Eh...I assume with R you mean the set of real numbers.
You need to check the following:
1.
2. ?
3.
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 so we present X as for example and to prove this point we need to observe that X' we can present likewise, I mean for example
Therefore i tried use De Morgan's laws but I get 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 contains all open sets and all closed sets in .
It is also clear that is closed with respect to finite intersection.
BUT, I can not show that is closed with respect to either finite union nor complementation. Given either of those two, it is easy to show that 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:
, which is the union of an open and a closed set. So if we show for any F open, G closed we're done.
To do this I was trying to show the existence of an open set such that
That is finding a such that
This seems intuitively clear to me, but I can't prove it...really
Besides, I was thinking about proving 3th;
, and so we need to prove that:
I use some algebra rule and I get: 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 is a -algebra. Note that since that for all and since is, by assumption, closed under countable unions we see that . But, this implies that where is closed and open. Note though that containing is dense, and since it's closed we see that so that . Thus, . But, this implies that is open which is clearly impossible.
Hee Drex! I like the way you work from Contraposition. You meant probably here.
However I don't think it's a good counterexample. You're contradiction is something like:
From assumption follows: is in . Then . Your contradiction is that
(which is indeed closed) so that , which then must be open. However, is also open...so you can take and . I don't see any contradiction here.
I think you missed the important part of my proof.
If we assume that is a -algebra then since for all then . But, this implies that where is open and is closed, right? That is by definition what it means for to be in . But I claim, in general, that cannot be written as the intersection of an open and a closed set. To see this suppose that are closed and open respectively and . Note first then that . But, this means that so that if were true we'd have to have that (in other words, any closed dense set is the full space). But, this would imply that . But, this says that is open, which is not true.
In your example you can't take since 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!