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Thread: Creating open sets from closed ones

  1. #1
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    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.
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  2. #2
    Super Member girdav's Avatar
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    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 $\displaystyle \{b\}$.
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    Re: Creating open sets from closed ones

    Quote Originally Posted by Jame View Post
    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 )} $.
    Last edited by Plato; Sep 21st 2011 at 10:30 AM.
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    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?
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    Re: Creating open sets from closed ones

    Quote Originally Posted by Jame View Post
    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 $\displaystyle X$.
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    Super Member girdav's Avatar
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    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 $\displaystyle U=f^{-1}\left(\left]-1,\frac 12\right[\right)$, $\displaystyle V=f^{-1}\left(\left]\frac 12,2\right[\right)$.
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    Re: Creating open sets from closed ones

    Quote Originally Posted by girdav View Post
    @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.
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  8. #8
    Super Member girdav's Avatar
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    Re: Creating open sets from closed ones

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