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Math Help - Creating open sets from closed ones

  1. #1
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    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.

    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 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\}.
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    Re: Creating open sets from closed ones

    Quote Originally Posted by Jame View Post
    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.
    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 \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 )} .
    Last edited by Plato; September 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, X is the universal set. So all sets are subsets of X.
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  6. #6
    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 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).
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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 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.
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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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