Closure Collects a Set and Its Nearby Points
The derived set records points approached by other points of a set, while the closure also retains every point already in the set, including isolated points. In the previous tutorial, we saw that \(\overline{E}=E\cup E'\). Here we use the neighborhood characterization of closure to develop another practical test and examine how closure behaves under basic set operations.
This is the Neighborhood Characterization of Closure established earlier in the course. It says that a point belongs to the closure if it is impossible to choose a ball around that point that misses the set. The point need not itself belong to \(E\): it may be included because points of \(E\) occur arbitrarily near it. The Closure as the Smallest Closed Superset Theorem gives a complementary description: \(\overline E\) is the smallest closed set containing \(E\). We will use these established facts without reproving them.
A useful new way to test closure uses sequences. A sequence drawn from \(E\) can approach a point whether or not that point is in \(E\). Conversely, if every neighborhood of a point meets \(E\), we can choose a point of \(E\) closer to it at each successive stage.
Proof. Suppose first that \(x\in\overline E\). For each positive integer \(n\), the ball \(B_{1/n}(x)\) meets \(E\). Choose \(x_n\in B_{1/n}(x)\cap E\). Then $$ |x_n-x|<\frac1n. $$ Given \(\varepsilon>0\), choose \(N\) so that \(1/N<\varepsilon\). For every \(n\geq N\), we have $$ |x_n-x|<\frac1n\leq\frac1N<\varepsilon. $$ Thus \(x_n\to x\). Conversely, suppose \(x_n\in E\) for every \(n\) and \(x_n\to x\). Given \(r>0\), convergence gives an \(N\) such that \(|x_n-x|<r\) for every \(n\geq N\). In particular, \(x_N\in B_r(x)\cap E\). Every ball around \(x\) therefore meets \(E\), so \(x\in\overline E\). If \(E=\varnothing\), there is no sequence with all its terms in \(E\), and the neighborhood definition also gives \(\overline E=\varnothing\). This covers that case as well. \(\square\)
The sequence in this theorem is allowed to repeat values. It is not required to consist of distinct points, so the theorem applies to isolated points of \(E\): for \(x\in E\), the constant sequence \(x_n=x\) converges to \(x\). This is one difference from the sequential characterization of limit points, which requires sequence terms different from the candidate limit.
Worked Example: The Closure of a Half-Open Interval
Let \(E=[-1,2)\). Every point \(x\in E\) belongs to \(\overline E\), since every ball centered at \(x\) contains \(x\), which is itself in \(E\). The right endpoint \(2\) is not in \(E\), but it is in \(\overline E\): for any \(r>0\), set \(t=\min\{r/2,1/2\}\). Then \(2-t\in[-1,2)\), \(2-t\neq2\), and $$ |(2-t)-2|=t\leq r/2<r. $$ Thus \(B_r(2)\) meets \(E\) for every \(r>0\).
Now let \(x<-1\). The ball of radius \((-1-x)/2>0\) around \(x\) lies strictly to the left of \(-1\), so it misses \(E\). If \(x>2\), the ball of radius \((x-2)/2>0\) lies strictly to the right of \(2\), and also misses \(E\). These cases account for every point outside \([-1,2]\). Therefore $$ \overline{[-1,2)}=[-1,2]. $$ The endpoint \(2\) is added because points of the interval approach it; no points are added beyond it.
Closure Behaves Locally Inside Open Sets
Restricting a set to an open region can remove points that are far away, but it does not change which points of that region are approached by the set. Openness is important: every point in an open set has a whole ball around it that remains in the set. That ball lets us keep the points witnessing closure inside the region.
Proof. First, \(A\cap U\subseteq A\). If a ball meets \(A\cap U\), it also meets \(A\), so the neighborhood definition gives \(\overline{A\cap U}\subseteq\overline A\). Hence $$ \overline{A\cap U}\cap U\subseteq\overline A\cap U. $$ For the reverse inclusion, take \(x\in\overline A\cap U\). Since \(U\) is open, there is a \(\rho>0\) such that \(B_\rho(x)\subseteq U\). Given any \(s>0\), set \(t=\min\{s,\rho\}/2>0\). Because \(x\in\overline A\), the ball \(B_t(x)\) meets \(A\); choose \(y\in B_t(x)\cap A\). Since \(t<\rho\), we also have \(y\in B_\rho(x)\subseteq U\), so \(y\in A\cap U\). Since \(t<s\), \(y\in B_s(x)\). We have shown that every ball \(B_s(x)\) meets \(A\cap U\), and therefore \(x\in\overline{A\cap U}\). Together with \(x\in U\), this proves \(x\in\overline{A\cap U}\cap U\), establishing the reverse inclusion and the equality. \(\square\)
Worked Example: Restricting to an Open Interval
Take \(A=(0,2)\) and \(U=(1,3)\). Then \(A\cap U=(1,2)\), so its closure is \([1,2]\). Consequently, $$ \overline{A\cap U}\cap U=[1,2]\cap(1,3)=(1,2]. $$ Also, \(\overline A=[0,2]\), and therefore $$ \overline A\cap U=[0,2]\cap(1,3)=(1,2]. $$ The two calculations agree, as the Closure Within an Open Set Theorem predicts. The endpoint \(1\) is in \(\overline{A\cap U}\), but not in \(U\), so it does not appear on either side of the equality.
