A Local Test for Openness
A basis breaks open sets into smaller building blocks. To use those blocks effectively, it helps to have a test that starts with a proposed set and examines its points one at a time. The test is local: at each point of the set, find a basis element that contains the point and stays inside the set. The basis element may vary from point to point.
Let \((X,\mathcal{T})\) be a topological space. A basis \(\mathcal{B}\) for \(\mathcal{T}\) is a collection of open subsets of \(X\) such that every open set is a union of members of \(\mathcal{B}\). In particular, the empty set is the union of no basis elements. A basis element is not required to be the whole open set it helps describe.
Proof. Suppose first that \(U\) is open. Since every open set is a union of basis elements, write \(U=\bigcup_{\alpha\in J}B_\alpha\), where each \(B_\alpha\in\mathcal{B}\). If \(x\in U\), then \(x\) belongs to at least one \(B_\alpha\) in this union. That basis element satisfies \(x\in B_\alpha\subseteq U\).
Conversely, suppose that for every \(x\in U\) there is a basis element \(B_x\) with \(x\in B_x\subseteq U\). Then \(U=\bigcup_{x\in U}B_x\): every \(B_x\) is contained in \(U\), and every \(x\in U\) belongs to its chosen \(B_x\). A union of basis elements is open, so \(U\) is open. If \(U=\varnothing\), the condition has no points to check, and \(U\) is open as the empty union. \(\square\)
The test gives a practical two-way procedure. To prove that a set is open, provide a suitable basis element for every point in it. To prove that it is not open, find one point for which no basis element containing that point lies entirely inside the set. In the second task, checking only some of the points cannot establish failure; the obstructing point must be identified.
Applying the Test in the Real Line
Worked Example: Finding Local Intervals
Give a pointwise basis proof that \(U=(0,2)\cup(3,5)\) is open in \(\mathbb{R}\) with its standard topology. Use the basis \(\mathcal{B}\) consisting of all open intervals \((a,b)\) with \(a<b\).
Take any \(x\in U\). It lies in exactly one of the two intervals \((0,2)\) or \((3,5)\). Write that interval as \((a,b)\), so \((a,b)\) is either \((0,2)\) or \((3,5)\). Since \(a<x<b\), both \(x-a\) and \(b-x\) are positive. Set \(r=\min\{x-a,b-x\}/2\), which is positive. Then \(r\le (x-a)/2\) and \(r\le (b-x)/2\), so \(x-r>a\) and \(x+r<b\). Consequently, \(x\in(x-r,x+r)\subseteq(a,b)\subseteq U\). The interval \((x-r,x+r)\) is a basis element. As this works for every \(x\in U\), the pointwise test proves that \(U\) is open.
Check: for \(x=1\), the containing component is \((0,2)\), and \(r=\min\{1-0,2-1\}/2=1/2\); the witness is \((1/2,3/2)\subseteq U\). For \(x=4\), the component is \((3,5)\), and \(r=\min\{4-3,5-4\}/2=1/2\); the witness is \((7/2,9/2)\subseteq U\). Each computed interval contains its selected point and stays in the corresponding component.
The pointwise nature of the test matters even when the set is not open. A single point can block the test; finding such a point is enough to show that the set fails to be open.
Worked Example: A Boundary Point Blocks Openness
Test \(V=[0,1)\) in \(\mathbb{R}\) with the standard topology and the basis of open intervals. The point \(0\) belongs to \(V\). Any basis element containing \(0\) has the form \((a,b)\) with \(a<0<b\). Because \(a<0\), the interval contains negative real numbers, so it is not a subset of \(V\). Thus there is no basis element \(B\) with \(0\in B\subseteq V\), and the pointwise test shows that \(V\) is not open.
For contrast, at \(x=1/2\) the interval \((1/4,3/4)\) is a basis element with \(1/2\in(1/4,3/4)\subseteq V\). This witness at \(1/2\) does not rescue the set: openness requires a witness at every point of \(V\), including \(0\).
Check: the interval \((-1,1)\) contains \(0\), but it is not contained in \(V\), since \(-1/2\in(-1,1)\) and \(-1/2\notin[0,1)\). More generally, every interval basis element containing \(0\) includes some negative number, so the failure is not limited to this one attempted choice.
Testing a Basis on a Finite Space
The same reasoning applies when basis elements are not intervals. For a collection to be a basis, its elements must cover the space, and when two basis elements overlap, their intersection must be covered locally by basis elements as well. This is the basis criterion for a collection of subsets.
