From Neighborhoods to Open Sets
A neighborhood describes a set locally, at one specified point: it must contain a ball around that point. We now use this idea to describe a property of an entire set. A set is open when every point it contains has room around it that remains inside the set. The ball may be different at different points; openness does not require one radius to work throughout the set.
The definition has two parts: the point \(x\) must belong to \(U\), and some whole ball around \(x\) must also lie in \(U\). The radius is allowed to depend on \(x\). In particular, a set can be open even when points close to its edge require very small radii.
The empty set \(\varnothing\) is open: there are no points in it for which the condition could fail. The real line \(\mathbb{R}\) is open as well, since for any \(x\in\mathbb{R}\), every ball around \(x\) is contained in \(\mathbb{R}\). These cases are consistent with the definition and are useful to keep in mind when applying general results.
Finding a Ball at Each Point
To prove that a set is open, start with an arbitrary point in the set and find a positive radius that keeps the entire ball inside it. The radius need not be the largest possible one. Often it is enough to measure how far the point is from a boundary and choose a smaller positive number.
Worked Example: An Open Interval
Let \(U=(-3,5)\). We show that \(U\) is open. Choose any \(x\in U\), so \(-3<x<5\). Both distances to the endpoints are positive: \(x+3>0\) and \(5-x>0\). Set
The minimum of two positive numbers is positive, so \(r>0\). If \(z\in B_r(x)\), then \(|z-x|<r\), which gives \(x-r<z<x+r\). Since \(r\leq(x+3)/2<x+3\), we have \(x-r>-3\). Since \(r\leq(5-x)/2<5-x\), we have \(x+r<5\). Thus \(-3<z<5\), so \(z\in U\). We have shown \(B_r(x)\subseteq U\) for an arbitrary \(x\in U\); hence \(U\) is open.
The radius in this argument depends on \(x\). For example, a point near \(-3\) has less room on its left than a point near the center of the interval. Requiring a single radius for all points would be stronger than the definition of open.
Worked Example: A Union of Two Open Rays
Consider \(V=(-\infty,-2)\cup(4,\infty)\). Take any \(x\in V\). If \(x<-2\), choose \(r=(-2-x)/2\), which is positive. For \(z\in B_r(x)\), we have \(z<x+r=(-2+x)/2<-2\), so \(z\in V\). If \(x>4\), choose \(r=(x-4)/2\), also positive. For \(z\in B_r(x)\), we have \(z>x-r=(x+4)/2>4\), so again \(z\in V\).
These cases cover every point of \(V\), and each has a positive-radius ball contained in \(V\). Therefore \(V\) is open. The set has two separate pieces, but openness is checked point by point and does not require the set to be one interval.
There is a useful connection with the result from “Open Balls” called A Smaller Ball Around Each Interior Point. If \(y\in B_r(a)\), that result gives a positive-radius ball around \(y\) contained in \(B_r(a)\). Consequently every open ball is open: each of its points has a smaller ball that stays inside it. This also shows why balls are the basic local pieces used in the definition.
Worked Example: A Set That Is Not Open
Let \(W=[0,2)\). The point \(0\) belongs to \(W\). However, \(W\) is not open because no positive-radius ball around \(0\) is contained in \(W\). Indeed, for any \(r>0\), the point \(z=-r/2\) satisfies \(|z-0|=r/2<r\), so \(z\in B_r(0)\), while \(z<0\) implies \(z\notin W\). Thus every ball around \(0\) includes a point outside \(W\).
The obstruction is at the included endpoint \(0\), not at a lack of nearby points in general. The set contains points immediately to the right of \(0\), but a ball on the real line extends on both sides of its center. This example reinforces that membership alone does not establish openness.
Unions of Open Sets
A union combines sets by including a point whenever it belongs to at least one of them. This operation preserves openness even for an arbitrary collection of open sets, not just a finite one. At any point of the union, it is enough to use a ball supplied by one open set containing that point.
Proof. If \(A\) is empty, the union is \(\varnothing\), which is open. Otherwise, let \(x\in\bigcup_{\alpha\in A}U_\alpha\). By the definition of union, there is some \(\alpha_0\in A\) such that \(x\in U_{\alpha_0}\). Since \(U_{\alpha_0}\) is open, there exists \(r>0\) such that \(B_r(x)\subseteq U_{\alpha_0}\). Because \(U_{\alpha_0}\subseteq\bigcup_{\alpha\in A}U_\alpha\), we have \(B_r(x)\subseteq\bigcup_{\alpha\in A}U_\alpha\). Thus every point in the union has a ball contained in the union, proving it is open. \(\square\)
The proof works for infinitely many sets because it does not try to find one radius that works simultaneously for every set. It selects one set containing the point in question, then uses that set’s radius. This is different from an intersection, where a ball must meet all the requirements at once.
