Inside, Outside, and the Boundary Between
For a subset \(A\) of a topological space, points can be securely inside \(A\), securely outside it, or unable to be separated from both sides by an open neighborhood. The exterior describes the second possibility. The boundary consists of the points in the third: every open neighborhood of such a point meets both \(A\) and its complement. These descriptions depend on the topology, not only on the underlying set.
Throughout, let \((X,\mathcal{T})\) be a topological space and \(A\subseteq X\). An open neighborhood of \(x\in X\) is an open set \(U\in\mathcal{T}\) with \(x\in U\). Recall that \(\operatorname{int}A\) is the union of all open subsets of \(X\) contained in \(A\). Thus \(x\in\operatorname{int}A\) exactly when some open neighborhood of \(x\) is contained in \(A\).
The word “both” is essential. A neighborhood must contain at least one point of \(A\) and at least one point not in \(A\); it need not contain points of both sets other than \(x\). In particular, whether \(x\) itself belongs to \(A\) does not decide whether \(x\) is a boundary point.
Recognizing Exterior and Boundary Points
Proof. By definition, \(\operatorname{ext}A=\operatorname{int}(X\setminus A)\). Membership in the interior means that there is an open set \(U\) with \(x\in U\subseteq X\setminus A\). This is exactly an open neighborhood of \(x\) disjoint from \(A\). Conversely, if such a neighborhood \(U\) exists, then \(U\subseteq X\setminus A\), so \(x\in\operatorname{int}(X\setminus A)=\operatorname{ext}A\). \(\square\)
Proof. The first condition is the definition of boundary. For the equivalent description, suppose first that \(x\in\partial A\). If \(x\in\operatorname{int}A\), then there is an open neighborhood \(U\) with \(U\subseteq A\). This neighborhood does not intersect \(X\setminus A\), contradicting the boundary condition. Therefore \(x\notin\operatorname{int}A\). If \(x\in\operatorname{ext}A\), the exterior proposition gives an open neighborhood of \(x\) disjoint from \(A\), again contradicting the boundary condition. Thus \(x\notin\operatorname{ext}A\).
Conversely, suppose \(x\notin\operatorname{int}A\) and \(x\notin\operatorname{ext}A\). If an open neighborhood \(U\) of \(x\) missed \(A\), then \(U\subseteq X\setminus A\), so the exterior proposition would imply \(x\in\operatorname{ext}A\). Hence every open neighborhood of \(x\) intersects \(A\). If an open neighborhood \(V\) missed \(X\setminus A\), then \(V\subseteq A\), which would imply \(x\in\operatorname{int}A\). Hence every open neighborhood of \(x\) also intersects \(X\setminus A\). Both requirements in the definition of boundary hold, so \(x\in\partial A\). \(\square\)
This test gives a practical way to classify points. A point is in the interior if some neighborhood stays entirely in \(A\); it is in the exterior if some neighborhood stays entirely outside \(A\); it is on the boundary if neither kind of neighborhood exists. This does not say that every point must be in exactly one of those three sets: that conclusion would require proving an additional set identity. For present purposes, the neighborhood tests give the classification directly.
Worked Examples
Worked Example: An Interval in the Real Line
Let \(\mathbb{R}\) have its standard topology and take \(A=[0,1)\). Find \(\operatorname{int}A\), \(\operatorname{ext}A\), and \(\partial A\).
Interior. Every \(x\in(0,1)\) has an open interval around it contained in \([0,1)\): choose \(r=\min(x,1-x)/2>0\), so \((x-r,x+r)\subseteq(0,1)\subseteq A\). Neither endpoint is interior. Every open interval around \(0\) contains negative numbers, and every open interval around \(1\) contains numbers greater than or equal to \(1\); those points are not in \(A\). Points outside \([0,1]\) are not interior because they are not in \(A\). Thus \(\operatorname{int}A=(0,1)\).
Exterior. The complement is \(X\setminus A=(-\infty,0)\cup[1,\infty)\). Each \(x<0\) has an open interval contained in \((-\infty,0)\), and each \(x>1\) has an open interval contained in \((1,\infty)\). The point \(1\) is not in the interior of the complement, because every open interval around \(1\) includes points less than \(1\) from \(A\). Therefore \(\operatorname{ext}A=(-\infty,0)\cup(1,\infty)\).
Boundary. At \(0\), every open interval around \(0\) contains points of \(A\), such as a sufficiently small positive number, and points of \(X\setminus A\), such as a negative number. Thus \(0\in\partial A\). At \(1\), every open interval around \(1\) contains points of \(A\) just below \(1\), and points of \(X\setminus A\) at or above \(1\), so \(1\in\partial A\).
If \(x\in(0,1)\), an interval around \(x\) small enough to stay in \((0,1)\) misses the complement, so \(x\) is not a boundary point. If \(x<0\) or \(x>1\), an interval around \(x\) small enough to stay outside \([0,1]\) misses \(A\), so \(x\) is not a boundary point. Hence \(\partial A=\{0,1\}\).
Check: \(0\in A\), whereas \(1\notin A\), yet both are boundary points. The result is consistent with the neighborhood test: at each endpoint, every open neighborhood meets both \(A\) and its complement.