Worked Example: A Nontrivial Finite Basis
Let \(X=\{a,b,c,d\}\), and consider \(\mathcal{B}=\{\{a,b\},\{b\},\{b,c\},\{d\}\}\). Give \(X\) the topology whose open sets are unions of members of \(\mathcal{B}\). First check that this collection is a basis: its members cover \(X\), since \(a,b,c,d\) each belong to at least one member. The only nonempty intersection of distinct members that is not one of the members is \(\{a,b\}\cap\{b,c\}=\{b\}\), which is itself a basis element. The other intersections of distinct members are either \(\varnothing\) or already members. Thus the basis criterion applies.
Consider \(U=\{a,b,d\}\). It is the union \(\{a,b\}\cup\{d\}\), so it is open. The pointwise test gives explicit witnesses: for \(a\), use \(\{a,b\}\subseteq U\); for \(b\), use \(\{b\}\subseteq U\); and for \(d\), use \(\{d\}\subseteq U\). Each listed basis element contains the point it witnesses.
Now consider \(V=\{a,d\}\). The only basis element containing \(a\) is \(\{a,b\}\), and \(\{a,b\}\nsubseteq V\) because \(b\notin V\). Therefore \(a\) has no suitable basis element inside \(V\), so \(V\) is not open. Equivalently, no union of members of \(\mathcal{B}\) can equal \(\{a,d\}\): including \(a\) in such a union requires including \(\{a,b\}\), which also includes \(b\).
Check: \(\{a,b\}\cup\{d\}=\{a,b,d\}=U\), with no extra points. For \(V\), the basis element forced by \(a\) adds \(b\), so any union of basis elements containing \(a\) contains a point outside \(V\).
This example also illustrates why the basis test concerns containment, not equality. A set can be open even though it is not itself one of the basis elements: \(U=\{a,b,d\}\) is open as a union of two of them. The basis elements provide local pieces; an open set need not be one such piece.
Worked Example: The Smallest Basis for an Indiscrete Space
Let \(X=\{p,q\}\) and \(\mathcal{B}=\{X\}\). The open sets in the resulting topology are \(\varnothing\) and \(X\). For \(U=X\), each point has the basis element \(X\) as a witness, since \(p\in X\subseteq X\) and \(q\in X\subseteq X\). For \(U=\varnothing\), there are no points requiring witnesses, and the set is open.
The singleton \(\{p\}\) is not open. Its only possible basis element containing \(p\) is \(X\), and \(X\nsubseteq\{p\}\) because \(q\in X\) but \(q\notin\{p\}\). The pointwise test therefore fails at \(p\).
Check: the only unions of members of \(\mathcal{B}\) are the empty union \(\varnothing\) and the union \(X\). Thus the open sets obtained by the pointwise test agree exactly with the stated topology.
Unions and a Common Mistake
Proof. If \(U\) is open, the pointwise basis test supplies, for each \(x\in U\), a basis element \(B_x\) with \(x\in B_x\subseteq U\). These elements have union \(U\), since each is contained in \(U\) and every point of \(U\) belongs to one of them. Conversely, any union of basis elements is open because each basis element is open and arbitrary unions of open sets are open. The empty set is included as the empty union. \(\square\)
Common mistake. One incorrect approach is to demand that the proposed open set itself be a basis element. For instance, in the standard topology on \(\mathbb{R}\), \(U=(0,1)\cup(2,3)\) is not a single open interval, but the first worked example proves that it is open. Another incorrect approach is to choose a set that contains the point but extends outside \(U\): such a set is not a witness, because the required condition is \(x\in B\subseteq U\). The test works only when the containing basis element stays entirely inside the proposed set.
When using the criterion, keep its quantifiers in view: for every point \(x\) in the proposed set, there must exist a basis element \(B\), and that element must satisfy both \(x\in B\) and \(B\subseteq U\). The choice of \(B\) may depend on \(x\). For a proof of openness, exhibit those choices; for a proof of non-openness, show that at least one point has no valid choice.
Check Your Understanding
Use the pointwise basis test in each question. State which basis elements witness openness, or identify a point where the test fails.
- In \(\mathbb{R}\) with its standard topology and the basis of open intervals, explain why \((1,4)\) is open by giving a suitable interval for an arbitrary point \(x\in(1,4)\).
- For the finite basis \(\mathcal{B}=\{\{a,b\},\{b\},\{b,c\},\{d\}\}\) on \(X=\{a,b,c,d\}\), decide whether \(\{b,c,d\}\) is open.
- For the same finite basis, decide whether \(\{b,d\}\) is open and justify the decision point by point.
- If \(\mathcal{B}=\{X\}\), which subsets of \(X\) pass the pointwise test?
- Explain why the condition “there exists a basis element contained in \(U\)” is not by itself enough to prove that \(U\) is open.