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Definition of closure from here

Boundary (Topology): The points in the closure of a set which are not in the interior of that set.

Definition of Exterior from here

Let $T$ be a topological space. Let $H \subseteq T$. The exterior of $H$ is the complement of the closure of $H$ in $T$. Alternatively, the exterior of $H$ is the interior of the complement of $H$ in $T$.

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In general, if $T$ is a topological space, then any set $H$ in $T$ partitions $T$ into three disjoint sets: the interior of $H$, the boundary of $H$, and the exterior of $H$. So the exterior only equals the boundary when both are empty, that is when all of $T$ is in the interior of $H$, i.e. when $H$=$T$.

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you mean the first definition is wrong? Should it be like "The points in the closure of a set which are not in the interior of that set and not in the exterior of that set."? –  hhh Feb 9 '11 at 23:55
    
and is the second definition also wrong? Should "The exterior of H is the complement of the closure of H in T." be "The exterior of H is the complement of the closure of H in T and the complement of the interior of H in T"? –  hhh Feb 9 '11 at 23:59
    
No, both definitions are correct. The boundary is disjoint from the interior by the definition of the boundary. The exterior is disjoint from the boundary because it's disjoint from the closure, which contains the boundary. The exterior is disjoint from the interior because the interior is contained in $H$ by the definition of the interior, and hence is contained in the closure of $H$ by the definition of the closure. –  Chris Eagle Feb 10 '11 at 0:16
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@hhh: Well, for one thing, above the boundary is defined in terms of the closure (it's the closure minus the interior). A direct definition would look like this: a point $x \in T$ lies in the boundary of $H$ provided that every open neighborhood of $x$ has nonempty intersection with $H$, but $x$ has no open neighborhood entirely contained in $H$. –  Justin Campbell Feb 10 '11 at 0:52
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@hhh If $A$ is a subset, and $T$ is the whole space, $T = cl(A) \cup ext(A) = int(A) \cup bd(A) \cup ext(A)$. $cl(A) = int(A) \cup bd(A)$, where all unions are disjoint. –  Henno Brandsma Feb 10 '11 at 18:47

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