# Thread: Creating open sets from closed ones

1. ## Creating open sets from closed ones

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

I am trying to use the whole hypothesis.

Certainly, I can say

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

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

But these choices of U and V dont work.

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

What set do I want to consider?

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

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

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

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

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

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

EDIT: See reply #7 below for correction.

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

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, $X$ is the universal set. So all sets are subsets of $X$.

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 $C=\left\{\left(x,\frac 1x\right),x>0\right\}$ and $D=\left\{(x,y),y=0\right\}$?
To solve the problem, take $f(x)=\frac{d(x,D)}{d(x,C)+d(x,D)}$ and $U=f^{-1}\left(\left]-1,\frac 12\right[\right)$, $V=f^{-1}\left(\left]\frac 12,2\right[\right)$.

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 $C=\left\{\left(x,\frac 1x\right),x>0\right\}$ and $D=\left\{(x,y),y=0\right\}$?
To solve the problem, take $f(x)=\frac{d(x,D)}{d(x,C)+d(x,D)}$ and $U=f^{-1}\left(\left]-1,\frac 12\right[\right)$, $V=f^{-1}\left(\left]\frac 12,2\right[\right)$.
I was sloppy with that positive distance bit.
Trying to simplify the notation I slipped up.

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

The $U = \bigcup\limits_{x \in C} {B(x;c_x )} \;\& \;V = \bigcup\limits_{y \in D} {B(y;d_y )}$ 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.