Two Closures Can Meet Without the Sets Meeting
For two sets \(A\) and \(B\), any point approached by points of \(A\cap B\) is necessarily approached by points of \(A\) and by points of \(B\). The converse need not hold: the points of \(A\) near a given point and the points of \(B\) near it may be different, with no point belonging to both sets. This distinction is the reason closure respects one direction of the intersection comparison but not always the other.
Throughout, \((X,\mathcal{T})\) is a topological space. Recall that \(\overline{C}\) is the smallest closed subset of \(X\) containing \(C\). The open-neighborhood criterion says that \(x\in\overline{C}\) exactly when every open neighborhood of \(x\) intersects \(C\). In particular, this criterion applies whether or not \(x\) itself belongs to \(C\).
Proof. Let \(x\in\overline{A\cap B}\). Every open neighborhood \(U\) of \(x\) intersects \(A\cap B\). Since \(A\cap B\subseteq A\), the same neighborhood \(U\) intersects \(A\); the neighborhood criterion gives \(x\in\overline{A}\). Since \(A\cap B\subseteq B\), it also gives \(x\in\overline{B}\). Thus \(x\in\overline{A}\cap\overline{B}\), proving the inclusion. \(\square\)
The proof identifies what it does not establish. From \(x\in\overline{A}\cap\overline{B}\), we know that each neighborhood of \(x\) meets \(A\), and that each neighborhood of \(x\) meets \(B\). These facts do not say that any neighborhood meets \(A\) and \(B\) at the same point. Membership in the closure of the intersection requires precisely that joint condition.
The Shared Endpoint of Disjoint Intervals
Worked Example: Two Open Intervals on Opposite Sides of 0
Work in \(\mathbb{R}\) with its standard topology. Let \(A=(-1,0)\) and \(B=(0,1)\). We compute the intersection and both closures, then compare the two sides of the theorem.
First, \(A\cap B=\varnothing\), because a real number cannot be both less than \(0\) and greater than \(0\). The closure of the empty set is empty: the empty set is itself closed and contains \(\varnothing\). Therefore \(\overline{A\cap B}=\varnothing\).
Next, \(\overline{A}=[-1,0]\). To check the endpoints, every open neighborhood of \(-1\) contains a point in \((-1,0)\): if the neighborhood contains \((-1-\varepsilon,-1+\varepsilon)\) for some \(\varepsilon>0\), choose \(y=-1+\min(\varepsilon,1)/2\). Then \(-1<y<0\) and \(|y-(-1)|=\min(\varepsilon,1)/2<\varepsilon\). Every open neighborhood of \(0\) contains a point of \((-1,0)\), for example \(y=-\min(\varepsilon,1)/2\) when it contains \((-\varepsilon,\varepsilon)\). Points strictly between \(-1\) and \(0\) belong to \(A\), so every neighborhood of each such point intersects \(A\). Points outside \([-1,0]\) have neighborhoods missing \(A\): for \(x<-1\), use a sufficiently small interval around \(x\) lying below \(-1\); for \(x>0\), use one lying above \(0\). This verifies \(\overline{A}=[-1,0]\).
By the corresponding endpoint and neighborhood checks on the right, \(\overline{B}=[0,1]\). Hence \(\overline{A}\cap\overline{B}=[-1,0]\cap[0,1]=\{0\}\). The two sides are therefore \(\overline{A\cap B}=\varnothing\) and \(\overline{A}\cap\overline{B}=\{0\}\), so the inclusion is strict.
Check: the only point in both interval closures is \(0\). Every neighborhood of \(0\) contains points of \(A\) to its left and points of \(B\) to its right, but \(A\cap B\) has no points at all. Thus \(0\) is in both closures separately, not in the closure of the intersection.
A Neighborhood Test for Equality
Proof. By the open-neighborhood criterion for closure, \(x\in\overline{A\cap B}\) if and only if every open neighborhood of \(x\) intersects \(A\cap B\). The theorem already proved gives \(\overline{A\cap B}\subseteq\overline{A}\cap\overline{B}\). Thus equality holds exactly when every point of the right-hand side also belongs to the left-hand side, which, by the criterion, is exactly the stated neighborhood condition. \(\square\)
This test separates two different kinds of information. The condition \(x\in\overline{A}\cap\overline{B}\) tests whether neighborhoods meet each set separately. To conclude \(x\in\overline{A\cap B}\), one must test whether every neighborhood meets their intersection. The interval example fails at the second test.
