# Proof of CLopen sets

1. Jan 28, 2008

### Hummingbird25

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.

2. Jan 28, 2008

### HallsofIvy

Staff Emeritus
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.

3. Jan 28, 2008

### Hurkyl

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