Worked Example: The Sierpinski Topology
Let \(X=\{a,b\}\) with topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\), and let \(A=\{a\}\). Compute the exterior and boundary.
The complement is \(X\setminus A=\{b\}\). The only open subset of \(X\) contained in \(\{b\}\) is \(\varnothing\), because \(\{b\}\notin\mathcal{T}\). Therefore \(\operatorname{int}(X\setminus A)=\varnothing\), and \(\operatorname{ext}A=\varnothing\).
To test \(a\), note that \(\{a\}\) is an open neighborhood of \(a\) and does not meet \(X\setminus A=\{b\}\). Therefore \(a\notin\partial A\). The only open neighborhood of \(b\) is \(X\): the open sets \(\varnothing\) and \(\{a\}\) do not contain \(b\). The set \(X\) meets \(A\) at \(a\) and meets \(X\setminus A\) at \(b\). Therefore \(b\in\partial A\), and \(\partial A=\{b\}\).
Check: The boundary point \(b\) is outside \(A\). Its only open neighborhood is forced to contain both points, which is why it meets both sides of the set.
Worked Example: One Set, Two Topologies
Let \(X=\{1,2,3,4\}\) and \(A=\{1,2\}\). First give \(X\) the discrete topology \(\mathcal{T}_{d}=\mathcal{P}(X)\), in which every subset is open. Then give \(X\) the indiscrete topology \(\mathcal{T}_{i}=\{\varnothing,X\}\). Find the exterior and boundary in each case.
Discrete topology. The complement of \(A\) is \(\{3,4\}\), which is open, so its interior is itself and \(\operatorname{ext}A=\{3,4\}\). For each \(x\in X\), the singleton \(\{x\}\) is an open neighborhood. If \(x\in A\), this neighborhood misses \(X\setminus A\); if \(x\notin A\), it misses \(A\). No point has every neighborhood meeting both sets, so \(\partial A=\varnothing\).
Indiscrete topology. The only open sets are \(\varnothing\) and \(X\). Since \(\{3,4\}\) is not open, its interior is \(\varnothing\), and \(\operatorname{ext}A=\varnothing\). For any \(x\in X\), the only open neighborhood is \(X\). It meets \(A\), since \(1\in A\), and it meets the complement, since \(3\in X\setminus A\). Thus every point is a boundary point, so \(\partial A=X\).
Check: The underlying set \(A\) and its complement are unchanged, but the boundary differs: it is empty in the discrete topology and all of \(X\) in the indiscrete topology. The open neighborhoods, and therefore the boundary test, depend on the topology.
Common Pitfalls and Why the Distinctions Matter
A common mistake is to assume that the boundary must lie outside the set. In the interval example, \(0\) is both in \(A=[0,1)\) and on its boundary. The Sierpinski example has a boundary point outside \(A\). A boundary is therefore not defined by membership or nonmembership in \(A\); it is defined by what happens in every open neighborhood.
Another tempting but incorrect shortcut is to identify the boundary with \(A\setminus\operatorname{int}A\). That set only records points of \(A\) that are not interior points. For the real-line example, this calculation gives \(\,[0,1)\setminus(0,1)=\{0\}\,\), so it misses the boundary point \(1\). The correction is to test both sides: every neighborhood of \(1\) meets \(A\) just below \(1\) and meets \(X\setminus A\) at and above \(1\). Therefore \(1\) belongs to the boundary even though it is not in \(A\).
The distinction between exterior and boundary is useful whenever a point’s local behavior matters. An exterior point has at least one neighborhood entirely outside the set; a boundary point has no neighborhood that avoids either side. This local test works in any topological space, including spaces where neighborhoods do not look like intervals or metric balls.
| Question about \(x\) | Neighborhood test |
|---|---|
| Is \(x\) an interior point of \(A\)? | Some open neighborhood of \(x\) is contained in \(A\). |
| Is \(x\) an exterior point of \(A\)? | Some open neighborhood of \(x\) is contained in \(X\setminus A\). |
| Is \(x\) a boundary point of \(A\)? | Every open neighborhood of \(x\) meets both \(A\) and \(X\setminus A\). |
The definitions also handle the extreme cases without modification. If \(A=\varnothing\), no neighborhood meets \(A\), so \(\partial A=\varnothing\), and \(\operatorname{ext}A=\operatorname{int}X=X\). If \(A=X\), no neighborhood meets \(X\setminus A\), so \(\partial A=\varnothing\), and \(\operatorname{ext}A=\operatorname{int}\varnothing=\varnothing\). These checks show why the condition “meets both” must be applied literally.
Check Your Understanding
Use open neighborhoods in the stated topology to justify each answer.
- In \(\mathbb{R}\) with the standard topology, find the exterior and boundary of \(A=(0,1)\).
- In the discrete topology on any set \(X\), show that every subset \(A\subseteq X\) has empty boundary.
- In the indiscrete topology on a nonempty set \(X\), find the boundary of a nonempty proper subset \(A\subset X\).
- Give an example of a boundary point that belongs to \(A\), and verify the neighborhood condition.
- Explain why a point with an open neighborhood disjoint from \(A\) cannot belong to \(\partial A\).