Closed separated sets in disjoint open sets

[modnarandom]

Hi,

I was reading over a solution after working on a problem and got confused about some parts:

http://nweb.math.berkeley.edu/sites/default/files/pages/f10solutions.pdf
(first problem)

First, how do we know that there are disjoint open sets U and V for each of the separated sets? (does this have anything to do with the covers for A and B?) Also, I'm not sure why the intersection of the sets F_i is empty.

Thanks!

 

[algebrat]

Hi,

I was reading over a solution after working on a problem and got confused about some parts:

http://nweb.math.berkeley.edu/sites/default/files/pages/f10solutions.pdf
(first problem)

First, how do we know that there are disjoint open sets U and V for each of the separated sets? (does this have anything to do with the covers for A and B?) ...

I think that's by definition of $\bigcap X_i$ not being connected (mostly, you may have to justify $A,B$ are each closed).

...Also, I'm not sure why the intersection of the sets F_i is empty.

Thanks!...

$\bigcap F_i=\bigcap (X_i-U\cup V)=\bigcap X_i-(U\cup V)=\emptyset.$

I'll let you do the hard work, you need to check the second and third equalities, I'm lazy (exercise left to the reader).

 

[HallsofIvy]

The usual definition of "separated" is that A and B are separated if and only if $\overline{A}\cup B= A\cup\overline{B}= \phi$. Let U be the interior of A and V the interior of B.

 

[algebrat]

The usual definition of "separated" is that A and B are separated if and only if $\overline{A}\cup B= A\cup\overline{B}= \phi$. Let U be the interior of A and V the interior of B.

But how can you guarantee that $A\subset U$?

 

[algebrat]

The usual definition of "separated" is that A and B are separated if and only if $\overline{A}\cup B= A\cup\overline{B}= \phi$. Let U be the interior of A and V the interior of B.

I have seen [STRIKE]lots of[/STRIKE] other "usual" definitions of separated.

OP modnarandom, which definition of separated would you like to use?

 

[modnarandom]

Thanks for the replies! I think I understand why the intersection of F_i is empty. But I'm still confused about the existence of the disjoint open sets. I think the interior of a set A is contained in A, so I don't think that can be used for an open set containing A. I already showed that A and B are closed sets. Can this be used for anything?

algebrat: I usually use the definition that HallsofIvy gave, but I've seen others too.

 

[algebrat]

Thanks for the replies! I think I understand why the intersection of F_i is empty. But I'm still confused about the existence of the disjoint open sets. I think the interior of a set A is contained in A, so I don't think that can be used for an open set containing A. I already showed that A and B are closed sets. Can this be used for anything?

algebrat: I usually use the definition that HallsofIvy gave, but I've seen others too.

Then, once we have $A,B$ are closed, my guess is that we may have to use the metric space condition (we have not used yet right?) to find $U,V$.

 

[modnarandom]

I tried to use the fact that X is a compact space but I'm not sure how to do that. Since A and B are closed subsets of a compact space, they are also compact. So, maybe U and V might be the open covers of A and B. But I don't think they're necessarily disjoint.

 

[algebrat]

I tried to use the fact that X is a compact space but I'm not sure how to do that. Since A and B are closed subsets of a compact space, they are also compact. So, maybe U and V might be the open covers of A and B. But I don't think they're necessarily disjoint.

Right, $U,V$ will need some work. Also, be careful about the "the" when you say open cover. I'm not sure what you meant by that, leaving me with little to give feedback on. I don't mean to be discouraging.

What about the condition of $X$ being a metric space, has that been used yet? Remember, when proving an exercise, we should check that we've used all the hypotheses (unlike in physics problems, where they often give red herrings).

Here is a big hint, for any $x\in A$, find open sets $U,V$ that separate $x$ from $B$.

The last hint is not easy to prove, but it is a good foothold. You should figure out how to prove it, and how it's truth helps your problem. (Both are tricky, and both use compactness.)

 

[modnarandom]

Just to clarify, do you mean two open sets U and V such that one contains x and the other doesn't? Also the "the" was a mistake (wasn't supposed to be there).

 

[algebrat]

Just to clarify, do you mean two open sets U and V such that one contains x and the other doesn't?

Right. More specifically, I suggested they should separate $x$ from $B$.

Also the "the" was a mistake (wasn't supposed to be there).

Sounds good.