# Thread: Creating open sets from closed ones

1. ## Creating open sets from closed ones

If $\displaystyle C$ and $\displaystyle D$ are closed subsets in $\displaystyle X$ with $\displaystyle C \cap D = \emptyset$, then $\displaystyle \exists$ open sets $\displaystyle U$ and $\displaystyle V$, s.t $\displaystyle U \supset C$ and $\displaystyle V \supset D$ with $\displaystyle U \cap V = \emptyset$

I am trying to use the whole hypothesis.

Certainly, I can say

$\displaystyle X-D$ is open and $\displaystyle X-D \supset C$

$\displaystyle X-C$ is open and $\displaystyle X-C \supset D$

But these choices of U and V dont work.

$\displaystyle (X-D) \cap (X-C) \neq \emptyset$

What set do I want to consider?

$\displaystyle C \cap D$ and $\displaystyle C \cup D$ are both closed.

Thank you for your help.

2. ## Re: Creating open sets from closed ones

What do we assume about the topology on $\displaystyle X$? For example, if we take $\displaystyle X=\{a,b,c\}$ and the topology $\displaystyle \{\emptyset, \{b,c\},\{a,c\},\{c\},X\}$, then $\displaystyle \{a\}$ and $\displaystyle \{b\}$ are closed disjoint subsets and the open subsets which contain $\displaystyle \{a\}$ are $\displaystyle \{a,c\}$ and $\displaystyle X$, the open subsets which contain $\displaystyle \{b\}$ are $\displaystyle \{b,c\}$ and $\displaystyle X$. Therefore, we can't find two disjoint open subsets which contain respectively $\displaystyle \{a\}$ and $\displaystyle \{b\}$.

3. ## Re: Creating open sets from closed ones

Originally Posted by Jame
If $\displaystyle C$ and $\displaystyle D$ are closed subsets in $\displaystyle X$ with $\displaystyle C \cap D = \emptyset$, then $\displaystyle \exists$ open sets $\displaystyle U$ and $\displaystyle V$, s.t $\displaystyle U \supset C$ and $\displaystyle V \supset D$ with $\displaystyle U \cap V = \emptyset$
@Jame, as you can see from reply @2 this statement is not true in a general topological space. In fact, this often taken as the definition of normal spaces.
So in future postings, please provide context for your questions.

I had a look at some of your more resent posts. From them it appears that you are working with metric spaces, such as $\displaystyle \mathbb{R}^n$.
We can show that every metric space is normal.

EDIT: See reply #7 below for correction.

Now define $\displaystyle \mathcal{U} = \bigcup\limits_{x \in C} {\mathcal{B}(x;\delta )}~\&~\mathcal{V} = \bigcup\limits_{x \in D} {\mathcal{B}(x;\delta )}$.

4. ## Re: Creating open sets from closed ones

My apologies, X is a metric space. I type to fast!

Thank you both for your replies!

I have a question though. I see that by making the radius that number, we guarantee U and V are disjoint. But do the open balls stay contained in X? Does that even matter?

5. ## Re: Creating open sets from closed ones

Originally Posted by Jame
I have a question though. I see that by making the radius that number, we guarantee U and V are disjoint. But do the open balls stay contained in X? Does that even matter?
Here is perhaps the most fundamental principle in all topology:
The union of any collection of open sets is an open set.
Moreover, $\displaystyle X$ is the universal set. So all sets are subsets of $\displaystyle X$.

6. ## Re: Creating open sets from closed ones

We can show that every metric space is normal.
To do so, note that disjoint nonempty closed sets are at a positive distance apart.
@plato: there is a thing I don't understand. What hapens if we take $\displaystyle C=\left\{\left(x,\frac 1x\right),x>0\right\}$ and $\displaystyle D=\left\{(x,y),y=0\right\}$?
To solve the problem, take $\displaystyle f(x)=\frac{d(x,D)}{d(x,C)+d(x,D)}$ and $\displaystyle U=f^{-1}\left(\left]-1,\frac 12\right[\right)$, $\displaystyle V=f^{-1}\left(\left]\frac 12,2\right[\right)$.

7. ## Re: Creating open sets from closed ones

Originally Posted by girdav
@plato: there is a thing I don't understand. What hapens if we take $\displaystyle C=\left\{\left(x,\frac 1x\right),x>0\right\}$ and $\displaystyle D=\left\{(x,y),y=0\right\}$?
To solve the problem, take $\displaystyle f(x)=\frac{d(x,D)}{d(x,C)+d(x,D)}$ and $\displaystyle U=f^{-1}\left(\left]-1,\frac 12\right[\right)$, $\displaystyle V=f^{-1}\left(\left]\frac 12,2\right[\right)$.
I was sloppy with that positive distance bit.
Trying to simplify the notation I slipped up.

Here is the usual proof.
If $\displaystyle x\in C$ then $\displaystyle c_x=\frac{\mathcal{D}(D;c)}{3}>0$.
If $\displaystyle y\in D$ make a similar statement for $\displaystyle d_y$.

The $\displaystyle U = \bigcup\limits_{x \in C} {B(x;c_x )} \;\& \;V = \bigcup\limits_{y \in D} {B(y;d_y )}$ gives the required open sets.

8. ## Re: Creating open sets from closed ones

OK, I guessed that was what you meant, but I wanted to be sure.