# Complement of nonempty open set

• Nov 20th 2012, 11:13 PM
mrjaytski
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!
• Nov 21st 2012, 12:46 AM
ModusPonens
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.
• Nov 21st 2012, 12:54 AM
mrjaytski
Re: Complement of nonempty open set
R^1 connected is quite easy to show. How about R^n?
• Nov 21st 2012, 03:29 AM
Plato
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.
Post what you come up with.
• Nov 21st 2012, 05:46 AM
mrjaytski
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.
• Nov 21st 2012, 07:15 AM
Plato
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?
• Nov 22nd 2012, 01:57 PM
ModusPonens
Re: Complement of nonempty open set
If a space is path connected, it is connected. Clearly R^n is path connected.
• Nov 23rd 2012, 02:15 AM
ansarey
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.
• Nov 23rd 2012, 05:06 AM
Plato
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?
• Nov 23rd 2012, 08:38 AM
ansarey
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
• Nov 23rd 2012, 09:54 AM
Plato
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.
• Nov 23rd 2012, 01:00 PM
ansarey
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.
• Nov 24th 2012, 03:50 PM
Plato
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.