- #1
Hummingbird25
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Homework Statement
Hi
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.
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
Proof
By the definition of uniform continuity then
[tex]g: \mathbb{R}^n \rightarrow R[/tex] be continuous 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
Hummingbird.