If a set $A\subset\mathbb{R}^k$ satisfying $B\subset A\subset\overline{B}$, then we call $A$ a $k$-dimensional standard rectangle.

Here $B$ is an open $k$-dimensional rectangle, that's to say,

$$B=(a_1,b_1)\times(a_2,b_2)\times\cdots\times(a_k,b_k)$$ We can see


Using my topology knowledge, I have proved that for a $k$-dimensional standard rectangle $A$, we have $\operatorname{Int}\,\overline{A}=\operatorname{Int}\,A$, and $\overline{\operatorname{Int}\,A}=\overline{A}$.

But if we let $P$ be a finite union of some $k$-dimensional standard rectangles, (I call it a "simple figure") I can only prove $\overline{\operatorname{Int}\,P}=\overline{P}$.

Question: I believe that $\operatorname{Int}\,\overline{P}=\operatorname{Int}\,P$ (or $\partial\overline{P}=\partial P$) is also correct, but I cannot give a strict proof. Any help would be appreciated.


That is actually untrue. Let us work in $\mathbb {R} $. Take $ P = ]0,1 [ \cup ]1,2 [$. This clearly is a union of open rectangles. Clearly, $\overline{P} = [0,2]$. Thus, $ {Int}\overline {P} = ]0,2 [ \ \neq ]0,1 [ \cup ]1,2 [ = {Int} P $.

  • $\begingroup$ Thank you very much. I have restricted myself to the high dimensional conditions, and ignored some simple examples. No wonder I cannot prove it. $\endgroup$ – painday May 6 '17 at 5:36

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