In Engelking (theorem 1.3.2 p.24) and Choquet (prop 6.6 p.17 and exercice 17 p.112), there are several facts about frontier (or boundary) in topological spaces, but I can't find counter-examples simple but not too trivial to some counter-intuitive results.
I guess it is because reasonning in the plane is too specific (metric space and separation…).
Let $X$ denoted a topological space. For instance, one surprising property is: $\bar{A}\backslash A \subset fr(A)$ without equality in general. A counter-example in $X=\mathbb{R}$ is $A=\mathbb{Q}$ ($\bar{\mathbb{Q}}\backslash \mathbb{Q}= \mathbb{R}\backslash \mathbb{Q}\varsubsetneq fr(\mathbb{Q})=\mathbb{R}$).
Other surprising fact is : $fr(\bar{A})\subset fr(A)$ without equality in general. (Same counter-example.)
In general, examples one can easily find involve dense subsets. For instance, a ball (open, closed, or neither the first nor the second) in the plane satisfies the equality of the inclusion above. I just can see it by calculating the sets, but what I want is to understand why it works in some case and not in others…
Can you help me, please ?
Is there explicit or exhaustive literature about that ?
Best Answer
Proof :
We define $fr(A)$ to be $\bar{A}\backslash int(A)$ where $int(A)$ is the interior of $A$ (the biggest open set contained in $A$). Hence we have $fr(A)=\bar{A}\backslash A$ if and only if $\bar{A}\backslash A=\bar{A}\backslash int(A)$.
By set-theoretic argument (using the fact that both $A$ and $int(A)$ are contained in $\bar{A}$) this happens if and only if $A=int(A)$. This happens if and only if $A$ is open (if $A$ is open then $A$ is the biggest open set contained in $A$, and if $A=int(A)$ then in particular $A$ is open).
Now $fr(\bar{A})=\bar{\bar{A}}\backslash int(\bar{A})=\bar{A}\backslash int(\bar{A})$ and $fr(A)=\bar{A}\backslash int(A)$. So both are equal if and only if $int(\bar{A})=int(A)$. I don't think there is a better caracterization of those last sets.