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A point $p$ in a metric space $X$ is a boundary point of the set $A$, if any neighbourhood of $p$ has points of both $A$ and $X-A$.Prove that the set of all boundary points of $A$ is closed.

My attempt: By definition of an open set this means that for every $x$ in the boundary there is an open ball centred at $x$ contained in the boundary. An open ball is a neighbourhood of $x$, which implies it contains points of $A$ and $X - A$, which in turn implies there are points in both $A$ and $X - A$ that are in the boundary of $A$.

If $A$ is open, then pick any such point in $A$ that is also in the boundary. This point cannot be in $X - A$ by definition of set subtraction. Further, because $A$ is open there exists an open ball centred around this point contained in $A$. Again, an open ball is a neighbourhood, which means a neighborhood of this point does not contain points of $X - A$, implying it cannot be in the boundary, a contradiction. If A is closed then $X - A$ is open and a symmetric argument holds. Hence the boundary is closed.

Is my work correct?

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The first paragraph of the attempt makes it seem like you're trying to prove that the boundary $\partial A$ is open. Note too that $A$ may be neither open nor closed. ["A subspace is not a door."] –  Dylan Moreland Jun 12 '12 at 6:03
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What if $A$ is neither open or closed? –  William Jun 12 '12 at 6:03
    
Also, note that this result and many proofs of it will not require that $X$ be a metric space, just a topological space. –  Dylan Moreland Jun 12 '12 at 6:10

3 Answers 3

up vote 9 down vote accepted

The boundary of a set $A$ is defined as $\overline{A} \cap \overline{X - A}$. It is the intersection of two closed sets and hence is closed.

By the way your proof is not correct because you assumed that $A$ is either open or closed. There are sets like $(0,1] \subset \Bbb{R}$ in the usual topology that are neither open nor closed.

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He gives a definition in the first paragraph. Of course, $p$ being in the closure of $B$ is the same as "all neighborhoods of $p$ intersect $B$", so your definition is the same. I don't know if that's clear to the OP. Anyway, +1 for a succinct, correct answer. –  Dylan Moreland Jun 12 '12 at 6:06

Your very first statement simply isn’t true: there need not be any non-empty open set contained in the boundary of $A$. Suppose that $A=[0,1]$ in the space $\Bbb R$: the boundary of $A$ is the set $\{0,1\}$, which does not contain any non-empty open subset of $\Bbb R$.

I suggest that you try to show that $X\setminus\operatorname{bdry}A$ is open, from which it will follow at once that $\operatorname{bdry}A$ is closed. To do this, pick a point $x\in X\setminus\operatorname{bdry}A$, and show that some open neighborhood of $x$ is disjoint from $\operatorname{bdry}A$. You’ll need to consider two cases: if $x\in X\setminus\operatorname{bdry}A$, either $x$ has an open neighborhood disjoint from $A$, or $x$ has an open neighborhood disjoint from $X\setminus A$.

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This is exactly what I was going to write (your second paragraph). Albeit, slightly modified (perhaps it isn't right). What do you think of using that the complement of the boundary is the union of the interior and the exterior where the exterior is the complement of the closure? Then the complement of the boundary is the union of two open sets and hence open. –  Matt N. Jun 16 '12 at 6:14
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@Matt: That works fine, provided that one already knows that $\operatorname{bdry}A=\operatorname{int}A\cup(X\setminus\operatorname{cl}A)$. –  Brian M. Scott Jun 16 '12 at 6:19
    
I guess one would have to show that and one doesn't have the beauty of a one sentence short answer anymore : ) –  Matt N. Jun 16 '12 at 6:20

It appears that your proof is not correct since you only consider the case when $A$ is either open or closed.

A set is closed if and only if it contains all its limit points.

Suppose $(x_n)$ is sequence of boundary points of $A$ which converges to some point $x$. One seeks to show that $x$ is also a boundary point. Let $U$ be an open set containing $x$. By definition of being the limit of $(x_n)$, there exists a $N$ such that $x_N \in U$. Since $X_N$ is a boundary point and $U$ is a neighborhood of $x_N$, $U$ contains a point of $A$ and $X - A$. Since $U$ is arbitrary, $x$ is a boundary point. The set of boundary points of $A$ is a closed set.

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