# Topological boundary identity

Let $X$ be a topological space and $N$ a subset of $X\newcommand{\int}{\operatorname{int}}$. since $\bar N$ is closed, then $\bar{ \bar N}=\bar N$. Knowing that by definition $\bar A=\int(A)\sqcup \partial A$, the identity $\bar{ \bar N}=\bar N$ is equivalent to $\int(N)\sqcup \partial N=\int(\bar N)\sqcup \partial \bar N$. It is known that $\partial \bar N\subset \partial N$ and $\int(N)\subset \int(\bar N)$ this means that when we "close" $N$ some of the boundary poins become interior points in the closure. Now suppose that $N$ is open, then $\partial \bar N=\partial N$ and $\int(\bar N)=\int(N)=N$ is my reasoning correct? thank you for your clarification!!

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To whom it may concern, I have added an \int command to the LaTeX code in the question, which means it is also available in comments and other answers on the page. E.g. $\int(\mathbb Q)=\varnothing$. – Asaf Karagila Oct 5 '12 at 22:42

## 1 Answer

$\newcommand{\cl}{\operatorname{cl}}\newcommand{\bdry}{\operatorname{bdry}}$Consider the open set $N=(0,1)\cup(1,2)$ in $\Bbb R$: $\cl N=[0,2]$, and $\int\cl N=(0,2)\ne N$. Moreover, $\bdry\cl N=\{0,2\}\ne\{0,1,2\}=\bdry N$.

An open set $U$ such that $U=\int\cl U$ is called a regular open set. Every $T_3$ (regular + $T_1$) space has a base of regular open sets, and the regular open sets of any topological space form a complete Boolean algebra that has applications in set theory.

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What sort of applications? – Asaf Karagila Oct 5 '12 at 23:20
@Asaf: Forcing via Boolean-valued models. It’s a matter of opinion how important that is; I may remove the adjective. – Brian M. Scott Oct 5 '12 at 23:26
Oh, I think that Boolean-valued models is a very important use. I had a discussion with some friends about it. While it's not very usable in everyday forcing arguments, it does give a lot of insight on the inner workings of forcing. E.g. every countable forcing is Cohen forcing; every intermediate model of ZFC between the ground is a generic extension of the ground, and the full extension is generic over it; and several other theorems of this manner. I suppose that you mean the completion of the poset to a Boolean algebra, right? – Asaf Karagila Oct 5 '12 at 23:29
@Asaf: Yep, exactly that. – Brian M. Scott Oct 5 '12 at 23:30