Complements of Closed Sets

What Lies Outside a Closed Set?

The previous tutorial showed that a set is closed exactly when its complement in \(\mathbb{R}\) is open. Here we use that fact to examine the complement more closely. A point outside a closed set does not merely fail to belong to the set: it has a whole interval around it that avoids the set. Moreover, the complement can be organized into maximal open intervals, whose finite endpoints lie in the closed set.

We will also establish a useful distance result. If \(F\) is a nonempty closed subset of \(\mathbb{R}\), then every real number has at least one nearest point in \(F\). This nearest point need not be unique, but its existence gives a precise way to measure how far a point outside \(F\) lies from the set.

A Nearest Point in a Closed Set

For a nonempty set \(F\subseteq\mathbb{R}\) and \(x\in\mathbb{R}\), the distance from \(x\) to \(F\) is the infimum of the distances from \(x\) to points of \(F\). The infimum is always a finite nonnegative real number: the distances are nonnegative, and choosing any one point of \(F\) gives a finite upper bound. The infimum need not be attained for an arbitrary set. Closedness is what ensures attainment.

Proof. Put \(\delta=d(x,F)\). For each \(n\in\mathbb{N}_0\), the definition of infimum gives a point \(y_n\in F\) such that

The upper bound implies \(|x-y_n|<\delta+1\), so the triangle inequality gives \(|y_n|\leq |x|+|x-y_n|<|x|+\delta+1\). Thus \((y_n)\) is bounded. By the Bolzano-Weierstrass Theorem, it has a convergent subsequence \((y_{n_k})\), say \(y_{n_k}\to p\). Since every \(y_{n_k}\) belongs to \(F\) and \(F\) is closed, the Sequential Characterization of Closed Sets gives \(p\in F\).

The displayed bounds show that \(|x-y_n|\to\delta\), because \(0\leq |x-y_n|-\delta<1/(n+1)\). Also, \(y_{n_k}\to p\) implies \(|x-y_{n_k}|\to|x-p|\), by continuity of the absolute value (equivalently, by the reverse triangle inequality). The same subsequence therefore has distance limit both \(\delta\) and \(|x-p|\), so \(|x-p|=\delta=d(x,F)\). This proves the theorem. \(\square\)

If \(x\in F\), then \(d(x,F)=0\), and \(x\) itself is a nearest point. If \(x\notin F\), the complement characterization from the previous tutorial says that \(\mathbb{R}\setminus F\) is open. Hence some ball around \(x\) lies in the complement, so every point of \(F\) is at least that radius away from \(x\). In particular, \(d(x,F)>0\). The theorem then supplies a point of \(F\) exactly that positive distance away.

Maximal Open Intervals in the Complement

An open subset of the real line can have many pieces, and those pieces may be bounded intervals, unbounded rays, or the whole line. For the complement of a closed set, we can describe each piece directly, without assuming in advance how many pieces there are. Fix a point \(x\notin F\). We will extend an interval around \(x\) as far as possible while keeping it inside \(\mathbb{R}\setminus F\).

Proof. Since \(U\) is open, choose \(r>0\) such that \((x-r,x+r)\subseteq U\). Define

The set \(A\) is nonempty because \(x-r/2\in A\), and \(B\) is nonempty because \(x+r/2\in B\). Let \(a=\inf A\), allowing \(a=-\infty\) if \(A\) is unbounded below, and let \(b=\sup B\), allowing \(b=+\infty\) if \(B\) is unbounded above. The ball around \(x\) ensures \(a<x<b\).

We first check that every point strictly between \(a\) and \(x\) lies in \(U\). If \(a<y\leq x\), the definition of infimum gives some \(t\in A\) with \(t<y\); otherwise \(y\) would be a lower bound for \(A\) larger than \(a\). Since \([t,x]\subseteq U\), it follows that \(y\in U\). This argument also applies when \(a=-\infty\), because \(A\) is then unbounded below. Likewise, every \(y\) with \(x\leq y<b\) lies in \(U\). Thus \((a,b)\subseteq U\), and this interval contains \(x\).

If \(a\) is finite, then \(a\notin U\). Indeed, if \(a\in U\), openness would give some \(s>0\) such that \((a-s,a+s)\subseteq U\). We already know that every point of \((a,x]\) lies in \(U\). Together these facts imply \([a-s/2,x]\subseteq U\), so \(a-s/2\in A\), contradicting that \(a\) is a lower bound for \(A\). Since \(a\notin U\), it belongs to \(F\). The same argument at the right endpoint shows that if \(b\) is finite, then \(b\in F\).

To verify maximality, let \(J\) be any open interval contained in \(U\) and containing \(x\). For each \(y\in J\), the interval property gives \([x,y]\subseteq J\) when \(y\geq x\), and \([y,x]\subseteq J\) when \(y\leq x\). Thus \(y\) belongs to \(B\) or \(A\), respectively, and so \(a\leq y\leq b\). Since \(J\) is open, it also contains a point \(y'\) beyond \(y\) on the same side of \(x\) (that is, \(y'>y\) if \(y\geq x\), or \(y'<y\) if \(y\leq x\)); the same argument puts \(y'\) in \(B\) or \(A\), so \(a<y<b\). Therefore \(J\subseteq(a,b)\). This proves that \((a,b)\) is maximal among open intervals in \(U\) containing \(x\).

Each point \(x\in U\) lies in its interval \(I_x\), so these intervals cover \(U\). Now suppose two such maximal intervals overlap. Their union is an open interval contained in \(U\): the overlap ensures that the union has no gap. It contains both original intervals, so maximality forces them to be equal. If they do not overlap, they are disjoint. This proves all the claims. \(\square\)

The result gives a concrete picture of the complement: it is partitioned into disjoint maximal open intervals, and every finite endpoint of any such interval is in the closed set. An endpoint cannot be missing from \(F\), because if it were in the open complement, the interval could be extended farther.

Pointwise Distance Is Not Uniform Distance

For a nonempty closed set \(F\), every point \(x\notin F\) has \(d(x,F)>0\). It is tempting to conclude that there is one positive number that works for every point outside \(F\). That conclusion is false: the distance can depend on \(x\) and can become arbitrarily small as \(x\) approaches an endpoint of a complementary interval.

A useful distinction is that closedness controls the boundary of each complementary interval, not a uniform gap between the whole set and all points outside it. The nearest-point theorem is pointwise: for each chosen \(x\), it identifies at least one point of \(F\) realizing \(d(x,F)\). The maximal-interval theorem is also local in this sense: each complementary interval has its own endpoints, and different intervals may have very different lengths.