Hi,
I have a lot of problems with proving that M is algebra.
$\displaystyle M=\{ F \cap G: G,F \subset R $ and F is open set and G is closed set$\displaystyle \} $
Any clues?
Eh...I assume with R you mean the set of real numbers.
You need to check the following:
1.$\displaystyle \emptyset\in M ?$
2. $\displaystyle X\in M \Rightarrow X^c\in M$?
3. $\displaystyle X_1,\cdots ,X_n\in M\Rightarrow \bigcup_{k=1}^n X_n\in M?$
These are all pretty straightforward to check:
So, I don't really see your problem. Any specific point you're having trouble with?
So lets start with 2nd.
Lets take $\displaystyle X \in M$ so we present X as for example $\displaystyle X=F_1 \cap G_1$ and to prove this point we need to observe that X' we can present likewise, I mean for example $\displaystyle X'=F_2 \cap G_2$
Therefore i tried use De Morgan's laws but I get $\displaystyle X'=F_1' \cup G_1'$ and I do not need the union but intersection and do not have an idea what is next
Frankly, I wonder if some bit of information is missing from the statement.
I can show that $\displaystyle M$ contains all open sets and all closed sets in $\displaystyle \mathbb{R} $.
It is also clear that $\displaystyle M$ is closed with respect to finite intersection.
BUT, I can not show that $\displaystyle M$ is closed with respect to either finite union nor complementation. Given either of those two, it is easy to show that $\displaystyle M$ 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.
I must admit, this is harder then I initially expected. for complementation I was thinking the following:
$\displaystyle (F\cap G)^c = F^c\cup G^c$, which is the union of an open and a closed set. So if we show $\displaystyle F\cup G \in M$ for any F open, G closed we're done.
To do this I was trying to show the existence of an open set $\displaystyle G^* \supset G$ such that
$\displaystyle (\overline{F}\cup G)\cap (F\cup G^*)= F\cup G$
That is finding a $\displaystyle G^*$ such that $\displaystyle (\overline{F}\setminus F)\cap (G^*\setminus G)\subset F\cap G$
This seems intuitively clear to me, but I can't prove it...really
Besides, I was thinking about proving 3th;
$\displaystyle A, B \in M => A \cup B \in M$, $\displaystyle A= F_1 \cap G_1, B=F_2 \cap G_2$ and so we need to prove that:
$\displaystyle (F_1 \cap G_1) \cup (F_2 \cap G_2) \in M$ I use some algebra rule and I get:$\displaystyle (F_1 \cup F_2) \cap (G_1 \cup G_2) \cap (F_1 \cup G_2) \cap (F_2 \cup G_1)$ 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.
Suppose that $\displaystyle M$ is a $\displaystyle \sigma$-algebra. Note that since $\displaystyle \{q\}=\{q\}\cap\mathbb{R}$ that $\displaystyle \{q\}\in m$ for all $\displaystyle q\in\mathbb{Q}$ and since $\displaystyle M$ is, by assumption, closed under countable unions we see that $\displaystyle \displaystyle \bigcup_{q\in\mathbb{Q}}\{q\}=\mathbb{Q}\in M$. But, this implies that $\displaystyle M=F\cap G$ where $\displaystyle G$ is closed and $\displaystyle F$ open. Note though that $\displaystyle G$ containing $\displaystyle \mathbb{Q}$ is dense, and since it's closed we see that $\displaystyle \mathbb{R}=\overline{\mathbb{Q}}\subseteq\overline {G}=G$ so that $\displaystyle G=\mathbb{R}$. Thus, $\displaystyle \mathbb{Q}=G\cap F=\mathbb{R}\cap F=F$. But, this implies that $\displaystyle F$ is open which is clearly impossible.
Hee Drex! I like the way you work from Contraposition. You meant $\displaystyle M=\mathbb{Q}$ probably here.
However I don't think it's a good counterexample. You're contradiction is something like:
From assumption follows: $\displaystyle \mathbb{Q} $ is in $\displaystyle M $. Then $\displaystyle \mathbb{Q} = F\cap G $. Your contradiction is that $\displaystyle G= \mathbb{R}$
(which is indeed closed) so that $\displaystyle F = \mathbb{Q}$, which then must be open. However, $\displaystyle \mathbb{R}$ is also open...so you can take $\displaystyle F= \mathbb{R}$ and $\displaystyle G= \mathbb{Q}$. I don't see any contradiction here.
I think you missed the important part of my proof.
If we assume that $\displaystyle M$ is a $\displaystyle \sigma$-algebra then since $\displaystyle \{q\}\in M$ for all $\displaystyle q\in\mathbb{Q}$ then $\displaystyle \displaystyle \bigcup_{q\in \mathbb{Q}}\{q\}=\mathbb{Q}\in M$. But, this implies that $\displaystyle \mathbb{Q}=F\cap G$ where $\displaystyle F$ is open and $\displaystyle G$ is closed, right? That is by definition what it means for $\displaystyle \mathbb{Q}$ to be in $\displaystyle M$. But I claim, in general, that $\displaystyle \mathbb{Q}$ cannot be written as the intersection of an open and a closed set. To see this suppose that $\displaystyle C,O$ are closed and open respectively and $\displaystyle C\cap O=\mathbb{Q}\quad (1)$. Note first then that $\displaystyle \mathbb{Q}\subseteq C$. But, this means that $\displaystyle \mathbb{R}=\overline{\mathbb{Q}}\supseteq \overline{C}=C$ so that if $\displaystyle (1)$ were true we'd have to have that $\displaystyle C=\mathbb{R}$ (in other words, any closed dense set is the full space). But, this would imply that $\displaystyle \mathbb{Q}=C\cap O=\mathbb{R}\cap O=O$. But, this says that $\displaystyle \mathbb{Q}$ is open, which is not true.
In your example you can't take $\displaystyle G=\mathbb{Q}$ since $\displaystyle \mathbb{Q}$ 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
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!