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?
Best Answer
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.