What the Interior Measures
A set can contain points that are not surrounded by other points of the set. For example, the endpoints of a closed interval belong to the interval, but every open interval around either endpoint reaches outside it. The interior records exactly the points that do have an open neighborhood entirely inside the set.
Let \((X,\mathcal{T})\) be a topological space, where \(\mathcal{T}\) is its collection of open sets, and let \(A\subseteq X\). An open subset of \(A\) means a set \(U\in\mathcal{T}\) such that \(U\subseteq A\). There is always at least one such set, namely \(\varnothing\), including when \(A=\varnothing\).
The topology axiom that arbitrary unions of open sets are open applies to this union, even if there are infinitely many open subsets of \(A\). The definition also immediately places \(\operatorname{int}A\) inside \(A\): every set being united is a subset of \(A\). The important point is that the definition gathers all open subsets of \(A\), not just the largest interval or the largest open set one happens to notice.
The Largest Open Subset
Proof. By definition, \(\operatorname{int}A\) is a union of open sets, so it is open by the arbitrary-union axiom for a topology. Every set in that union is contained in \(A\), so their union is contained in \(A\). Now let \(V\) be any open set with \(V\subseteq A\). Then \(V\) is one of the open subsets included in the union defining \(\operatorname{int}A\). Hence \(V\subseteq\operatorname{int}A\). These conclusions also hold when \(A=\varnothing\): the only subset of \(A\) is \(\varnothing\), and \(\operatorname{int}\varnothing=\varnothing\). \(\square\)
This characterization gives a useful way to recognize the interior without listing every open subset separately. It also distinguishes the interior from the original set: points in \(A\) need not be in \(\operatorname{int}A\), because belonging to \(A\) does not guarantee that a whole open neighborhood stays in \(A\).
Proof. Suppose \(x\in\operatorname{int}A\). Since the interior is the union of the open subsets of \(A\), membership in that union means that \(x\) belongs to at least one such open set \(U\). This \(U\) is open, contains \(x\), and is contained in \(A\). Conversely, suppose \(U\) is an open neighborhood of \(x\) and \(U\subseteq A\). Then \(U\) is one of the open subsets in the union defining \(\operatorname{int}A\), so \(x\in U\subseteq\operatorname{int}A\). \(\square\)
In \(\mathbb{R}\) with its standard topology, open intervals form a basis: a set is open if each of its points lies in an open interval contained in the set. The neighborhood test therefore says that \(x\in\operatorname{int}A\) exactly when some interval \((x-\varepsilon,x+\varepsilon)\), with \(\varepsilon>0\), is contained in \(A\). This local test is the main practical tool for computing interiors in the real line.
Computing Interiors in \(\mathbb{R}\)
Worked Example: A Half-Open Interval and an Isolated Point
In \(\mathbb{R}\) with the standard topology, let \(A=[0,1)\cup\{2\}\). We determine which points have an open interval around them contained in \(A\).
If \(x\in(0,1)\), choose \(\varepsilon=\min(x,1-x)/2\). Both \(x-\varepsilon>0\) and \(x+\varepsilon<1\), so \((x-\varepsilon,x+\varepsilon)\subseteq(0,1)\subseteq A\). Thus every \(x\in(0,1)\) lies in \(\operatorname{int}A\).
The point \(0\) is not interior: every interval around \(0\) contains negative numbers, which are not in \(A\). The point \(1\) is not in \(A\), so it cannot be in the interior, which is a subset of \(A\). The point \(2\) belongs to \(A\), but no interval around it is contained in \(A\): for any \(\varepsilon>0\), the point \(2+\min(\varepsilon,1)/2\) lies within distance \(\varepsilon\) of \(2\) and is not in \(A\). These are all the points of \(A\) not already considered, and the interior cannot contain points outside \(A\).
Conclusion: \(\operatorname{int}A=(0,1)\). As a check, this is open in \(\mathbb{R}\), is contained in \(A\), and the preceding neighborhood test shows that none of the other points of \(A\) can be added while keeping an open subset of \(A\).
Worked Example: The Rationals and the Irrationals
In the standard real line, compute the interiors of \(\mathbb{Q}\) and \(\mathbb{R}\setminus\mathbb{Q}\). We use the density of the rationals: every nonempty open interval contains a rational number. This follows from the Archimedean property by choosing a sufficiently large positive integer \(n\) and an integer \(m\) so that \(a<m/n<b\) for a given interval \((a,b)\).
