# Thread: Closure, Boundary, and Neighborhoods

1. ## Closure, Boundary, and Neighborhoods

Prove that:
Let E be a subset of R^d and x an element of R^d. Then:
c) x ∈ ∂E if and only if every neighborhood of x contains points of E and points of the complement of E.

(parts a and b of this theorem state: (a) x ∈ int(E) if and only if there is a neighborhood of x that is contained in E; (b) x ∈ E if and only if every neighborhood of x contains a point of E ... I've proven parts a and b though)

2. ## Re: Closure, Boundary, and Neighborhoods

Originally Posted by xsavedkt
Prove that:
Let E be a subset of Rd and x an element of Rd. Then:
c) x ∈ ∂E if and only if every neighborhood of x contains points of E and points of the complement of E.
(parts a and b of this theorem state: (a) x ∈ E◦ if and only if there is a neighborhood of x that is contained in E; (b) x ∈ E if and only if every neighborhood of x contains a point of E ... I've proven parts a and b though)
I hate to tell you, but what you have posted is incomprehensible.
The notation is completely non-standard.
What in the world is Rd?
You write that "c) x ∈ ∂E if and only if every neighborhood of x contains points of E and points of the complement of E."
But that is the very definition of a boundary point.
So what is the actual question?

3. ## Re: Closure, Boundary, and Neighborhoods

Originally Posted by Plato
I hate to tell you, but what you have posted is incomprehensible.
The notation is completely non-standard.
What in the world is Rd?
You write that "c) x ∈ ∂E if and only if every neighborhood of x contains points of E and points of the complement of E."
But that is the very definition of a boundary point.
So what is the actual question?
I made some minor changes to the notation. Rd was supposed to mean R^d (R being the real number line)... sorry!
However, you said that the question is the very definition of a boundary point.... that's not true though, in the definition of a boundary, we don't talk about the neighborhood of x. And a boundary POINT is made possible by the definition of a boundary and then proving this theorem (at least that's how it seems like to me). I think the question is to prove that every boundary point's neighborhoods contain a point in E and the complement in E... does this make more sense?

Thanks for trying to help...

4. ## Re: Closure, Boundary, and Neighborhoods

Originally Posted by xsavedkt
you said that the question is the very definition of a boundary point.... that's not true though, in the definition of a boundary, we don't talk about the neighborhood of x.
That is total nonsense!
For the last 100 years, a boundary point x of a set E has been defined as: if O is an open set containing x then O contains a point of E and a point of E complement.
Now we cannot be faulted if you are burdened with non-standard definitions. Maybe you should bring this to the instructor's attention.

5. ## Re: Closure, Boundary, and Neighborhoods

Originally Posted by Plato
That is total nonsense!
For the last 125 years, a boundary point x of a set E has been defined as...
Maybe the instructor is using one of the other equivalent definitions of "boundary".

I learned it as X(bar) toss X.

6. ## Re: Closure, Boundary, and Neighborhoods

Originally Posted by TheChaz
Maybe the instructor is using one of the other equivalent definitions of "boundary".

I learned it as X(bar) toss X.
The definition of boundary with this instructor is: the set E \ E◦ is called the boundary of E and is denoted ∂E.

7. ## Re: Closure, Boundary, and Neighborhoods

Originally Posted by Plato
That is total nonsense!
For the last 100 years, a boundary point x of a set E has been defined as: if O is an open set containing x then O contains a point of E and a point of E complement.
Now we cannot be faulted if you are burdened with non-standard definitions. Maybe you should bring this to the instructor's attention.
Hmm, that is interesting. First off, our instructor didn't really talk about a "boundary POINT" anyways but just defined boundary as "the set E \ E◦ is called the boundary of E and is denoted ∂E"... Second, based on the definition you presented, I'm taking O to be any neighborhood of x and therefore by this definition, every neigh of x contains a point in E and E's complement. So yeah, you would be right... I'm just getting more and more confused I guess, I will have to bring it up for sure.

8. ## Re: Closure, Boundary, and Neighborhoods

The statement that you are trying to prove is essentially that two or more of the common equivalent definitions of a boundary of a set are actually equivalent. See Boundary (topology) - Wikipedia, the free encyclopedia