Yes, all of that is true. Do you have a question?
A=(0,2) B=(1,3)
A and B are open.
The boundary points of A are 0 and 2, neither of which belong to A
The boundary points of B are 1 and 3, neither of which belong to B
A^B = (1,2)
A=(0,2) B=[1,3)
The boundary point 1 of B belongs to B
A^B = [1,2), The interior is (1,2)
The boundary point 1 in B is also a point of A so it belongs to the intersection A^B, where it is also a boundary point because to the right of 1 are points common to A and B and to the left of 1 are only points of A.
A=(0,2] B = [1,3)
A^B = [1,2]. The interior is (1,2)
P is an interior point of A if it has a neighborhood entirely in A .
Glad you asked.
I was responding to a question in another thread which was closed before I had a chance to respond.
But I do have a question.
Someone objected to my using "boundary," and I note Rudin dosen't use it , Rosenlicht relegates it to a problem, Shilov uses it in an advanced context at the end of the book, and Taylor uses it.
Rosenlicht and Taylor define it as:
(overline is closure, c is complement)
Which isn't very pleasant.
The question:
How would you rephrase the original post without using the term "boundary."
I guess you could write "intersection of the closure with the closure of the complement" instead of "boundary" in every case. But that would be silly. What's bothering you here, the set we call the boundary of S, or the use of the word "boundary" to name it?
Sorry, I was mistaken about the thread being closed, and I probably misunderstood the objection to use of "boundary."
That leaves only my curiousity about Rudin not using "boundary," (it's not in index of '64 edition) which is a question I should really ask Rudin, rather than post it.