Worked Example: Overlapping Open Intervals
In \(\mathbb{R}\) with the standard topology, let \(A=(-2,1)\) and \(B=(0,3)\). Here the intervals overlap, so their intersection is \((0,1)\). We compare its closure with the intersection of the individual closures.
The closure of \(A\) is \([-2,1]\): its interior points are already in \(A\), and every neighborhood of either endpoint \(-2\) or \(1\) meets \(A\); points outside this closed interval have a neighborhood missing \(A\). Likewise, \(\overline{B}=[0,3]\). The intersection of these closures is \([-2,1]\cap[0,3]=[0,1]\).
Since \(A\cap B=(0,1)\), its closure is \([0,1]\). Indeed, every neighborhood of \(0\) contains points strictly between \(0\) and \(1\), every neighborhood of \(1\) contains points strictly between \(0\) and \(1\), and every point of \((0,1)\) belongs to the set. A point outside \([0,1]\) has a neighborhood missing \((0,1)\). Thus \(\overline{A\cap B}=[0,1]=\overline{A}\cap\overline{B}\).
Check: the endpoints \(0\) and \(1\) are not in \(A\cap B=(0,1)\), but are in its closure because every neighborhood of either endpoint contains points of the intersection. In this example the two sets provide shared points near both endpoints, so the neighborhood condition for the intersection is satisfied.
Worked Example: A Finite Discrete Space
Let \(X=\{a,b,c\}\) with the discrete topology, in which every subset of \(X\) is open. Take \(A=\{a,b\}\) and \(B=\{b,c\}\). This example checks a contrasting case where equality holds.
In a discrete topology, every subset is also closed, since its complement is a subset and hence open. Therefore the smallest closed set containing any \(C\subseteq X\) is \(C\) itself, so \(\overline{C}=C\). Now \(A\cap B=\{b\}\), and \(\overline{A\cap B}=\overline{\{b\}}=\{b\}\). Also, \(\overline{A}\cap\overline{B}=A\cap B=\{a,b\}\cap\{b,c\}=\{b\}\). Equality holds.
Check: the open neighborhood \(\{b\}\) of \(b\) intersects \(A\cap B\), as required for \(b\in\overline{A\cap B}\). The other points of \(\overline{A}\cap\overline{B}\) do not need to be considered, because that intersection contains only \(b\).
Common Mistake: Reversing the Inclusion
A common error is to treat closure as if it distributed over intersections, writing \(\overline{A\cap B}=\overline{A}\cap\overline{B}\) without checking the reverse inclusion. The two open intervals in the first example provide a concrete correction. There, \(A\cap B=\varnothing\), so its closure is \(\varnothing\), whereas \(\overline{A}\cap\overline{B}=\{0\}\). The endpoint \(0\) is reached by points of each set separately, but no point of the intersection can approach it because the intersection is empty.
The valid general conclusion is the inclusion \(\overline{A\cap B}\subseteq\overline{A}\cap\overline{B}\). Equality needs an additional check: every point in the intersection of the two closures must have every open neighborhood meet \(A\cap B\). The overlapping-interval example and the finite discrete example both pass this check; the disjoint-interval example does not.
Check Your Understanding
For each question, distinguish between neighborhoods meeting the two sets separately and neighborhoods meeting their intersection.
- In the standard topology on \(\mathbb{R}\), compute \(\overline{(-3,0)\cap(0,2)}\) and \(\overline{(-3,0)}\cap\overline{(0,2)}\).
- For \(A=(-1,2)\) and \(B=(1,4)\) in \(\mathbb{R}\), compute both sides of the inclusion and decide whether equality holds.
- Give the neighborhood-criterion argument that proves \(\overline{A\cap B}\subseteq\overline{A}\cap\overline{B}\).
- State the neighborhood condition that must hold at each point of \(\overline{A}\cap\overline{B}\) for equality to hold.
- In a discrete topology, explain why \(\overline{A\cap B}=\overline{A}\cap\overline{B}\) for all subsets \(A\) and \(B\).