Turning Openness into Closedness
The previous tutorial compared openness, which is tested by neighborhoods around points of a set, with closedness, which is tested by limits of sequences in a set. Taking a complement links these two conditions: a set is closed precisely when the points outside it form an open set. This gives two ways to test the same property, and each can be more convenient in a particular example.
Throughout this tutorial, complements are taken in \(\mathbb{R}\). Thus the complement of \(U\subseteq\mathbb{R}\) is \(\mathbb{R}\setminus U\), the set of all real numbers that do not belong to \(U\). The ambient set matters: a complement is always taken relative to some specified set, and here that set is the real line.
The key point is that the neighborhood around a point of the complement is not required to avoid the complement. Instead, to show that \(U^{\mathrm c}\) is closed, consider a convergent sequence of points that all lie outside \(U\). If its limit were inside \(U\), openness would force all sufficiently late terms of the sequence into \(U\), a contradiction.
The Complement Characterization of Closed Sets
Proof. First suppose \(U\subseteq\mathbb{R}\) is open, and let \(F=\mathbb{R}\setminus U\). To prove \(F\) is closed, use the Sequential Characterization of Closed Sets. Let \((x_n)\) be a convergent sequence with \(x_n\in F\) for every \(n\), and suppose \(x_n\to x\). We must show \(x\in F\).
If \(x\notin F\), then \(x\in U\). Since \(U\) is open, there is an \(r>0\) such that \(B_r(x)\subseteq U\). Convergence gives an \(N\) such that \(|x_n-x|<r\) for every \(n\geq N\). Hence \(x_n\in B_r(x)\subseteq U\) for those \(n\), contradicting \(x_n\in F\). Therefore \(x\in F\), and \(F\) is closed.
For the converse, suppose \(F\) is closed, and put \(U=\mathbb{R}\setminus F\). We show \(U\) is open. Take any \(x\in U\). If there were no \(r>0\) with \(B_r(x)\subseteq U\), then for every positive integer \(n\), the ball \(B_{1/n}(x)\) would contain a point outside \(U\), hence a point \(x_n\in F\). These choices give \(|x_n-x|<1/n\), so \(x_n\to x\). Since \(F\) is closed, the Sequential Characterization of Closed Sets implies \(x\in F\), contradicting \(x\in U\). Thus some ball around \(x\) is contained in \(U\). Since \(x\) was arbitrary, \(U\) is open. This proves both directions. \(\square\)
The proof covers the edge cases as well. If \(U=\varnothing\), its complement is \(\mathbb{R}\), which is closed; if \(U=\mathbb{R}\), its complement is \(\varnothing\), also closed. In the reverse direction, if \(F=\varnothing\) or \(F=\mathbb{R}\), its complement is open. No special nonempty-set assumption is needed.
Worked Example: The Complement of Two Open Intervals
Let \(U=(-5,-1)\cup(2,7)\). Each interval is open, so their finite union is open. Its complement consists of the points at or to the left of \(-5\), the points from \(-1\) through \(2\), and the points at or to the right of \(7\):
The endpoints belong to the complement because neither open interval contains its endpoints. For instance, \(-5\notin U\), \(-1\notin U\), \(2\notin U\), and \(7\notin U\). Every real number strictly between \(-5\) and \(-1\), or strictly between \(2\) and \(7\), belongs to \(U\); the remaining real numbers belong to the displayed complement. By the complement characterization, \(\mathbb{R}\setminus U\) is closed.
This example also illustrates why the complement is not obtained by simply changing every interval endpoint from parentheses to brackets without checking the gaps. The gap \([-1,2]\) is entirely outside \(U\), so it belongs to the complement as well as the four boundary points and the two unbounded rays.
Infinite Unions and Accumulating Points
The complement characterization is especially useful when an open set is given as an infinite union. Earlier in the course, the Arbitrary Unions of Open Sets theorem established that any union of open sets is open, even when there are infinitely many sets. The theorem above then immediately shows that the complement of such a union is closed.
Worked Example: Intervals Between Consecutive Reciprocals
For each positive integer \(n\), let \(I_n=(1/(n+1),1/n)\), and define \(U=\bigcup_{n=1}^{\infty} I_n\). Every \(I_n\) is open, so the arbitrary-union theorem shows that \(U\) is open. The intervals fill the points strictly between \(0\) and \(1\) except for the reciprocal endpoints. More precisely,
To see why no points between \(0\) and \(1\) have been missed apart from the reciprocals, take \(x\in(0,1)\) that is not \(1/n\) for any positive integer \(n\). Consecutive terms of the decreasing sequence \(1,1/2,1/3,\ldots\) enclose \(x\), so \(1/(n+1)<x<1/n\) for some positive integer \(n\). Thus \(x\in I_n\). The point \(0\) is not in any interval, every \(1/n\) is an endpoint and is omitted, and every \(x\geq1\) or \(x<0\) lies outside every \(I_n\). This verifies the stated complement. Since \(U\) is open, that complement is closed.
