# A set which is the closure of its interior points

I am trying to give a sufficient condition for a set in $$\mathbb{R}^n$$ which is the closure of its interior points. A priori, such a set has to be a closed set. A closed set in general is not the closure of its interior point. A trivial counter example is a set with empty interior, e.g the segment $$[0,1]\times \{0\}$$ in $$\mathbb{R}^2$$. So the next candidate is one with non empty interior. One can come up with a counter example is a sphere with hair, e.g the union of $$\{x^2+y^2 \le 1\} \cup [1,2]\times \{0\}$$ in $$\mathbb{R}^2$$. So to avoid these kinds of counter example, I propose the following question ("conjecture" may be?)

Question 1: Let $$K$$ be a closed connected subset of $$\mathbb{R}^n , n \ge 2$$ with nonempty interior $$Int(K)$$. Suppose that the boundary $$\partial K$$ is a connected submanifold of $$\mathbb{R}^n$$. Then $$K = \overline{Int(K)}$$.

Note: By submanifold of $$\mathbb{R}^n$$ I mean an embedded submanifold of $$\mathbb{R}^n$$. In particular, $$\partial K$$ is orientable.

The connectivity assumption on $$K$$ is to avoid the case that $$K$$ has an isolated point. I am not sure that I need the connectivity of the boundary $$\partial K$$. The reason I put it there because the fact that $$K$$ is connected does not implies that $$\partial K$$ is connected. This is due to this question Connectedness of the boundary which asserts that (if I understood correctly) a connected set $$K$$ has connected boundary if and only if the complement $$K^c= \mathbb{R}^n \setminus K$$ is connected. Again, the counter example to this situation (connected set with non-connected boundary) does not satisfy the assumption that the boundary itself is a submanifold of $$\mathbb{R}^n$$. But even with this question I don't know how to tackle yet so just assume that $$\partial K$$ is a connected submanifold of $$\mathbb{R}^n$$.

My attempt: I am able to give an argument in the case when $$K$$ is compact. My argument goes as follows:

1. Since $$K$$ connected, $$Int(K)$$ has no isolated point and $$\overline{Int(K)} = Int(K) \cup \partial Int(K)$$. Note that $$K = Int(K) \cup \partial K$$ thus $$\partial Int(K) \subset \partial K$$.
2. Since $$K$$ is compact, $$\partial K$$ is a compact, connected submanifold of $$\mathbb{R}^n$$. Denote by $$k$$ the codimension of $$\partial K$$. It is enough to prove that $$\partial Int(K)= \partial K$$.
3. If $$k \ge 2$$: consider a point $$x \in \partial Int(K)$$ and a local chart $$(U,\varphi)$$ of $$\mathbb{R}^n$$ so that $$\varphi(U) =\mathbb{R}^n$$ and $$\varphi(U \cap \partial K) = \mathbb{R}^{n-k} \times \{0\}$$ Since $$k \ge 2$$, it follows that $$U \setminus \partial K$$ is connected. On one hand, since $$x \in \partial Int(K)$$, $$U \cap Int(K) \neq \emptyset$$. On another hand, we have a decomposition $$U \setminus \partial K = (U \cap Int(K)) \cup (U \cap K^c)$$ of $$U$$ as two disjoint open set. Thus the connectivity of $$U \setminus \partial K$$ implies that $$U \cap K^c = \emptyset$$, i.e. $$U \subset K$$. Since $$U$$ open, $$U \subset Int(K)$$ which yields a contradiction since $$U$$ contains a boundary point of $$Int(K)$$.
4. We deduce that $$\partial K$$ is a compact hypersurface of $$\mathbb{R}^n$$ which is orientable. By

"Lima, Elon L. "The Jordan-Brouwer separation theorem for smooth hypersurfaces." The American Mathematical Monthly 95.1 (1988): 39-42.",

I can deduce that $$\mathbb{R}^n \setminus \partial K$$ has two connected component $$U_1,U_2$$ whose boundaries are exactly $$\partial K$$. As $$K^c$$ is connected, without loss of generality, I assume that $$K^c \subset U_1$$. It is enough to prove that $$Int(K) \subset U_2$$, hence $$Int(K) = U_2$$ and it follows that $$\partial U_2 = \partial Int(K) = \partial K$$.

Suppose that $$Int(K) \setminus U_2 \neq \emptyset$$, thus $$Int(K) \cap U_1 \neq \emptyset$$. Note that $$U_2 = (U_2 \cap Int(K)) \cup (U_2 \cap \partial Int(K)) \cup (U_2 \cap \overline{Int(K)}^c).$$ Hence by connectivity of $$U_2$$, we can deduce that $$U_2 \cap \partial Int(K) \neq \emptyset$$ which is a contradiction since $$\partial Int(K) \subset Int(K) \cap U_2 =\emptyset$$.

(This is in fact an argument for a general fact: If $$U$$ is an open connected subset of $$\mathbb{R}^n$$ then for every $$V \subset U$$ and $$V \neq U$$, we have $$\partial V \cap U \neq \emptyset$$.)

Is my argument correct? I am really grateful if anyone can come up with a proof or a counter example of Question 1, with or without the hypothesis of connectivity of $$\partial K$$. I prefer arguing by not so complicated machinery (not homology, cohomology, Mayer-Viertoris sequence, etc..) Thank you.

• The set $[0,1]$ is the closure of its interior points. The interior is $(0,1)$ and the closure of that set is $[0,1]$. I don't understand the statement you make in the beginning. – Tony S.F. Jun 19 at 12:17
• "A closed set in general is never the closure of its interior point." is certainly false. – Randall Jun 19 at 12:54
• @TonyS.F. I edited. I refer to the segment $[0,1]\times \{0\}$ in $\mathbb{R}^2$. – Curiosity Jun 19 at 14:27
• FYI, a set that is the closure of its interior points, is called a "regular closed set"; their complements are called "regular open sets" (and they are the interior of their closures). They occur often in general topology.. – Henno Brandsma Jun 19 at 21:55
• @HennoBrandsma Thank you for the keyword. But after a while googling, I still don't find any result relating this regularity and the geometry of the boundary. In fact, I found on MO that convex set in topological vector spaces is indeed a regular closed set but it is still not a satisfied conclusion for me. – Curiosity Jun 23 at 9:50