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Open & Closed Sets

  • Thread starter kingwinner
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  • #1
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Hi, I have some questions regarding open and closed sets.

Definitions: Let S be a subset of R^n. S is called "open" if it contains none of its boundary points and S is "closed" if it contains all of its boundary points.

1) Let S={(x,y,z) E R^3 | z=0}.
1a) What is the boundary of S?
1b)Is S open, closed, both, or neither?

My attempts:
1a) The boundary of S is S itself, am I correct?
1b) S is closed since every point in the given plane S is a boundary point and S certainly contains every point in the given plane S, i.e. S contains ALL of its boundary points. Is this correct?

Now do I have to check separately that S is "not open" (how?), or can I conclude immediately that "S is closed implies S is not open"?

=================

2) "R^n and the null set are BOTH closed and open." I have no clue why this statement is true. How can a set be BOTH closed and open? I am just so lost...


Thanks for helping!
 

Answers and Replies

  • #3
matt grime
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It's good to start with a definition. One for the boundary of a set would help you.
 
  • #4
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A boundary point is in the closure of S and isn’t interior to S.

x is in the closure of S if every open ball about x contains at least one point in S.
x is in the interior of S if there exist an open ball about x which is a subset of S.

putting these together, the boundary points are all of those points where we can make a ball has at least one point in S, but no ball will be a subset of S.

Now try to visualize the set you described, what does it look like? Is it a cube? A plane? Is it a line?
 
  • #5
HallsofIvy
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"closed" contains all of its boudary points.

"open" contains none of its boundary points.

To be both open and closed both of those would have to be true: in other words "all of its boundary points" would have to be the same as "none of its boundary points".

How can that be true? Only if the set has no boundary points. What are the boundary points of the empty set? What are the boundary points of all of Rn itself?
 
  • #6
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"closed" contains all of its boudary points.

"open" contains none of its boundary points.

To be both open and closed both of those would have to be true: in other words "all of its boundary points" would have to be the same as "none of its boundary points".

How can that be true? Only if the set has no boundary points. What are the boundary points of the empty set? What are the boundary points of all of Rn itself?
The empty set and R^n have no boundary points, so they are both open and closed...I see

For #1b, how can I prove that S is "not open"?
 
  • #7
HallsofIvy
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For #1, you have S= {(x,y,0)}

Open means "does not contain any of its boundary points". You said in your first post:

"1a) The boundary of S is S itself, am I correct?"
Yes, that is correct. Now does S contain any of its boundary points?

(You correctly said that S is closed because it contains all of its boundary points, we have just agreed that a set can be both open and closed only if it has no boundary points. Is that the case here?)
 
  • #8
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it's both open and closed

your set is the entire xy-plane.

It is both open and closed because:
it contains all of its boundary points (is open)because it doesn't have any boundary points
and it contains none of its boundary points (open)because it doesn't have any.

So it is open and closed because
it contains all of its boundary points and none of its boundary points all at the same time
simply because it doesn't have any boundary points.
 
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  • #9
Hey,..
i think S is Closed because it contains all its boundary points but R^3 and the empty set are open and closed if u are considering this as a trivial topology,..

i define a set is closed if its complement is open,..

then if u consider the empty set as being closed then R^3 is open ,
and if u consider the empty set as being open then R^3 is closed,.

The empty set is both open and closed, u can see this because of mathematical logic,
false statement => true statement is a true logically true statement,..
so if you say let x be an element of the empty set,.. then it lies on the boundary,.. so the empty set is closed,... by starting with a false statement you can deduce whatever you like about the empty set regarding its elements becase it has no elements (since closure is defined in terms of elements),.. its wired hey,..
Its like ,,.. if pigs can fly then i am the king of england,... is logically true
 
  • #10
HallsofIvy
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your set is the entire xy-plane.

It is both open and closed because:
it contains all of its boundary points (is open)because it doesn't have any boundary points
and it contains none of its boundary points (open)because it doesn't have any.

