Complement of nonempty open set

I've tried to think this, help would be appreciated to get me going a gain.

If we have two nonempty open set $\displaystyle A,B \subset {R}^{n}$ so that $\displaystyle A \cap B = \emptyset$ *(two separate sets)*. How could I show that complement of union A and B is also nonempty, $\displaystyle \complement (A \cup B) \neq \emptyset$.

I'm thinking something like, because the complement is $\displaystyle {R}^{n} \setminus (A \cup B)$ so because its R^n there must be some points that belong to that complement. But basicly I think that don't work (just to say because it's R^n) because I have show that it (complement of the union is nonempty) follows from the notion that A,B are open nonempty sets.

Any ideas and help? Thanks!

Re: Complement of nonempty open set

R^n is connected so it can't be the union of two components. Thus R^n\(AUB) is non empty.

Re: Complement of nonempty open set

R^1 connected is quite easy to show. How about R^n?

Re: Complement of nonempty open set

Quote:

Originally Posted by

**mrjaytski** R^1 connected is quite easy to show. How about R^n?

It is also easy.

Start with $\displaystyle \mathbb{R}^2$. Then expand.

Post what you come up with.

Re: Complement of nonempty open set

Is there, by the way, any other way to show this complement to be non empty? Without using connectivity.

Re: Complement of nonempty open set

Quote:

Originally Posted by

**mrjaytski** Is there, by the way, any other way to show this complement to be non empty? Without using connectivity.

Why do you want to do that?

Quote:

Originally Posted by

**mrjaytski** If we have two nonempty open set $\displaystyle A,B \subset {R}^{n}$ so that $\displaystyle A \cap B = \emptyset$ *(two separate sets)*. How could I show that complement of union A and B is also nonempty, $\displaystyle \complement (A \cup B) \neq \emptyset$.

The OP is an equivalent way of asking for a proof that $\displaystyle \mathbb{R}^n$ is connected. In fact, in Moore spaces that is the very definition of connectivity.

Go back to my first reply. Can you think of $\displaystyle \mathbb{R}^1$ as a subset of $\displaystyle \mathbb{R}^2~?$

If $\displaystyle \mathbb{R}^2$ is not connected the do we have a contradiction?

Re: Complement of nonempty open set

If a space is path connected, it is connected. Clearly R^n is path connected.

Re: Complement of nonempty open set

if the complement is empty, then it means as "ModusPonens" stated, R^n is union of two distinct sets A & B.

consider a point P on boundary of A. the boundary of A is not belongs to A (cause it is open). but if the complement is empty, this boundary must belong to B. the point P can not belong to interior of B (because of in each neighbor of P, at least one point of A exist), since it must belong to boundary of B. this is contradiction, since it means that P is not belong to B.

So at least P is not belong to A & B and since belongs to complement.

i'm not a professional math expert, but p.s somone evaluate this prove.

Re: Complement of nonempty open set

Quote:

Originally Posted by

**ansarey** if the complement is empty, then it means as "ModusPonens" stated, R^n is union of two distinct sets A & B.

consider a point P on boundary of A.

How do you know that the boundary of $\displaystyle A$ is not empty?

Re: Complement of nonempty open set

you are right.

i suppose another proof.

the complement of open set is closed. so union of A and its complement equal to R^n. also the intersection of A and its complement is empty. now if the R^n equal to union of A & B, then B must be complement of A. B is open but complement of A is closed. contradiction

Re: Complement of nonempty open set

Quote:

Originally Posted by

**ansarey** the complement of open set is closed. so union of A and its complement equal to R^n. also the intersection of A and its complement is empty. now if the R^n equal to union of A & B, then B must be complement of A. B is open but complement of A is closed. contradiction

There are typologies in which there are sets which are both open and closed.

They are often called clopen sets.

If fact, a space is connected if and only if the only clopen sets are the empty set and the whole set.

In your proofs, you must use the properties of $\displaystyle \mathbb{R}^n$.

You have done that so far.

Re: Complement of nonempty open set

dear friends, please introduce some ref about set, vector space, and especially properties of spaces (i.e connectivity, etc). i am phd student of engineering and working on dynamical system and nonlinear control. i am familiar with subject like open, closed, closure, clopen, etc. but i think i dont have enough insight and intuition in this field. for example, i can think about above problem only in level of definition (and try to prove it by searching a contradiction), and can not think in higher level (think about connectivity, and other properties of spaces like R^n as Mr plato and ModusPonens can do) to prove or understand this type of problems.

Re: Complement of nonempty open set

Quote:

Originally Posted by

**ansarey** for example, i can think about above problem only in level of definition (and try to prove it by searching a contradiction), and can not think in higher level (think about connectivity, and other properties of spaces like R^n as Mr plato and ModusPonens can do) to prove or understand this type of problems.

The book __Elementry Theory of Metric Spaces__ by Robert B Reisel is written for a Moore style course. That makes it ideal for self-study. Do not let the title fool you. I used it as a text for a course in topology designed for pre-university teachers. The subtitle is "*A Course in Constructing Mathematical Proofs*.