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Proof of CLopen sets

  1. Jan 28, 2008 #1
    1. The problem statement, all variables and given/known data


    I have been working with sets which are both closed and open the socalled clopen sets. I have some question.

    Lets say that [tex]T \subseteq \mathbb{R}^n[/tex] is a subset which is both closed and open, and then if [tex]T = \mathbb{R}^n[/tex] or [tex]T = \emptyset[/tex]. Assume that [tex]\{T \neq \mathbb{R}^n|T \neq \emptyset\}[/tex]

    Proving this results in a contradiction.

    (1) Let [tex]U = \mathbb{R}^n \setminus T[/tex] and show that U is open and closed and not-empty.

    3. The attempt at a solution

    If [tex]U = \mathbb{R}^n \setminus (T = \mathbb{R}^n) = \emptyset.[/tex] since [tex]T = \mathbb{R}^n[/tex] which is non-empty and both closed and open according to above. Thus [tex]U = \mathbb{R}^n\setminus \emptyset = \mathbb{R}^n.[/tex] which therefore upholds the claim in (1).

    (2) Let [tex]g: \mathbb{R}^n \rightarrow \mathbb{R}[/tex]
    be defined as:

    [tex]g(x) = \left( \begin{array}{cc}1 \ \ \mathrm{for \ t \ \in \ T}\\ 0 \ \ \mathrm{for \ u \ \in U} \end{array}[/tex]

    Prove that g is continious at every point [tex]t_{0} \in T[/tex]. is it something which uniform continouity which I need to use here?

    If yes then I need to show here that for any t in g converges towards [tex]t_0[/tex]??

    If yes then


    By the definition of uniform continuity then

    [tex]g: \mathbb{R}^n \rightarrow R[/tex] be continous at every [tex]t_0[/tex] if and only if there for every [tex]\epsilon > 0[/tex] exists a [tex]\delta > 0[/tex] such that [tex]|g(t) - g(t_0)| < \epsilon \Leftrightarrow \|t - t_0 \| < \delta.[/tex]

    Is the trick then to show that g upholds the defintions above??

    Sincerely Yours
  2. jcsd
  3. Jan 28, 2008 #2


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    What are you allowed to use? If R is a "clopen" set such that it is neither the empty set nor all of Rn, then S= complement (R) is neither the empty set nor Rn. S is closed because R is open- and R is also closed, of course. Then Rn= R union S where both R and S are closed: Rn is not a connected set. If you are allowed to use the fact that Rn is a connected set, that's a contradiction.
  4. Jan 28, 2008 #3


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    Staff Emeritus
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    Gold Member

    I confess that I am unable to figure out what you meant to say. Could you restate it?
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