So it is open and closed because
it contains all of its boundary points and none of its boundary points all at the same time
simply because it doesn't have any boundary points.
No, that is not true. Yes, the set is the entire xy-plane, as a subset of R3! It does, in fact, have boundary points- as kingwinner said in the initial post, the boundary points of S are all of the points in S itself. Since kingwinner was using as his definition of "closed", "contains all of its boundary points", clearly S is closed. Since kingwinner was using as his definition of "open", "contains none of its boundary points", but S does contain boundary points, it is NOT open.

tommygun101, your post doesn't have anything to do with the original question.
 
  • #11
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your set is the entire xy-plane.

It is both open and closed because:
it contains all of its boundary points (is open)because it doesn't have any boundary points
and it contains none of its boundary points (open)because it doesn't have any.

So it is open and closed because
it contains all of its boundary points and none of its boundary points all at the same time
simply because it doesn't have any boundary points.
A common definition of boundary point is: A point x in a set S is a boundary point of S if every neighborhood of x intersects both S and S compliment.

Here the xy-plane DOES have boundary points - and as the original poster right claimed the set of boundary points of the xy-plane in 3-space IS the xy-plane. So the set of boundary points of the xy-plane is NONEMPTY, moreover the set CONTAINS not only SOME of it's boundary points but ALL of them. From this we see that the set is closed and not open.

A different but equivalent definition of open is one wer every point has a neighborhood that is contained in the set. Take ANY point in the xy plane and since a neighborhood in 3-space is a 3-dimensional sphere any neighborhood of any point will NOT be contained in a single plane.
 
  • #12
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Actually, his orginal statement was false, so it doesn't matter

Originally Posted by HallsofIvy
"No, that is not true. Yes, the set is the entire xy-plane, as a subset of R3! It does, in fact, have boundary points- as kingwinner said in the initial post, the boundary points of S are all of the points in S itself. Since kingwinner was using as his definition of "closed", "contains all of its boundary points", clearly S is closed. Since kingwinner was using as his definition of "open", "contains none of its boundary points", but S does contain boundary points, it is NOT open.

tommygun101, your post doesn't have anything to do with the original question".
What I say:

The xy plane is 2 dimensions. The third dimension was taken out by setting z equal to zero.It doesn't matter that the xy plane is bounded in three space by the z axis. The xy plane in 2 dimensions extends infinitely. The third dimension was thrown out when z was set equal to zero. Not to mention that z = 0 is the equation for the xy plane.

and so, I argue that this statement is false:
"It does, in fact, have boundary points...the boundary points of S are all of the points in S it's self"

The definition of a boundary point is as follows:
A point (xo,yo) in a region [set] R in the xy-plane is an interior point of R if it is the center of a disk of positive radius that lies entirely in R. A point (xo,yo) is a boundary point of R if every disk centerd at (xo,yo) contains points that lie outside of R as well as points that lie in R. (The boundary point itself need not belong to R.)


Since S is the entire xy-plane, there cannot be a disk of positive radius centered at (xo,yo) in the xy plane that contains points outside of the xy-plane as well as points inside the xy-plane, and so by definition of boundary point, the xy-plane doesn't have any boundary points. Thus the statement "the boundary points of S are all of the points in S it's self" is a false statement.

As tommygun fortunately pointed out for all of us, if you start out with a false statement, anything implied from that false statement is automatically true. Thus, anyone can assume that since the boundary points of S are all of the points in S it's self then he is God for all I care, and the statement would be vacuously true by defalt.
 
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  • #13
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What SiddharthM says:
"A common definition of boundary point is: A point x in a set S is a boundary point of S if every neighborhood of x intersects both S and S compliment.

Here the xy-plane DOES have boundary points - and as the original poster right claimed the set of boundary points of the xy-plane in 3-space IS the xy-plane. So the set of boundary points of the xy-plane is NONEMPTY, moreover the set CONTAINS not only SOME of it's boundary points but ALL of them. From this we see that the set is closed and not open.

A different but equivalent definition of open is one wer every point has a neighborhood that is contained in the set. Take ANY point in the xy plane and since a neighborhood in 3-space is a 3-dimensional sphere any neighborhood of any point will NOT be contained in a single plane."

