A Local Test for Openness
Openness is defined for a whole subset of a space, but on the real line it can be checked one point at a time. The key question is whether every point of a set has some open interval around it that stays inside the set. The interval may depend on the point: there is no requirement that one interval, or one fixed radius, work throughout the set.
Throughout, \(\mathbb{R}\) has its standard topology. Its basis consists of the bounded open intervals \((a,b)\), where \(a,b\in\mathbb{R}\) and \(a<b\). Thus a set is open when it is a union of basis intervals. An interval \((a,b)\) contains exactly the real numbers strictly between \(a\) and \(b\); neither endpoint is included.
Proof. Suppose first that \(U\) is open, and let \(x\in U\). Since \(U\) is a union of basis intervals, there is some basis interval \((a,b)\) in that union with \(x\in(a,b)\subseteq U\). Membership \(x\in(a,b)\) means \(a<x<b\), as required.
Conversely, suppose each \(x\in U\) lies in some interval \((a_x,b_x)\subseteq U\). Then \[ U=\bigcup_{x\in U}(a_x,b_x). \] Indeed, each interval on the right is contained in \(U\), so the union is contained in \(U\); and each \(x\in U\) belongs to its chosen interval, so \(U\) is contained in the union. A union of basis intervals is open in the standard topology, hence \(U\) is open. If \(U=\varnothing\), the condition has no points to check and the displayed union is the empty union, which is \(\varnothing\), also open.
To see the equivalent symmetric-interval formulation, given \(a<x<b\), choose \(r=\min(x-a,b-x)/2\). Both quantities in the minimum are positive, so \(r>0\); and \(x-r>a\), \(x+r<b\), giving \((x-r,x+r)\subseteq(a,b)\). Conversely, an interval \((x-r,x+r)\) is itself an open interval containing \(x\). \(\square\)
The quantifiers matter. The condition says for each point \(x\), there exists a suitable interval. The interval can change as \(x\) changes. For an unbounded set, for example, it would generally be unreasonable to ask for one bounded interval to contain every point. The local test instead asks for a small interval around each individual point.
Applying the Test to Intervals
Worked Example: An Open Interval
Show directly that \(U=(-2,3)\) is open, and check the local condition at \(x=1\).
Take any \(x\in(-2,3)\). Then \(x+2>0\) and \(3-x>0\). Set \[ r=\frac{\min(x+2,3-x)}{2}. \] This gives \(r>0\), \(x-r>-2\), and \(x+r<3\). Therefore every \(y\in(x-r,x+r)\) satisfies \(-2<y<3\), so \((x-r,x+r)\subseteq(-2,3)\). The local criterion proves that \(U\) is open.
At the particular point \(x=1\), the two distances to the endpoints are \(1-(-2)=3\) and \(3-1=2\). Thus \(r=\min(3,2)/2=1\), and \((0,2)\subseteq(-2,3)\). This verifies the test at \(1\). As a check, \(1\in(-2,3)\) because \(-2<1<3\), and both endpoints of the chosen neighborhood, \(0\) and \(2\), lie strictly between \(-2\) and \(3\).
Worked Example: A Union of Open Intervals
Show that \(U=(-4,-1)\cup(0,2)\cup(5,\infty)\) is open in \(\mathbb{R}\). Here \((5,\infty)\) denotes the set of real numbers greater than \(5\); it is not itself a bounded basis interval.
Let \(x\in U\). If \(-4<x<-1\), set \(r=\min(x+4,-1-x)/2\). Both entries in the minimum are positive, and the resulting interval \((x-r,x+r)\) lies inside \((-4,-1)\). If \(0<x<2\), set \(r=\min(x,2-x)/2\); then \(r>0\) and \((x-r,x+r)\subseteq(0,2)\).
In the third part, \(x>5\). Set \(r=(x-5)/2\). This is positive, and \(x-r=(x+5)/2>5\), so every point of \((x-r,x+r)\) is greater than \(5\). Hence this neighborhood lies in \((5,\infty)\). In every case, \(x\) has an open interval around it contained in \(U\). The local criterion shows that \(U\) is open.
For a basis check on the unbounded part, \((5,\infty)=\bigcup_{n\in\mathbb{N}}(5,5+n)\): every point \(x>5\) belongs to one such interval by choosing an integer \(n>x-5\), and each interval in the union contains only numbers greater than \(5\). Thus the unbounded part, and consequently \(U\), is also a union of bounded basis intervals.
Worked Example: Half-Open and Closed Intervals
Determine which of \([0,2)\), \((0,2]\), and \([0,2]\) are open in \(\mathbb{R}\).
The set \([0,2)\) is not open. Its point \(0\) belongs to the set. Any open interval \((a,b)\) containing \(0\) must satisfy \(a<0<b\), so \(a/2\in(a,b)\) and \(a/2<0\). Thus \((a,b)\) contains a point outside \([0,2)\), and no open interval around \(0\) is contained in the set.