Finite Intersections of Open Sets
An intersection contains only points that belong to every set being intersected. If there are finitely many open sets, each containing a particular point, take a ball around that point from each set and use the smallest of those radii. The resulting ball fits inside them all.
Proof. Let \(x\in\bigcap_{j=1}^{m}U_j\). Then \(x\in U_j\) for each \(j\in\{1,\ldots,m\}\). Since each \(U_j\) is open, for each \(j\) there is a radius \(r_j>0\) such that \(B_{r_j}(x)\subseteq U_j\). Define \(r=\min\{r_1,\ldots,r_m\}\). Since this is the minimum of finitely many positive numbers, \(r>0\). If \(z\in B_r(x)\), then \(d(x,z)<r\leq r_j\) for every \(j\), so \(z\in B_{r_j}(x)\subseteq U_j\) for every \(j\). Therefore \(z\in\bigcap_{j=1}^{m}U_j\), and \(B_r(x)\subseteq\bigcap_{j=1}^{m}U_j\). This supplies a positive-radius ball around every point in the intersection, so the intersection is open. If the intersection has no points, it is \(\varnothing\), which is open by definition. \(\square\)
Worked Example: Intersecting Two Open Intervals
Let \(U_1=(-4,3)\) and \(U_2=(1,7)\). Their intersection is \(U_1\cap U_2=(1,3)\). To verify openness directly, take \(x\in(1,3)\) and set
Both entries in the minimum are positive, so \(r>0\). If \(z\in B_r(x)\), then \(x-r<z<x+r\). The inequalities \(r\leq(x-1)/2<x-1\) and \(r\leq(3-x)/2<3-x\) give \(x-r>1\) and \(x+r<3\). Therefore \(z\in(1,3)=U_1\cap U_2\). This verifies directly that the intersection is open, in agreement with the finite-intersection theorem.
Why Infinite Intersections Are Different
The finite-intersection theorem depends on having only finitely many radii to consider. An infinite collection of positive radii can become arbitrarily small, so there may be no positive radius that works for every set in the collection. In fact, an infinite intersection of open sets need not be open.
Worked Example: An Infinite Intersection That Is Not Open
For each positive integer \(n\), let \(U_n=(-1/n,1/n)\). Each \(U_n\) is an open interval and therefore is open. We claim that
First, \(0\in U_n\) for every \(n\), since \(-1/n<0<1/n\). Now take \(x\neq0\). Choose a positive integer \(n>1/|x|\), which is possible because the positive integers are unbounded above. Then \(1/n<|x|\), so \(x\notin(-1/n,1/n)=U_n\). Thus \(x\) does not belong to the intersection. This proves the claimed equality.
The singleton \(\{0\}\) is not open: for any \(r>0\), the point \(r/2\) belongs to \(B_r(0)\) but is not \(0\), so \(B_r(0)\nsubseteq\{0\}\). Hence this infinite intersection of open sets is not open. The example does not contradict the finite-intersection theorem; its finite-radius minimum argument has no positive analogue for all the radii \(1,1/2,1/3,\ldots\).
Why Openness Matters
Openness is a way to express that a set has no included point pressed directly against its boundary, when viewed from within the set. More precisely, it says that every included point has some local ball that stays in the set. The size of that ball can vary from point to point, so openness is a local condition rather than a claim about the overall shape or size of a set.
The union and intersection theorems provide practical ways to build open sets. Combining any collection of open sets by taking their union preserves openness; imposing finitely many open conditions at once also preserves openness. The distinction between arbitrary unions and finite intersections is important: an infinite union is always open, while an infinite intersection can collapse to a set with no room around its points.
A common pitfall is to test only whether the endpoints of an interval are included. That can be a quick warning for intervals, but the definition applies to general sets and asks about every point. A reliable proof begins with an arbitrary point of the set and produces a positive radius; a reliable disproof identifies one point for which every positive radius fails. These methods apply beyond intervals and will be useful as the course develops more examples of open sets.
Check Your Understanding
Use the definition and the results proved here to answer the following questions.
- In the definition of an open set, why may the radius depend on the point?
- Explain why \(\mathbb{R}\) and \(\varnothing\) are open.
- In the proof about arbitrary unions, why is it enough to choose one set in the family containing the point?
- Where does the finiteness assumption enter the proof that finite intersections of open sets are open?
- Why does \(\bigcap_{n=1}^{\infty}(-1/n,1/n)\) fail to be open?