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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$\begingroup$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.
$\endgroup$ 1 $\begingroup$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 $X\setminus A$.
$\endgroup$ 3 $\begingroup$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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