to prove this point we need to observe that X' we can present likewise, I mean for example $X'=F_2 \cap G_2$
Therefore i tried use De Morgan's laws but I get $X'=F_1' \cup G_1'$ and I do not need the union but intersection and do not have an idea what is next

6. Frankly, I wonder if some bit of information is missing from the statement.
I can show that $M$ contains all open sets and all closed sets in $\mathbb{R}$.
It is also clear that $M$ is closed with respect to finite intersection.

BUT, I can not show that $M$ is closed with respect to either finite union nor complementation. Given either of those two, it is easy to show that $M$ is an algebra of sets. Is either the missing piece?

One the other hand, I know that all of this type of problem involve very clever tricks. I may not have found the trick for this one. BUT nor have I found a counter-example.

7. I must admit, this is harder then I initially expected. for complementation I was thinking the following:

$(F\cap G)^c = F^c\cup G^c$, which is the union of an open and a closed set. So if we show $F\cup G \in M$ for any F open, G closed we're done.

To do this I was trying to show the existence of an open set $G^* \supset G$ such that

$(\overline{F}\cup G)\cap (F\cup G^*)= F\cup G$

That is finding a $G^*$ such that $(\overline{F}\setminus F)\cap (G^*\setminus G)\subset F\cap G$

This seems intuitively clear to me, but I can't prove it...really

8. Besides, I was thinking about proving 3th;
$A, B \in M => A \cup B \in M$, $A= F_1 \cap G_1, B=F_2 \cap G_2$ and so we need to prove that:
$(F_1 \cap G_1) \cup (F_2 \cap G_2) \in M$ I use some algebra rule and I get: $(F_1 \cup F_2) \cap (G_1 \cup G_2) \cap (F_1 \cup G_2) \cap (F_2 \cup G_1)$ and again it is easy to prove that 1st factor is open and 2nd is closed but I don know what is next...

If we won't find solution I will ask my teacher next week, mayby there is mistake as Plato said.

9. I've read recently that you can prove that M is not sigma algebra and I 've found hint: it is essential that there are real numbers!
Unfortunately it doesn't help me much ;/

10. Originally Posted by lohengrin
Hi,
I have a lot of problems with proving that M is algebra.
$M=\{ F \cap G: G,F \subset R$ and F is open set and G is closed set $\}$

Any clues?
Suppose that $M$ is a $\sigma$-algebra. Note that since $\{q\}=\{q\}\cap\mathbb{R}$ that $\{q\}\in m$ for all $q\in\mathbb{Q}$ and since $M$ is, by assumption, closed under countable unions we see that $\displaystyle \bigcup_{q\in\mathbb{Q}}\{q\}=\mathbb{Q}\in M$. But, this implies that $M=F\cap G$ where $G$ is closed and $F$ open. Note though that $G$ containing $\mathbb{Q}$ is dense, and since it's closed we see that $\mathbb{R}=\overline{\mathbb{Q}}\subseteq\overline {G}=G$ so that $G=\mathbb{R}$. Thus, $\mathbb{Q}=G\cap F=\mathbb{R}\cap F=F$. But, this implies that $F$ is open which is clearly impossible.

11. Originally Posted by Drexel28
But, this implies that $M=F\cap G$ where $G$ is closed and $F$ open.
Hee Drex! I like the way you work from Contraposition. You meant $M=\mathbb{Q}$ probably here.

However I don't think it's a good counterexample. You're contradiction is something like:

From assumption follows: $\mathbb{Q}$ is in $M$. Then $\mathbb{Q} = F\cap G$. Your contradiction is that $G= \mathbb{R}$

(which is indeed closed) so that $F = \mathbb{Q}$, which then must be open. However, $\mathbb{R}$ is also open...so you can take $F= \mathbb{R}$ and $G= \mathbb{Q}$. I don't see any contradiction here.

12. Originally Posted by Dinkydoe
Hee Drex! I like the way you work from Contraposition. You meant $M=\mathbb{Q}$ probably here.

However I don't think it's a good counterexample. You're contradiction is something like:

From assumption follows: $\mathbb{Q}$ is in $M$. Then $\mathbb{Q} = F\cap G$. Your contradiction is that $G= \mathbb{R}$

(which is indeed closed) so that $F = \mathbb{Q}$, which then must be open. However, $\mathbb{R}$ is also open...so you can take $F= \mathbb{R}$ and $G= \mathbb{Q}$. I don't see any contradiction here.
I think you missed the important part of my proof.

If we assume that $M$ is a $\sigma$-algebra then since $\{q\}\in M$ for all $q\in\mathbb{Q}$ then $\displaystyle \bigcup_{q\in \mathbb{Q}}\{q\}=\mathbb{Q}\in M$. But, this implies that $\mathbb{Q}=F\cap G$ where $F$ is open and $G$ is closed, right? That is by definition what it means for $\mathbb{Q}$ to be in $M$. But I claim, in general, that $\mathbb{Q}$ cannot be written as the intersection of an open and a closed set. To see this suppose that $C,O$ are closed and open respectively and $C\cap O=\mathbb{Q}\quad (1)$. Note first then that $\mathbb{Q}\subseteq C$. But, this means that $\mathbb{R}=\overline{\mathbb{Q}}\supseteq \overline{C}=C$ so that if $(1)$ were true we'd have to have that $C=\mathbb{R}$ (in other words, any closed dense set is the full space). But, this would imply that $\mathbb{Q}=C\cap O=\mathbb{R}\cap O=O$. But, this says that $\mathbb{Q}$ is open, which is not true.

In your example you can't take $G=\mathbb{Q}$ since $\mathbb{Q}$ is not closed. I hope I didn't sound mean in the above, I was just trying to be explicit.

P.S. Hi! I haven't talked to you in a while

13. Right. I guess I missed the important part here. I was under the assumption that Q is closed.

However, reconsidering the definition of closed, it means that R-Q is open which is not true, since any open ball B contains rational numbers. Q seemed intuitively closed to me.

Nice counterexample btw.

P.S.: Yeah! Where have you been!