Examples of Open Sets

Many Shapes Can Be Open

The definition of an open set is local: at each point of the set, some open ball around that point must remain in the set. It does not require the set to be one interval, bounded, or connected. In the previous tutorial, “Open Sets,” we verified that open intervals and open rays are examples, and we proved that arbitrary unions of open sets are open. We now use these facts to build and recognize a broader range of examples.

The basic construction is to gather intervals that fit inside the set. An open set may have several separate pieces, or infinitely many. Conversely, a set obtained by removing finitely many points from the real line is still open: around any remaining point, a sufficiently small ball avoids all the removed points. The two results below formalize these useful ways to produce examples.

Open Sets as Unions of Intervals

Every point of an open set comes with a ball contained in that set. On the real line, a ball \(B_r(x)\) is the interval \((x-r,x+r)\). Thus an open set can be covered by open intervals that are themselves contained in the set. The next theorem shows that this idea gives an exact description, not merely a cover.

Proof. First suppose \(U\) is open and nonempty. For each \(x\in U\), the definition of openness gives a number \(r_x>0\) such that \(B_{r_x}(x)\subseteq U\). Since \(B_{r_x}(x)=(x-r_x,x+r_x)\), each of these is a bounded open interval. Every point \(x\in U\) belongs to its own interval \(B_{r_x}(x)\), so \(U\) is contained in their union. Each interval is contained in \(U\), so their union is contained in \(U\). Therefore

If \(U=\varnothing\), the equality holds when the union is taken over the empty family. This proves that every open set is a union of open intervals.

For the reverse direction, let \(U\) be a union of open intervals. Each open interval is open: if \(y\in(a,b)\), then \(y-a>0\) and \(b-y>0\), and a positive radius smaller than both distances gives a ball around \(y\) contained in \((a,b)\). The same argument applies to an open ray, using the distance to its finite endpoint; the whole real line is open as well. By the Arbitrary Unions of Open Sets theorem from the previous tutorial, a union of open sets is open. The empty union is \(\varnothing\), which is open. Hence \(U\) is open. \(\square\)

The theorem can be used in either direction. To prove a set open, it may be simpler to express it as a union of familiar open intervals than to find a radius directly at every point. To understand an arbitrary open set, the theorem says it is assembled from local intervals, even when its overall form is complicated.

Removing Finitely Many Points

A second useful source of open sets is the complement of a finite set. Although the points that have been removed may be scattered across the real line, there are only finitely many of them. At any point that remains, the distances to the removed points have a positive minimum. A ball with radius smaller than that minimum avoids every removed point.

Proof. If \(F=\varnothing\), then \(\mathbb{R}\setminus F=\mathbb{R}\), which is open. Now suppose \(F\) is nonempty, and take an arbitrary \(x\in\mathbb{R}\setminus F\). For every \(a\in F\), we have \(a\neq x\), so \(|x-a|>0\). Since \(F\) is finite and nonempty, the finitely many positive numbers \(|x-a|\), for \(a\in F\), have a positive minimum. Define

Then \(r>0\). We claim \(B_r(x)\subseteq\mathbb{R}\setminus F\). If a point \(z\in B_r(x)\) belonged to \(F\), then \(|x-z|\geq\delta\), by the definition of \(\delta\). But \(z\in B_r(x)\) means \(|x-z|<r=\delta/2<\delta\), a contradiction. Thus every \(z\in B_r(x)\) lies outside \(F\). We have found such a ball for an arbitrary \(x\in\mathbb{R}\setminus F\), so \(\mathbb{R}\setminus F\) is open. \(\square\)

Finiteness matters in this proof because a finite collection of positive distances has a positive minimum. For an infinite removed set, the distances from a point to the removed points may have no positive lower bound. The proof then gives no positive radius that avoids them all.

Unions with Infinitely Many Pieces

The Arbitrary Unions of Open Sets theorem applies to infinite collections as well as finite ones. This allows open sets to have infinitely many separated pieces. For such examples, first describe each piece as an open interval, then apply the theorem to the whole family.

How to Choose a Proof

These examples suggest two efficient strategies. If a set is already written as a union of open intervals, use the Arbitrary Unions of Open Sets theorem rather than repeating the ball calculation for every point. If the set is the complement of finitely many points, use the minimum-distance argument. In either case, the reasoning ultimately guarantees a ball around each included point.

A common pitfall is to confuse “open” with “one open interval.” The set in the polynomial example and the infinite union above are open without being single intervals. Another pitfall is to assume that removing any collection of points preserves openness. The finite-complement theorem relies on the positive minimum distance to a finite set; when infinitely many points are removed, that argument may fail. The next tutorial studies closed sets, which provide a complementary way to describe subsets of the real line.