The set \((0,2]\) is not open either. At its point \(2\), any open interval \((a,b)\) containing \(2\) has \(a<2<b\). The point \((2+b)/2\) lies strictly between \(2\) and \(b\), so it belongs to \((a,b)\) but not to \((0,2]\).
Finally, \([0,2]\) is not open: the same argument at \(0\) shows that every open interval containing \(0\) contains a negative point. These failures each violate the local criterion at a specific point of the set. By contrast, every point \(x\in(0,2)\) does have an interval contained in each of these three sets: for example, choose \(r=\min(x,2-x)/2\). The obstruction is at the included endpoints, not at the interior points.
Check: The endpoints really are included in the three sets being tested: \(0\in[0,2)\), \(2\in(0,2]\), and both \(0,2\in[0,2]\). The criterion requires a suitable interval at every included point, so failure at just one of these points is enough to show that each set is not open.
Why Unions Work Locally
Proof. Let \(\{U_\alpha:\alpha\in J\}\) be a collection of open subsets of \(\mathbb{R}\), where \(J\) is any index set, and put \(U=\bigcup_{\alpha\in J}U_\alpha\). If \(x\in U\), the definition of union gives an index \(\alpha\in J\) with \(x\in U_\alpha\). Since \(U_\alpha\) is open, the local interval criterion gives an interval \((a,b)\) with \(x\in(a,b)\subseteq U_\alpha\). Because \(U_\alpha\subseteq U\), this interval is contained in \(U\). The criterion now shows \(U\) is open. If \(J\) is empty, the union is \(\varnothing\), which is open because it has no points for which the local condition must be checked. \(\square\)
This proof explains the role of the union: a point in the union belongs to at least one member, and that member supplies a suitable interval. It does not require the same member to contain every point, and it does not require the same interval to work at every point.
Worked Example: An Infinite Union and an Infinite Intersection
Let \(U_n=(-1/n,1/n)\) for \(n\in\mathbb{N}\). Each \(U_n\) is open by the local criterion: if \(x\in U_n\), then the positive number \(r=\min(x+1/n,1/n-x)/2\) gives \((x-r,x+r)\subseteq U_n\). Therefore their union is open by the proposition. In fact, \(U_1=(-1,1)\), and \(U_n\subseteq U_1\) for every \(n\geq1\), so \(\bigcup_{n\in\mathbb{N}}U_n=(-1,1)\). The reverse inclusion in this equality follows because every point of the union lies in some \(U_n\subseteq U_1\), and the forward inclusion follows because \(U_1\) is one of the sets in the union.
The intersection behaves differently. We claim that \(\bigcap_{n\in\mathbb{N}}U_n=\{0\}\). The number \(0\) lies in every \(U_n\), since \(-1/n<0<1/n\). If \(x\ne0\), choose \(n\in\mathbb{N}\) with \(n>1/|x|\); then \(1/n<|x|\), so \(x\notin(-1/n,1/n)\). Hence such an \(x\) is not in the intersection, proving the claim.
The singleton \(\{0\}\) is not open: every open interval \((a,b)\) containing \(0\) has \(a<0<b\), and \(a/2\) is a nonzero point in that interval. Thus this explicit family shows that an infinite intersection of open sets need not be open, even though arbitrary unions of open sets are open.
A Common Pitfall
Common mistake: “The set \([0,2)\) is open because it contains the interval \((0,2)\).” The inclusion \((0,2)\subseteq[0,2)\) is true, but it checks neighborhoods only for points strictly between \(0\) and \(2\). It says nothing about the included point \(0\). The local criterion must be checked at every point of the proposed open set. At \(0\), any interval \((a,b)\) containing it has \(a<0\), and then \(a/2\in(a,b)\) lies outside \([0,2)\). The correction is to test boundary points that are included as well as points in the interior; here that test shows \([0,2)\) is not open.
For the same reason, the fact that a set contains many open intervals does not by itself prove it is open. The useful question is whether each of its points lies in at least one open interval contained in the set. That local condition is both necessary and sufficient.
Check Your Understanding
Use the local interval criterion in each question. When testing whether a set is open, pay particular attention to any included endpoints.
- Is \((-3,5)\) open? For an arbitrary \(x\in(-3,5)\), give a positive \(r\) for which \((x-r,x+r)\subseteq(-3,5)\).
- Which of \((1,4]\), \([1,4)\), and \((1,4)\) are open in \(\mathbb{R}\)? Identify a point that decides each answer.
- Let \(U=(-2,0)\cup(3,\infty)\). Explain why each point has an open interval around it contained in \(U\).
- Why does the proof that arbitrary unions of open sets are open also cover the empty union?
- For \(V_n=(-1/n,1/n)\), compute \(\bigcap_{n\in\mathbb{N}}V_n\) and use the local criterion to decide whether that intersection is open.