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## Relative Boundaries in General Topology

Hi,

I was trying to help a student with an assignment in topology when I was stumped by a symbol that I had not seen before. Here's the problem.

a.) Let $(X,\square)$ be a topological space with $A\subseteq X$ and $U\subseteq A$. Prove that $Bd_A(U)\subseteq A\cap Bd_X(U)$.

The first thing that has got me stumped here is the subscripted boundaries. I have never seen this before, but I tried to reason it out as follows. The "ordinary" boundary of a set A is $Bd(A)=[ext(A)]^c\cap[int(A)]^c$, the intersection of all the points that are neither in the exterior of $A$ nor in the interior of $A$. The first problem is how to relate the boundary of a set to a second set (and thus introduce the subscripts), so I went back to the definition of the complement of a set $A$, which is the difference $\mathbb{U}-A$, where $\mathbb{U}$ is the universal set. This led me to conjecture that:

$Bd_A(U)=[A-ext(U)]\cap[A-int(U)]$
$Bd_X(U)=[X-ext(U)]\cap[X-int(U)]$

Before I move on, can someone tell me if that is correct? Thanks.

 Recognitions: Gold Member Science Advisor Staff Emeritus Yes, [tex]Bd_A(U) is the "boundary of U relative to A" which means the boundary of U in the "relative topology". If X is a topological space and A is a subset of A, then all open sets "relative to A" are open sets in X, intersect A. The interior of U "relative to A" is simply the interior of U (relative to X) intersect A and exterior of U "relative to A" is the exterior of U (relative to X) intersect A. Since boundary points of U are points of U that are neither interior nor exterior to U, but points in the boundary of U "relative to A" must be in A, they must be neither interior nor exterior "relative to X": that is must be in the boundary of U "relative to X": in boundary of U relative to X intersect A, just as the formla says.
 Recognitions: Homework Help Science Advisor http://www.ornl.gov/sci/ortep/topology/defs.txt contains some definitions. As an example, consider [0,1] as a subset of R: it's boundary os {0,1}, but in the subspace topology its boundary is empty. If you use the definition there it should become clear how to prove this result.