# A closed set which is the closure of its interior points

This is a cross-post to MSE to this question

https://math.stackexchange.com/questions/3267497/a-set-which-is-the-closure-of-its-interior-points

There is a related question in MO: Closure of the interior of a convex set in a topological vector space. Unfortunately, in my problem, I don't have the convexity and yet the convexity and the smoothness of the boundary are fairly different assumptions (a priori, closed set with smooth boundary can be non-convex). Here is the question.

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 https://math.stackexchange.com/questions/170337/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 $$\partial K$$. Thank you

• I do not have time to do much more than say that your approach seems ok, but might be simplified. Consider the set of points of the boundary with a neighborhood homeomorphic to a small ball cut out by a hyperplane with K on one side and its complement in the other. This set is clearly open and using the assumption it should be easy to prove it to be closed. Then it is all of the boundary, and thus all boundary points are limit of interior points. Jun 23, 2019 at 16:39
• @BenoîtKloeckner so you're saying that the question should be true even without the compactness assumption ? Jun 23, 2019 at 20:39
• yes, I do not feel the compactness assumption should be needed (I also feel that connexity of K alone should be sufficient, by showing that every component of its boundary then meets the set of "nice" points). Jun 24, 2019 at 7:48