Twas a nice try, but a real solution exists.

Alright, I'll give a hint, and if no one can get it by Sunday, I'll give out the answer.

The only thing that can be open and closed is something that doesn't exist.

Ricky, if I am thinking right, its the set of complex numbers.

What's on second.

Darthradius got it!

And Friday is pitching.

This requires a bit of knowledge of sets and logic.  I'll try to be brief.

First off, whenever we talk about open and closed sets, we are referring to sets of reals (R) unless otherwise noted.

A set is simply a collection of things, normally things with a similar property.  For example, the set of things you wear include, {shoes, socks, pants, sunglasses, ...}.  Similarly, the set of even numbers is {..., -4, -2, 0, 2, 4, ...}.

Now there is a special set called the empty set or the null set.  It contains 0 things.  So it is written {}, normally using phi

to denote it.

Also, if you have an if then statement like, "If it is tuesday, then I like cheese", if the first part is false, then no matter what the 2nd part is, the statement is true.  Using the same example, if it is Wednesday and I don't like cheese, the statement is still true.  If it is Thursday and I do like cheese, the statement is true.  The only time it is false if it is Tuesday, and I don't like cheese.

The definition for an open set is:

A set O is open if for all points a∈O there exists an ∈-neighborhood of a.

Don't worry about understanding much of that if you don't.  The first part is "if for all points a∈O."  But there are no points in O (the empty set)!  So like the statments above, this statement must be true.

Closed set:

A set C is closed if the complement of C (everything that isn't in C) is open.

The complement of C (the empty set still) is R, the set of all reals.  This set is open, and thus, C is closed.

Well good time understanding yourself darthradius and thanks to the both of you. I definetly understand better now!

hmmm...I believe the exact same argument would work for the entire set of real numbers, wouldn't it?

And I just re-read your explanation, Ricky, and I'm having a little bit of trouble with your statement of the definition of an open set as
"A set O is open if for all points a∈O there exists an ∈-neighborhood of a."

What did you mean by that?...You can create an epsilon neighborhood around any point, whether the set is open or closed.  I think that you meant to say something about the intersection of that neighborhood and the set...

The definitions of open and closed that I remember have to do with inclusion of boundary points.
A boundary point of a set S is defined as a point, say x (in R), such that for any epsilon>0, the neighborhood N(x;epsilon)
intersected with S is nonempty, and that same neighborhood N intersected with R\S is nonempty...

Now the definition that I know for a closed set is a set that contains all of those boundary points.  (And thus, an open set is one whose complement is closed)

For example, consider the set of integers Z.
the set of boundary points of Z (denoted bd Z) is actually all of the integers.  [Does that make sense?...because any neighborhood that you take about an integer includes both an integer (in Z), and all of the other real numbers between integers (in R\Z)]
So since Z contains bd Z (in fact, it IS bd Z), the set Z of integers is closed.

I hope I'm making sense...
So then, for this particular question, I would say (R="the set of real numbers")
bd R = empty set (because there are no points in R that would fit the definition of a boundary point of R)
then, bd R is a subset of R  ----> R is closed
but bd R is also a subset of R\R (which is the empty set) -----> R is open.
Thus, we have that R is both open and closed....

Last edited by darthradius (2006-03-21 10:33:24)

I was using those theorems...
By my definition, a set is open iff its complement is closed.
The empty set was both open and CLOSED, per our earlier "proof"
Then the set R is open since its complement (R\R...aka the empty set) is closed.

That's what I was trying to say...And I just realized that your definition is actually the same as mine, but you are using that every point in an open set (you called them 'a') are interior points.

As for your kicker...I have no idea how to do it...I'll need to think it over for awhile...(Are we certain that it can be done?)

Well, yes...it is fairly obvious
I, for one, would assume that there was such a set...
It should be possible to derive a contradiction...
But I don't know how just yet...
I'll be back...**said in Arnold's Austrian accent**