# Sets and boundaries for several variables

1. Feb 25, 2008

### goatsebear

1. The problem statement, all variables and given/known data
Specify the boundary of the following sets. State whether the sets are open, closed, or neither.

1. {1 / n : n $$\in$$ N}

2. { (x,y,z): x^2 + y^2 = z }

3. { (x,y): 0 < x $$\leq$$ 4, 0 < y $$\leq$$ 4 }

2. The attempt at a solution

1. I want to say that the boundary is where n =/= 0 because then that would make the fraction an error. I think that the set is open because the equation does not include the boundary.

2. I think that the boundary is the equation of x^2 + y^2 = z I'm not really sure on this one. Its more of a guess since I can't find any examples similar to it. Set would be closed since the equation includes the boundary.

3. The boundary would be x = 0, x=4, y = 0, y =4. Would be neither since half of it includes the boundary and the other half does not.

Thanks for any help.

Last edited: Feb 25, 2008
2. Feb 25, 2008

### Dick

What do the words 'open' and 'closed' mean to you? 1) should be easy and you got it wrong (assuming you meant as a subset of the reals). I think you are thinking too hard about 'boundaries'. BTW you should be careful to state which space these sets are to be regarded as subsets of before you can answer the open/closed question.

3. Feb 25, 2008

### goatsebear

A set is closed if the set contains its boundary right? A set is open if it contains a neighborhood of each of its points. My professor never really taught this subject well and it is really hard to try and teach this from the book.

For the first problem, is my boundary right? I would actually write them as {n : n=/= 0 }

4. Feb 25, 2008

### Dick

Basically right. But then {1/n} can't be open since any neighborhood of 1/2 e.g. contains points which are NOT of the form 1/n. It's also true that a set is closed if it's complement is open. 0 is in the complement of {1/n}. Now what?

5. Feb 26, 2008

### goatsebear

What do you exactly mean by complement? I have never heard that term before. And looking back at the other questions, I think that the boundary of 2 would be
x^2 + y^2 - z $$\leq$$ 0 which would include all the points on and in the figure.

6. Feb 26, 2008

### Dick

The complement of a set is the set of all points that are NOT in the set. So I'm saying 0 is not in {1/n}. But every neighborhood of 0 contains a points in {1/n}.

7. Feb 26, 2008

### goatsebear

So that would make {1/n} open since it does not its boundary but it does contain the neighborhood of points right?

8. Feb 26, 2008

### Dick

I'll give you a hint. {1/n} is neither open nor closed. You can't really understand what a boundary means until you understand what open and closed mean. So please forget about boundary for at least 5 minutes and try to explain to me the true fact that {1/n} is neither open nor closed. Start with why it's not open. That's the easy part.

9. Feb 26, 2008

### goatsebear

Alright. So it cannot be open because a circle of points around a point in the set includes points outside of the set. Some of the points are not interior points in it. I guess the point of n = 3 where then it would be 1/3.

For closed, since 0 is not in the set, it is not closed. The limit of the fcn goes to 0 and since it is not included, it is not closed.

10. Feb 27, 2008

### Dick

Yes. Good job! The boundary of {1/n} is {1/n}+{0}. Your answers for 2) and 3) are correct. For 2) consider an (x,y,z) such that z is not equal to x^2+y^2. Does it make sense there is neighborhood of (x,y,z) such that z is not equal to x^2+y^2? That means the complement is open. So the set is closed. I'm not sure how rigorously you are supposed to present a proof for these...

11. Feb 27, 2008

### Dick

BTW. {1/n} has NO interior.

12. Feb 27, 2008

### goatsebear

I basically just have to write down the boundary and if the set is open, closed or neither. Thanks for the help.