PDA

View Full Version : Proof of CLopen sets


Hummingbird25
Jan28-08, 03:56 PM
1. The problem statement, all variables and given/known data

Hi

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

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

Proving this results in a contradiction.

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

3. The attempt at a solution

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

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

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

Prove that g is continious at every point t_{0} \in T. 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 t_0??

If yes then

Proof

By the definition of uniform continuity then

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

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

Sincerely Yours
Hummingbird.

HallsofIvy
Jan28-08, 04:56 PM
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.

Hurkyl
Jan28-08, 05:24 PM
Lets say that T \subseteq \mathbb{R}^n is a subset which is both closed and open, and then if T = \mathbb{R}^n or T = \emptyset. Assume that \{T \neq \mathbb{R}^n|T \neq \emptyset\}
I confess that I am unable to figure out what you meant to say. Could you restate it?