Open subset topology

In all but the last section of this wiki, the setting will be a general metric space $[X,d].$ Those readers who are not completely comfortable with abstract metric spaces may think of $X$ as being Rn,{\mathbb R}^n,Rn, where $n=2$ or $3$ for concreteness, and the distance function $d[x,y]$ as being the standard Euclidean distance between two points.

An open set in a metric space $[X,d]$ is a subset $U$ of $X$ with the following property: for any $x\in U,$ there is a real number $ϵ>0$ such that any point in $X$ that is a distance 0 such that $B[x,ϵ]$ is completely contained in $U.$

So the intuition is that an open set is a set for which any point in the set has a small "halo" around it that is completely contained in the set. The idea is that this halo fails to exist precisely when the point lies on the boundary of the set, so the condition that $U$ is open is the same as saying that it doesn't contain any of its boundary points. With the correct definition of boundary, this intuition becomes a theorem.

The boundary of a set $S$ inside a metric space $X$ is the set of points $s$ such that for any $ϵ>0,$ $B[s,ϵ]$ contains at least one point in $S$ and at least one point not in $S.$

A subset $U$ of a metric space is open if and only if it does not contain any of its boundary points.

It is clear that an open set $U$ cannot contain any of its boundary points since the halo condition would not apply to those points. On the other hand, if a set $U$ doesn't contain any of its boundary points, that is enough to show that it is open: for every point $x\in U,$ since $x$ is not a boundary point, that implies that there is some ball around $x$ that is either contained in $U$ or contained in the complement of $U.$ But every ball around $x$ contains at least one point in $U,$ namely $x$ itself, so it must be the former, and $x$ has a halo inside $U.$ □_\square□​