The infinitely many points \(1/n\) in the complement approach \(0\), and \(0\) is also in the complement. Indeed, \(1/n\to0\). This is consistent with closedness: a closed set must contain such limits. An omitted accumulation point would contradict the theorem.
Worked Example: The Complement of a Finite Set
Let \(F=\{-2,0,5\}\) and take \(U=\mathbb{R}\setminus F\). The result from “Examples of Open Sets” says that the complement of a finite set is open, so \(U\) is open. Its complement is exactly \(F\), and the complement characterization therefore shows that \(F\) is closed.
The sequence test gives a direct check of the same conclusion. Suppose \((x_n)\) is a convergent sequence whose terms all belong to \(F\). At least one of the three values \(-2,0,5\) must occur infinitely often; otherwise each value would occur only finitely often, leaving only finitely many terms in total. The subsequence consisting of that infinitely repeated value is constant, so its limit is that value in \(F\). Since a subsequence of a convergent sequence has the same limit, the original sequence has its limit in \(F\). Thus \(F\) contains the limits of its convergent sequences.
Openness as Positive Distance from the Complement
There is also a point-by-point metric interpretation. If \(U\) is open and \(x\in U\), then some ball around \(x\) stays inside \(U\). Every point of the complement must therefore be at least that radius away from \(x\). Conversely, if \(x\) has positive distance from the complement, a sufficiently small ball around \(x\) cannot meet the complement and must lie in \(U\).
Proof. Suppose first that \(U\) is open, and take \(x\in U\). There is an \(r>0\) such that \(B_r(x)\subseteq U\). For every \(y\in F\), we must have \(|x-y|\geq r\), since any point with \(|x-y|<r\) belongs to \(B_r(x)\subseteq U\). Taking the infimum over \(y\in F\) gives \(d(x,F)\geq r>0\).
Conversely, suppose \(d(x,F)>0\) for every \(x\in U\). Given \(x\in U\), write \(\delta=d(x,F)>0\). If \(y\in B_{\delta/2}(x)\) belonged to \(F\), then the definition of distance would give \(d(x,F)\leq|x-y|<\delta/2\), contradicting \(d(x,F)=\delta\). Hence \(B_{\delta/2}(x)\subseteq U\). Every point of \(U\) has such a ball, so \(U\) is open. \(\square\)
Worked Example: Measuring the Gap to the Complement
Take \(U=(4,9)\), so \(F=\mathbb{R}\setminus U=(-\infty,4]\cup[9,\infty)\). For any \(x\in(4,9)\), the nearest points of \(F\) are the endpoints \(4\) and \(9\). Therefore
For example, at \(x=6\), the two distances are \(6-4=2\) and \(9-6=3\), so \(d(6,F)=2\). The ball \(B_1(6)=(5,7)\) lies inside \(U\), as the distance characterization predicts. As \(x\) approaches either endpoint from within the interval, the distance to \(F\) approaches \(0\). For example, if \(x=4+1/n\), then \(d(x,F)=1/n\). Each such \(x\) is still an interior point, but there is no single positive radius that works for every point in the interval.
A Useful Distinction: Being Outside Is Not Being Away
A common mistake is to think that every point outside an open set has a ball around it that stays outside the set. Openness says the opposite kind of thing: points inside the set have a ball contained in it. The complement of an open set is closed, but it need not be open. For example, the complement of \((0,1)\) is \((-\infty,0]\cup[1,\infty)\). Every ball around \(0\) meets \((0,1)\), even though \(0\) belongs to the complement. Closedness allows this; it requires limits of points in the complement to remain in it, not a neighborhood around every point of the complement to stay there.
The distance characterization makes the distinction precise. For a point inside an open set, its distance to the complement is positive. For a point on the edge, that distance can be zero. In the example \(U=(0,1)\), \(d(0,\mathbb{R}\setminus U)=0\), because \(0\) itself belongs to the complement. This does not make the complement fail to be closed; it shows why the positive-distance condition is a test for points in \(U\), not for points in its complement.
When a set is given by a complicated union of open pieces, it may be easier to identify its complement and use the complement characterization than to check every sequence directly. Conversely, if a candidate complement is difficult to describe, the sequential test for closedness may be more efficient. The two viewpoints express the same property, but they suggest different proof strategies.
Check Your Understanding
Use the complement characterization and the distance interpretation to answer the following questions.
- What does the complement characterization say about the complement of an open subset of the real line?
- In the union of intervals between consecutive reciprocals, why does the complement include every point \(1/n\)?
- In the distance characterization, why must the complement be nonempty when distance to it is used?
- Why does a point of an open set have positive distance from its complement?
- Does closedness of a complement guarantee that every point in it has a ball contained in the complement? Explain.