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Let $E := \{ p \in \mathbb{Q} : 0 < p < \sqrt 2 \}$.

(1) $\sqrt 2$ is not an inner point of $E$ because $\forall \delta \in \mathbb{R} : \sqrt 2 \not \in \bigcup_{x \in I} B(x, \delta) \subset E \subset \mathbb{Q} \subset \mathbb{R}$ and $\sqrt 2$ is not an outer point because $\forall \delta \in \mathbb{R} : \sqrt 2 \not \in \bigcup_{x\in I_{2}} B(x, \delta) \subset E^{c} \subset \mathbb{Q} \subset \mathbb{R}$. So $\sqrt 2$ is a boundary point of $\mathbb{Q}$ in $\mathbb{R}$.

(2) Notice that $0 \in \mathbb{R}$ and $0 \in \mathbb{Q}$. Since $\forall \delta > 0 : 0 \in \bigcup_{x\in I_{3}} B(x, \delta) \subset E \subset \mathbb{Q} \subset \mathbb{R}$ so $0$ is an inner point. $0$ is also an outer point because $\forall \delta > 0 : 0 \in \bigcup_{x \in I_{4}} B(x, \delta) \subset E^{c} \subset \mathbb{Q} \subset \mathbb{R}$. No boundary point.

[conlclusion, investigating]

Can you declare the borders in some other numbers such as complex numbers or some other way? I think one cannot do it but is there some result about it?

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The boundary of $E=\{p\in\mathbb{Q}:0<p<\sqrt{2}\}\subseteq\mathbb{R}$ is $[0,\sqrt{2}]$ not the empty set. –  LostInMath Sep 2 '11 at 20:46
But I'm a bit confused over this: if we define a boundary point p of a subset S, as a point $p$ with $B(p,r); r>0$ intersects both $intS$ and $ExtS$, then , since $intS$ is empty, p could not be a boundary point, right? –  gary Sep 2 '11 at 21:00
@gary: By definition a boundary point of a subset $S\subseteq X$ is a point whose every neighborhood intersects both $S$ and $X\setminus S$ (not $\operatorname{int}(S)$ and $\operatorname{ext}(S)$). –  LostInMath Sep 2 '11 at 21:04
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4 Answers

up vote 4 down vote accepted

Associated with a subset $E$ in a topological space $X$ there are three subsets of $X$ which are pairwise disjoint and cover $X$. They are

  • the set of interior points of $E$ or $\operatorname{int}E$, which consists of the points having a neighborhood which is contained in $E$,
  • the set of exterior points of $E$ or $\operatorname{ext}E$, which consists of the points having a neighborhood which is contained in $X\setminus E$, and
  • the set of boundary points of $E$ or $\partial E$, which consists of the points whose every neighborhood intersects $E$ and $X\setminus E$.

Let's see what these sets are in the case of $E=\{p\in\mathbb{Q}:0<p<\sqrt{2}\}\subseteq\mathbb{R}$.

First take any point $x<0$. Then $(x-|x|,x+|x|)\subseteq(-\infty,0)\subseteq\mathbb{R}\setminus E$, which means $x$ is an exterior point of $E$.

Then take any point $x>\sqrt{2}$ and denote $r=x-\sqrt{2}>0$. Then $(x-r,x+r)\subseteq(\sqrt{2},\infty)\subseteq\mathbb{R}\setminus E$. Again $x$ is an exterior point of $E$.

Finally take any point $x\in[0,\sqrt{2}]$ and a neighborhood $U$ of $x$. The neighborhood $U$ contains an interval $(x-r,x+r)$ for some (small) $r>0$. Because $x$ is in $[0,\sqrt{2}]$, this interval $(x-r,x+r)$ intersects the interval $(0,\sqrt{2})$, and we know that if the intersection of two open intervals is nonempty, then it (being an open interval) contains both a rational and an irrational number. Furthermore the rational numbers in $(x-r,x+r)\cap(0,\sqrt{2})$ are in $E$ and the irrational numbers are in $\mathbb{R}\setminus E$. We have shown that every neighborhood of $x$ intersects both $E$ and $\mathbb{R}\setminus E$, which means $x$ is a boundary point of $E$.

