Decomposition of Topological Space Into Interior, Boundary and Exterior

Naturally, given the way that the interior, boundary and exterior of a set is defined in topological spaces, one may assume that something cannot be simultaneously in any two of these sets, and furthermore that they cover the entire space.

That is, if we consider some topological space \(X\):

and the interior of some set \(S\) within it:

then the interior of the complement will be disjoint from the interior of \(S\):

and the remaining space not in either set will be their common boundary:

This result formalises that idea.

Theorem

Given any topological space \((X, \mathcal{T})\) for any set \(S \subseteq X\), the following disjoint union holds:

\[ \text{int}\ S \sqcup \partial S \sqcup \text{int}\ S^c.\]
Proof

This result follows very cleanly from the definitions of \(x\) being an interior point, interior to the complement, or a boundary point. Specifically consider that:

  1. \(x \in \mathrm{int}\ S\) if and only if there exists a neighbourhood \(N\) of \(x\) such that:

    \(\)
    N \subseteq S.
    \(\)

  2. \(x \in \mathrm{int}\ S^c\) if and only if there exists a neighbourhood \(N\) of \(x\) such that:

    \(\)
    N \subseteq S^c.
    \(\)

  3. \(x \in \partial S\) if and only if for all neighbourhoods \(N\) of \(x\):

    \(\)
    N \cap S \neq \varnothing \quad \text{and} \quad N \cap S^c \neq \varnothing.
    \(\)

These cases are then clearly mutually exclusive, and cover all such possible values of \(x\). That is, either there is a neighbourhood around \(x\) contained in \(S\) (implying \(x \in S\)), there is a neighbourhood around \(x\) contained in \(S^c\) (implying \(x \in S^c\)) or there is no such neighbourhood contained in either set, and hence any neighbourhood around \(x\) overlaps with both.