Every open neighborhood of a real number contains an open interval. Since that interval contains a rational, no open neighborhood can be contained in \(\mathbb{R}\setminus\mathbb{Q}\). The neighborhood test implies that no point belongs to the interior of the irrationals, so \(\operatorname{int}(\mathbb{R}\setminus\mathbb{Q})=\varnothing\).
For \(\mathbb{Q}\), take any open interval \((a,b)\). By density, choose a rational \(r\in(a,b)\). Choose \(n\in\mathbb{N}\) large enough that \(\sqrt{2}/n<b-r\). Then \(r+\sqrt{2}/n\) lies strictly between \(a\) and \(b\), and it is irrational: if it were rational, subtracting the rational \(r\) would make \(\sqrt{2}/n\) rational, and multiplying by \(n\) would make \(\sqrt{2}\) rational. Thus every open interval contains an irrational, so no open neighborhood is contained in \(\mathbb{Q}\). Therefore \(\operatorname{int}\mathbb{Q}=\varnothing\).
Check: both computed interiors are open, since the empty set is open, and neither \(\mathbb{Q}\) nor \(\mathbb{R}\setminus\mathbb{Q}\) contains a nonempty open interval. The neighborhood test rules out every possible point of either interior.
Worked Example: Two Closed Intervals That Meet
Let \(A=[0,1]\cup[1,2]\subseteq\mathbb{R}\). We claim \(\operatorname{int}A=(0,2)\).
If \(x\in(0,2)\), set \(\varepsilon=\min(x,2-x)/2\). Then \(x-\varepsilon>0\) and \(x+\varepsilon<2\), so the interval around \(x\) lies in \((0,2)\). Also, \((0,2)\subseteq A\), because points in \((0,1]\) belong to \([0,1]\), and points in \([1,2)\) belong to \([1,2]\). Thus every point of \((0,2)\) is interior.
The endpoints \(0\) and \(2\) are not interior: any interval around \(0\) contains negative points, and any interval around \(2\) contains points greater than \(2\). Points outside \(A\) cannot be interior because the interior is contained in \(A\). Hence \(\operatorname{int}A=(0,2)\).
Check: \((0,2)\) is open and is contained in \(A\). The endpoint argument excludes the only points of \(A\) not in \((0,2)\), so it is the largest open subset.
Worked Example: The Integers
Let \(A=\mathbb{Z}\subseteq\mathbb{R}\). Since the interior is contained in \(A\), it is enough to check that no integer is interior. Fix \(k\in\mathbb{Z}\) and \(\varepsilon>0\), and let \(\delta=\min(\varepsilon,1)/2\). Then \(0<\delta\leq 1/2\), so \(k+\delta\) lies in \((k-\varepsilon,k+\varepsilon)\) but is not an integer. Therefore no open interval around \(k\) is contained in \(\mathbb{Z}\). The neighborhood test gives \(\operatorname{int}\mathbb{Z}=\varnothing\).
Check: the empty set is open and contained in \(\mathbb{Z}\); the calculation for each integer shows that no nonempty open subset of \(\mathbb{Z}\) can contain any point.
A Common Mistake
Pitfall: When computing the interior of \([0,1]\cup[1,2]\), one might treat the two intervals separately and write \((0,1)\cup(1,2)\). That calculation wrongly omits \(1\). The point \(1\) has the open neighborhood \((1/2,3/2)\), and every point of this neighborhood lies in \([0,1]\cup[1,2]\). So \(1\) is interior, and the correct answer is \((0,2)\). The test concerns neighborhoods contained in the whole set, not neighborhoods contained in one chosen piece.
The same principle explains why the interior is a useful summary: it retains precisely the points that can be included in an open set without leaving \(A\). A point may be in the original set but not in its interior, as \(0\) is in \([0,1]\); a point where two pieces meet may still be interior, as \(1\) is in \([0,1]\cup[1,2]\). In every case, the definition and the neighborhood test give a direct way to decide.
Check Your Understanding
Use the largest-open-subset property and the neighborhood test to justify each answer.
- Compute the interior of \([0,1]\) in \(\mathbb{R}\), and explain why neither endpoint is interior.
- Find the interior of \((-\infty,0)\cup(0,\infty)\) in the standard real line.
- Show that the interior of \(\mathbb{Z}\) in \(\mathbb{R}\) is empty by finding a point outside \(\mathbb{Z}\) in every interval around an integer.
- For \(A=(0,1)\cup[1,2]\), decide whether \(1\) is in \(\operatorname{int}A\), and give an open neighborhood that verifies your answer.
- In any topological space \(X\), prove that if \(A\) is open, then \(\operatorname{int}A=A\).