The theorem describes closure only inside \(U\), which is why both sides are intersected with \(U\). The closures themselves need not be equal: points on the edge of \(U\) can be approached by points of \(A\) outside \(U\), but not by points of \(A\cap U\). The openness hypothesis supplies a neighborhood contained in \(U\) at each point where the equality is being tested.
Intersections: Inclusion, Not Always Equality
If a point is in the closure of an intersection, every neighborhood meets both sets, because it meets their intersection. Thus it must belong to the closure of each set. The reverse implication can fail: a neighborhood might meet each set at different points, without meeting their intersection at all.
Proof. Let \(x\in\overline{A\cap B}\). Every ball around \(x\) meets \(A\cap B\), and hence meets \(A\) and \(B\). By the neighborhood characterization of closure, \(x\in\overline A\) and \(x\in\overline B\). Therefore \(x\in\overline A\cap\overline B\), proving the inclusion. \(\square\)
Worked Example: The Inclusion Can Be Strict
Let \(A=(0,1)\) and \(B=(1,2)\). They are disjoint, so \(A\cap B=\varnothing\) and $$ \overline{A\cap B}=\overline\varnothing=\varnothing. $$ However, \(\overline A=[0,1]\) and \(\overline B=[1,2]\), so $$ \overline A\cap\overline B=\{1\}. $$ In particular, \(1\) belongs to both closures: points of \(A\) approach it from the left, and points of \(B\) approach it from the right. But no point belongs to both \(A\) and \(B\). Hence the inclusion in the theorem is strict in this example. This contrasts with the Closure of a Finite Union Theorem, established earlier, which gives equality for finite unions.
A Sequence Set and Its Closure
The sequential characterization can make a closure calculation intuitive, but excluding all other candidate points still requires care. For a set consisting of a convergent sequence, the sequence limit is a natural candidate to add. We must also check that the individual sequence terms and the limit account for every point in the closure.
Worked Example: A Sequence and Its Limit
Let \(S=\{3+1/n:n\geq1\}\). The point \(3\) is in \(\overline S\), since for every \(r>0\) the Archimedean property gives \(n>1/r\), and then $$ 3+\frac1n\in S,\qquad \left|3+\frac1n-3\right|=\frac1n<r. $$ Every point of \(S\) is in its closure as well. We show that any \(x\notin S\cup\{3\}\) is not in \(\overline S\).
If \(x<3\), all elements of \(S\) are at least \(3\), and the ball \(B_{(3-x)/2}(x)\) lies below \(3\), so it misses \(S\). If \(x>4\), all elements of \(S\) are at most \(4\), and \(B_{(x-4)/2}(x)\) lies above \(4\), so it misses \(S\). The point \(4\) itself belongs to \(S\). It remains to consider \(3<x<4\) with \(x\notin S\). Choose \(N\) so large that \(1/n<(x-3)/2\) whenever \(n>N\). For such \(n\), $$ x-\left(3+\frac1n\right)>(x-3)-\frac{x-3}{2}=\frac{x-3}{2}>0. $$ For the finitely many indices \(1\leq n\leq N\), every distance \(\left|x-(3+1/n)\right|\) is positive, since \(x\notin S\). Their minimum is therefore positive. Choose a radius smaller than both that minimum and \((x-3)/2\). The resulting ball around \(x\) misses the finitely many initial terms and the entire tail. Thus \(x\notin\overline S\), proving $$ \overline S=S\cup\{3\}. $$
A reliable closure calculation separates two tasks. To show a point is in the closure, use the neighborhood test or construct a sequence in the set converging to it. To show a point is not in the closure, exhibit one ball around it that misses the set. For intersections, check whether the sets meet each other near the candidate point; separate evidence that each set comes arbitrarily close is not enough to guarantee that their intersection does.
Include the points of the set and any points approached by its elements.
Show every ball meets the set, or use the Sequential Characterization of Closure.
For each remaining type of point, find a positive-radius ball that misses the set.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- What neighborhood condition characterizes membership in \(\overline E\)?
- State the Sequential Characterization of Closure. Does its sequence have to use distinct terms?
- Why does \(\overline{A\cap U}\cap U=\overline A\cap U\) require \(U\) to be open in the proof given here?
- State the general inclusion relating \(\overline{A\cap B}\) to \(\overline A\cap\overline B\).
- For \(A=(0,1)\) and \(B=(1,2)\), what are \(\overline{A\cap B}\) and \(\overline A\cap\overline B\)?
- What kind of ball proves that a point does not belong to the closure of a set?