Relative Boundaries in General Topology

by Tom Mattson
Tags: boundaries, relative, topology
Tom Mattson
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Apr13-05, 08:28 AM
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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:


Before I move on, can someone tell me if that is correct? Thanks.
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HallsofIvy is offline
Apr13-05, 09:27 AM
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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.
matt grime
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Apr13-05, 09:33 AM
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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.

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