What I say to SiddharthM:

You gave a definition of boundary point, and did nothing with it. You didn't use your definition to show that the xy plane does have boundary points. As I have shown previously, the xy-plane doesn't have boundary points.

Secondly, you abused the definition of open for a three dimensional space. The xy plane is 2 dimensions. The third dimension was taken out by setting z equal to zero.It doesn't matter that the xy plane is bounded in three space by the z axis. The xy plane in 2 dimensions extends infinitely. The third dimension was thrown out when z was set equal to zero. Not to mention that z = 0 is the equation for the xy plane.
 
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  • #14
HallsofIvy
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What I say:

The xy plane is 2 dimensions. The third dimension was taken out by setting z equal to zero.It doesn't matter that the xy plane is bounded in three space by the z axis. The xy plane in 2 dimensions extends infinitely. The third dimension was thrown out when z was set equal to zero. Not to mention that z = 0 is the equation for the xy plane.
No, it does matter. The xy-plane is 2 dimensional but you must still think of it as a subset of R3. In particular, any neighborhood of a point in the xy-plane is a ball centered on that point that includes both points in the xy-plane and points that are not in the xy-plane. The xy-plane, as a subset of R3, which is the case in this problem, has no interior points. All points in it are boundary points.

and so, I argue that this statement is false:
"It does, in fact, have boundary points...the boundary points of S are all of the points in S it's self"
Yes, I realize you are arguing that. You are wrong.

The definition of a boundary point is as follows:
A point (xo,yo) in a region [set] R in the xy-plane is an interior point of R if it is the center of a disk of positive radius that lies entirely in R. A point (xo,yo) is a boundary point of R if every disk centerd at (xo,yo) contains points that lie outside of R as well as points that lie in R. (The boundary point itself need not belong to R.)
Not "disk", "ball". The original post said "1) Let S={(x,y,z) E R^3 | z=0}."
S is given as a subset of R3.

Since S is the entire xy-plane, there cannot be a disk of positive radius centered at (xo,yo) in the xy plane that contains points outside of the xy-plane as well as points inside the xy-plane, and so by definition of boundary point, the xy-plane doesn't have any boundary points. Thus the statement "the boundary points of S are all of the points in S it's self" is a false statement.

As tommygun fortunately pointed out for all of us, if you start out with a false statement, anything implied from that false statement is automatically true. Thus, anyone can assume that since the boundary points of S are all of the points in S it's self then he is God for all I care, and the statement would be vacuously true by defalt.
One more time: the original post said "1) Let S={(x,y,z) E R^3 | z=0}. " S is a subset of R3. "Neighborhoods" are balls, not disks. They include points in the xy-plane and points not in the xy-plane.
 
  • #15
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Yes,
Halls of Ivy, I stand corrected. I am right in the case of 2 Dimensions. I believe I am wrong in the case of three dimensions. As I went over what you previously wrote (not your current post but the previous one) I realized that you pointed out that we were considering space, R^3.

I posted my arguements before I realized this, and you posted a reply before I could correct myself , as, I was in the shower:).

All justifications aside, we do consider solid balls in 3D, and a solid ball centered at (xo,yo) on the xy- plane would merely cut/intesect the sphere, it wouldn't engolf the sphere. And so it follows that the entire xy plane consists of boundary points because:

" a point (xo,yo,zo) is a boundary point of R if every sphere centered at (xo,yo,zo) encloses points that lie outside of R as well as points that lie inside R."

Since a region is closed if it contains its entire boundary and the xy plane certainly contains its boundary (it is its boundary), the xy plane is closed when considering 3D by definition.


I am greatful that you cared enough to point out what I didn't consider in the begining.
I'm glad we had this fun little debate. Otherwise, It would have taken me longer to figure out my mistake on my own; that is, if I would have found it at all.
 
  • #16
HallsofIvy
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It's a common error. If you are given a set A, in a topological space X, you can think of A as a topological space itself, with the inherited topology.

I was going to point out that if, given a set A, in a topological space X, we were to immediately assume that it was a topological space, everything would be easy! Every set would be both open and closed and every function would be continuous!
 

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