# Boundary question

1. Jul 10, 2006

### buddyholly9999

Is it possible for a set's boundaries to contain some interior points?
I can't think of an example off-hand...one would be nice.

2. Jul 10, 2006

### StatusX

Define your terms. From the way I use these words, the answer is trivially no, since the boundary of a set is the closure minus the interior.

3. Jul 10, 2006

### buddyholly9999

Ok..for an example...we have that bd (bd S) not equal to the bd S if bd S has interior points...meaning that it is not idempotent...what would an example of this be?

4. Jul 10, 2006

### HallsofIvy

Likewise. In what sense do you mean "boundary" and "interior". The standard definition of "interior point", is "x is an interior point of set A if and only if there exist an open set containing x that is a subset of A". The "interior of a set, A" is the set of all interior points of A.
Some texts define "exterior point of A" to be "an interior point of the complement of A" and "boundary point of A" to be "a point that is neither an interior point nor an exterior point of A". That gives precisely the same property "boundary is closure minus interior" that StatusX mentions and makes it clear that a boundary point is NOT an interior point.

5. Jul 10, 2006

### buddyholly9999

Exactly, which is why I can't come up with a reason that bd (bd S) is not equal to bd S. A boundary is point in which the neighborhood of that point and the intersection of both the set S and the R\S is both nonempty sets. So blatantly put....saying the topological boundary of sets that are a subset of the real numbers is idempotent is false...I want to know an example to prove this to me.

http://en.wikipedia.org/wiki/Boundary_(topology)

6. Jul 10, 2006

### HallsofIvy

Oh, I see. Note that "bd S" is NOT, in general, S, and so saying that the boundary of S "has interior points" does NOT mean that it contains interior points of S. Let S be the set of all rational numbers between 0 and 1. What is the boundary of S?

7. Jul 10, 2006

### buddyholly9999

I'm afraid I don't understand...aren't the boundaries {0, 1} for the open interval (0, 1) in the rational numbers. How does this give a counterexample of bd not being idempotent...i need further explanation.

Last edited: Jul 10, 2006
8. Jul 10, 2006

### StatusX

The confusion was that you wanted to know if a boundary can have any interior points of its own, where as we thought you were asking if it could have any interior points of the set whose boundary it is. The latter is clearly not possible, while the former is. HallsofIvy's clue was to look at the set of rational numbers between 0 and 1. What is this set's closure (the intersection of all closed sets containing the set)? What is its interior (the union of all open sets contained in the set)? And what is their difference? This will be the boundary, and it turns out to have a non-empty interior.

9. Jul 10, 2006

### mathwonk

or look at a disc. the boundary is the circle. what is the boundary of a circle?

10. Jul 10, 2006

### buddyholly9999

ok...so since bd S = (cl S)\(int S)... of all the rational numbers between 0 and 1...that bd S has interior points so that int (bd S) != emptyset...thus preventing bd (bd S) from being the same as bd S???

????
isn't the boundary of a circle surrounding the disk the circle itself? a closed set?

i keep editing this post because...for some reason..doesn't feel like i'm getting a straight answer right here...

that website i posted above it says that bd (bd S) not equal to bd S because there exists a set S that bd S has interior points....as already clarified...not interior points of S but it's own interior points...so what is the straight up example and that shows these two are not equal?

If anyone can explain this to me..without all the allusions to this and that...i'll be their friend...or something...put you on my christmas list..

Last edited: Jul 10, 2006
11. Jul 11, 2006

### HallsofIvy

Let S be the set of all rational numbers between 0 and 1 (in the topological space of all real numbers with the usual topology- NOT "in the rational numbers" as you said above). If p is in that set, then any neighborhood around it, $(p-\delta,p+\delta)$ contains some irrational numbers and so it is NOT an interior points. Since only points in the set can be interior points, this set has no interior points- its interior is empty. On the other hand, since any neighborhood of any point, rational or irrational, contains both rational and irrational points, every point in [0, 1] is a limit point: the closure of S is [0,1]. Therefore, the boundary of S= cl(S)- int(S)= [0,1]- $\phi$= [0, 1].
(Using the defiitions I gave before, no point in S is an interior point and, since any interval around an irrational number contains some rational number, no point in [0,1] is an exterior point: every point in [0,1], being neither interior nor exterior, is a boundary point- the boundary of S is [0,1].)
But, obviously, bd(bd(S))= {0, 1}$\ne$ bd(S).

12. Jul 11, 2006

### buddyholly9999

HallsofIvy...I love you.