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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 [itex](X,\square)[/itex] be a topological space with [itex]A\subseteq X[/itex] and [itex]U\subseteq A[/itex]. Prove that [itex]Bd_A(U)\subseteq A\cap Bd_X(U)[/itex].

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 [itex]Bd(A)=[ext(A)]^c\cap[int(A)]^c[/itex], the intersection of all the points that are neither in the exterior of [itex]A[/itex] nor in the interior of [itex]A[/itex]. 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 [itex]A[/itex], which is the difference [itex]\mathbb{U}-A[/itex], where [itex]\mathbb{U}[/itex] is the universal set. This led me to conjecture that:

[itex]Bd_A(U)=[A-ext(U)]\cap[A-int(U)][/itex]

[itex]Bd_X(U)=[X-ext(U)]\cap[X-int(U)][/itex]

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

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# Relative Boundaries in General Topology

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