Topological spaces · Definition

Boundary of a set

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Before you start you need to know

The boundary collects the points where a set and its complement meet: the «edge» separating inside from outside.

Definition

The boundary \(\partial A\) of a subset \(A\) of a topological space \(X\) is the set of points of the closure of \(A\) that do not belong to its interior.

\[\partial A=\overline{A}\setminus\operatorname{int}A\]

A point lies in \(\partial A\) exactly when every open set containing it meets both \(A\) and \(X\setminus A\). Hence the equivalent formula \(\partial A=\overline{A}\cap\overline{X\setminus A}\), which shows \(\partial A=\partial(X\setminus A)\). In usual \(\mathbb{R}\), \(\partial[0,1]=\partial(0,1)=\{0,1\}\) and \(\partial\mathbb{Q}=\mathbb{R}\).

The boundary is always a closed set and is disjoint from the interior of \(A\); the space splits into three disjoint parts, \(X=\operatorname{int}A\cup\partial A\cup\operatorname{int}(X\setminus A)\). A set is open if and only if it contains no point of its boundary, and closed if and only if it contains all of it. Therefore \(\partial A=\varnothing\) exactly when \(A\) is both open and closed.

Beware: the topological boundary is not what geometric intuition suggests. In \(\mathbb{R}\), the set \((0,1)\cup(1,2)\) has boundary \(\{0,1,2\}\), since the point \(1\) touches both sides. In the discrete topology every boundary is empty, and in the trivial one every nonempty proper subset has boundary \(X\).

Think about it

Can \(\partial A=X\) with \(A\) open and nonempty?

No: if \(A\) is open, \(\partial A=\overline{A}\setminus A\) contains no point of \(A\), so \(\partial A=X\) would force \(A=\varnothing\).

Describe \(\partial\mathbb{Z}\) in usual \(\mathbb{R}\).

\(\mathbb{Z}\): it is closed with empty interior, so its boundary is all of \(\mathbb{Z}\).

Exercises 10

References

  • James R. Munkres, Topology, 2nd edition, Pearson, 2000. §17
  • Stephen Willard, General Topology, Dover reprint, Dover Publications, 2004. §3