# Relative Boundaries in General Topology

by Tom Mattson
Tags: boundaries, relative, topology
 Emeritus Sci Advisor PF Gold P: 5,532 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.