# Interior and Closure of set proof

• May 25th 2011, 04:33 PM
rtplol
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.
• May 25th 2011, 07:36 PM
Tinyboss
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?
• May 25th 2011, 07:49 PM
rtplol
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 ><
• May 25th 2011, 07:50 PM
Tinyboss
Doesn't the second part of the question ask you to find a closed set that properly contains the closure of its interior?
• May 28th 2011, 03:04 PM
rtplol
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?
• May 28th 2011, 04:18 PM
Deveno
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?
• May 28th 2011, 05:29 PM
rtplol
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.