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.

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    $\begingroup$ 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. $\endgroup$ – Tony S.F. Jun 19 at 12:17
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    $\begingroup$ "A closed set in general is never the closure of its interior point." is certainly false. $\endgroup$ – Randall Jun 19 at 12:54
  • $\begingroup$ @TonyS.F. I edited. I refer to the segment $[0,1]\times \{0\}$ in $\mathbb{R}^2$. $\endgroup$ – Curiosity Jun 19 at 14:27
  • $\begingroup$ 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.. $\endgroup$ – Henno Brandsma Jun 19 at 21:55
  • $\begingroup$ @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. $\endgroup$ – Curiosity Jun 23 at 9:50

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