# Closure, interior and boundary point

Prove that a point belongs to $A^{-}$ if and only if it is either an interior or a boundary point of $A$; where $A^-$ is the closure.

To be an interior point: A point $x\in \mathbb{R}$ is the interior point of a set $A\subset \mathbb{R}$ if there is a neighborhood of $x$ which is entirely contained in $A$. A boundary point follows, which is the set of points with the property that every open set containing the point intersects the interior of $A$ and the interior of $A^c$. To be a closure, which is the intersection of all closed sets containing $A$; to be closed means there is a boundary point. But, how will it follow that it can also be an interior point?

• Part of the problem may be this: what do you mean by the point $x$ is contained inside and outside the set when you define boundary point?? I usually define it to be the set of points with the property that every open set containing the point intersects the interior of $A$ and the interior of $A^c$. – Clayton Feb 14 '13 at 2:58
• @Clayton Yes, you are right. I will edit to make it more clear. – Lays Feb 14 '13 at 3:01
• Can you also clarify what you mean when you say "to be closed means there is a boundary point"? A set $A$ is closed if and only if its complement $A^c$ is open. – Shaun Ault Feb 14 '13 at 3:07
• @Lays: When you say "to be closed means there is a boundary point" is false. Take the whole space $\Bbb R$. Then it is closed and also open, thus there is no boundary point. – Clayton Feb 14 '13 at 3:10
• If you are trying to have a line over $A$ the commands available are: $\bar{A}$ (\bar{A}) and $\overline{A}$ (\overline{A}). – Asaf Karagila Feb 14 '13 at 3:16

A boundary point of a set $A$ cannot be an interior point, but that’s not a problem: the theorem is that the closure of $A$ consists of those points that are in the interior of $A$ together with those that are in the boundary of $A$. In symbols,

$$\operatorname{cl}A=\operatorname{int}A\cup\operatorname{bdry}A\;,\tag{1}$$

and in fact the two sets on the righthand side of $(1)$ are disjoint. (By the way, a closed set need not have any boundary points at all: in $\Bbb R$ the only examples of this phenomenon are the closed sets $\varnothing$ and $\Bbb R$, but in more general topological spaces there can be many sets that are simultaneously open and closed and which therefore have empty boundary.)

You know that $x\in\operatorname{int}A$ if and only if $x$ has an open nbhd contained in $A$, and that $x\in\operatorname{bdry}A$ if and only if every open nbhd of $x$ intersects both $A$ and $\Bbb R\setminus A$. To prove $(1)$, I suggest that you first prove that

$\qquad\qquad\qquad\quad x\in\operatorname{cl}A$ if and only if for each open nbhd $V$ of $x$, $V\cap A\ne\varnothing$.

Once you have this, $(1)$ is pretty straightforward.

• Great, now I see what to do, thank you very much for your help! – Lays Feb 14 '13 at 6:07
• @Lays: You’re very welcome. – Brian M. Scott Feb 14 '13 at 6:08

First, being interior point or boundary point or exterior point are mutually exclusive possibilities. So, instead of showing your claim I will show any exterior point could not be an element of closure.Let $x$ be an exterior point of $A$, then $\exists V$ neighbourhood of $x$ so that $V\subseteq A^{c}$. Since every neighbourhood open, then $V^c$ should be close. Also, observe that $A\subseteq V^c$, but then $x\notin A^-$ since $A^-$ is the intersection of all closed set that contains $A$.