Thread: Interior and Closure of set proof

Hi There,

I'm currently studying for a deferred final, and understand the general concept of this question, but was hoping I could get a clear answer from someone knowledgable in order to reinforce my knowledge.

The Question: Let S be a set in R^n. Explain what is meant by the interior S^o of S. If the closure of S^o is T = [S^o with a bar above it, I'm horrible with LaTeX], and S is a closed set, prove that T is a subset of S. Give an example of a closed set in S where the inclusion is strict

My attempted solution: Well, I don't totally have a solution. Here's what I know though. The interior of a set is all points in a set that are not on the boundary. That is, all points in S that for any delta > 0 there exists a neighborhood N such that N is an open set in S. I feel my definition is weak, and I don't totally understand what a closure is and thus cannot go much further with this

If someone could help me with this question I'd greatly appreciate it.

The closure of S is the smallest closed set containing S, equivalently, the intersection of all closed sets containing S. Another equivalent definition is the union of S and the limit points of S. So, since a closed set contains all its limit points, you have the first result.

As for the strict inclusion, there are closed sets that have no interior. What happens with such a set?

Thanks tinyboss! A closed set which has no interior...wouldn't that set consist entirely of boundary points? I apologize for my absent mindedness but I don't totally see the relevance ><

Doesn't the second part of the question ask you to find a closed set that properly contains the closure of its interior?

Hi Tinyboss, thanks for your patience and help. I'm still having difficulty with this question. Here is my answer so far. A nudge in the right direction from this would be sufficient enough, as I realize I've taken up more time of yours than I should have. I am struggling a lot with the concept of closure, open and closed sets. Anyways, here it is:

The interior of a set S is the set of all points P such that for some (delta) > 0 there exists an open neighborhood N of radius (delta) centered at P. Suppose S is a closed set, and denote it's interior by I, and it's interior's closure by CL(I). By definition, CL(I) is the smallest closed set containing I, and I is an open subset of S. Since the question defines CL(I) as the complement of I, and CL(I) is a closed set, then CL(I) must subset S. The inclusion is is strict when S is a closed set whose interior is empty.

I don't think my proof is rigorous enough nor do I really understand it, it's mostly from what you've said and my textbook. Intuitively, wouldn't the union of CL(I) and I be equal to S? And in which case, wouldn't it's inclusion be strict iff I is non-empty?

a good set to test your intuition of this is the following set in R^2:

S = {(x,y) in R^2: x^2 + y^2 = 1}.

compare this with the following set:

T = {(x,y) in R^2: x^2 + y^2 ≤ 1}

and the set

U = {(x,y) in R^2: x^2 + y^2 < 1}.

two of these sets are closed (in the usual metric topology for R^2). one of them has the property that cl(int(A)) ⊂ A. which one?

Deveno, thanks for your help and useful example, it really helped me. Tinyboss, thanks for your help as well.

Managed to solve this to my own satisfaction.