The conclusion is that $\operatorname{int}E=\emptyset$, $\operatorname{ext}E=\mathbb{R}\setminus[0,\sqrt{2}]$ and $\partial E=[0,\sqrt{2}]$.

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The closure of $E$ in $\mathbb{R}$ is $[0,\sqrt 2]$ and the interior of $E$ in $\mathbb{R}$ is empty. So the boundary of $E$ in $\mathbb{R}$ is the closed interval $[0,\sqrt{2}]$.

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I find the last implication hard-to-understand bacause I cannot understand the quantifier between the last two statements. Is it right "The closure of $E$ in $R$ is $[0, \sqrt 2]$ so the interior of $E$ in $R$ is empty."? If yes, we have found only one of the three disjoint sets, namely the interior. You are apparently making a conclusion about the boundary without acknowledging the exterior. –  hhh Sep 3 '11 at 6:52
No. I said "$\bar{E}=[0,\sqrt{2}]$ and $E^\circ=\phi$", not "$\bar{E}=[0,\sqrt{2}] \Rightarrow E^\circ=\phi$". That $E^\circ=\phi$ should be clear, because $E$ contains no nonempty open ball in $\mathbb{R}$ as its proper subset. –  user1551 Sep 3 '11 at 9:00
the complement of open sets is closed. Exterior, interior and boundary are disjoint. Closure and interior do not need to be disjoint. Closure - Interior = Boundary. Yes, you are correct here. –  hhh Sep 3 '11 at 12:57
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I think the word you're looking for is boundary. The boundary of a set $A \subset X$ (in $X$) is defined as $\overline{A} \setminus A^\circ$ and it's sometimes written $\partial A$. In your example the boundary of $E$ in $\mathbb{Q}$ is $\{0\}$ as $(0,\sqrt{2})\cap \mathbb{Q}$ is open in $\mathbb{Q}$ and $E$ has $0$ as a limit point.

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yes but the question was about subsets of $Q$ in $R$, not about subsets of $Q$ in $Q$. –  hhh Sep 2 '11 at 21:19
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Edit: Here is a rewrite of my original reply, with corrections.

Every boundary set of an open or closed subset of $\mathbb Q$ in $\mathbb R$, with the boundary defined as $Cl(S)-Int(S)$, has empty interior in $\mathbb R$. In $\mathbb R^n$ in general, boundaries of open/closed sets--of any subset of $\mathbb R^n $-- do have empty interior, e.g., the boundary of the closed set {$z:|z|=1$} has empty interior in $\mathbb C$. Still, if the set is neither open nor closed in $\mathbb R^n $, the boundary is not necessarily empty; actually, the boundary of $\mathbb Q^n $ in $\mathbb R^n $ is the whole of $\mathbb R^n$; in general, the boundary of a set that is either open or closed in $\mathbb C$ has empty interior. I think there is a general characterization of subsets of $R^n$ that are boundaries of sets.

As "Lost in Math" posted (and proved; see below for his argument), this property of boundaries is true of open and of closed subsets in all topological spaces.

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The boundary of $\mathbb{Q}$ in $\mathbb{R}$ is $\mathbb{R}$, since every nonempty open subset of $\mathbb{R}$ contains a rational and an irrational number. –  LostInMath Sep 2 '11 at 20:57
Wow, another downvote : I understand the first downvote, but to also downvote me after I added a correction in the edit: why? I got the point and corrected, isn't that enough? Should I do penance or something? –  gary Sep 2 '11 at 21:27
Just to avoid any misunderstanding: I didn't downvote. But maybe the downvoters want to see the answer corrected since there still are some false claims. –  LostInMath Sep 2 '11 at 22:04
Thanks; I don't mind criticism if it is constructive. I rewrote and corrected. –  gary Sep 2 '11 at 22:21
gary: Perhaps it would be clearer if you placed the Edit at the beginning, or strike through the incorrect parts. I did not downvote either, but I was confused as I started reading your answer because of the false claims that remain. –  Jonas Meyer Sep 3 '11 at 5:16
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