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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Name a set that is both open and closed.

(Don't you just love the title? )

*Last edited by Ricky (2006-03-16 17:00:53)*

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**ganesh****Moderator**- Registered: 2005-06-28
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The set of twin primes.

Closed because it hasn't been proved that they go on endlessly.

Open because it hasn't been proved that they don't do on endlessly.

Character is who you are when no one is looking.

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**Ricky****Moderator**- Registered: 2005-12-04
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Twas a nice try, but a real solution exists.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**Jenilia****Member**- Registered: 2005-07-09
- Posts: 64

And whats that?

Ideas are funny little things, they won't work unless you do.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,569

Or a door with a hole.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 18,077

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

Character is who you are when no one is looking.

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**darthradius****Member**- Registered: 2005-11-28
- Posts: 97

I DO love the title...

And I believe that the empty set is the set that is both open and closed. It is sort of vacuously true that it contains its boundary points (also the empty set), so its closed. But the boundary points (empty set) are also contained in the complement of the empty set, so it is open...

Whew...that was tricky to say....

who's on first?

The greatest challenge to any thinker is stating the problem in a way that will allow a solution.

-Bertrand Russell

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**MathsIsFun****Administrator**- Registered: 2005-01-21
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What's on second.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**darthradius****Member**- Registered: 2005-11-28
- Posts: 97

I don't know's on third (I think...I can't remember too well)

The greatest challenge to any thinker is stating the problem in a way that will allow a solution.

-Bertrand Russell

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Darthradius got it!

And Friday is pitching.

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**Jenilia****Member**- Registered: 2005-07-09
- Posts: 64

I hardly understand Darthradius!

Ideas are funny little things, they won't work unless you do.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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.

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**darthradius****Member**- Registered: 2005-11-28
- Posts: 97

hooray!

I'm sorry Jenilia...I can hardly undertand myself...

Ricky's explanation is much better...

I especially love "If it's Tuesday then I like cheese"...

The greatest challenge to any thinker is stating the problem in a way that will allow a solution.

-Bertrand Russell

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**Jenilia****Member**- Registered: 2005-07-09
- Posts: 64

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

Ideas are funny little things, they won't work unless you do.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

There is one more possible answer. It has something to do with the null set.

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**darthradius****Member**- Registered: 2005-11-28
- Posts: 97

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 (2006-03-20 11:33:24)*

-Bertrand Russell

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Ah, you're right, I forgot the last part:

such that the ∈-neighborhood of a is a subset of O.

And yes, you are making sense and you are completely right. I'm not sure if I understand how you got open though:

but bd R is also a subset of R\R (which is the empty set) -----> R is open.

But what does the boundary points being in the empty set have to do with the set of reals being open?

If you want to show R to be open, just consider the defintion. Every ∈-neighborhood around any point a∈R is a subset of R. Thus, it must be open.

Of course, you can make it much easier using the theorems:

If A is closed, then A complement is open.

If A is open, then A complement is close.

The complement of R is the empty set, which was already shown to be both open and closed. Thus, both of the above theorems apply and R must be open and closed.

Now here's the kicker. Prove that there exists no other set which is both open and closed.

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**darthradius****Member**- Registered: 2005-11-28
- Posts: 97

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?)

-Bertrand Russell

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Yes, it can be shown that R and the empty set are the only sets which are both open and closed. Took me a little while to work this one out but I think I got it. If you or anyone else wants a hint, let me know.

Actually, I'll give you one which should be fairly obvious:

How do we go about proving the non-existance of something.

*Last edited by Ricky (2006-03-20 12:00:20)*

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**darthradius****Member**- Registered: 2005-11-28
- Posts: 97

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**

-Bertrand Russell

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