Half-Open Intervals as Building Blocks
A basis describes a topology through local building blocks: a set is open when it can be assembled as a union of basis elements. Here the building blocks are half-open intervals. Their left endpoints belong to the intervals, while their right endpoints do not. That endpoint convention determines exactly what happens when two intervals meet, so it matters in checking the basis criterion.
Throughout, the underlying set is \(\mathbb{R}\), with its usual order. For real numbers \(a<b\), write \([a,b)=\{x\in\mathbb{R}:a\le x<b\}\). In particular, \([a,b)\) is nonempty; the notation \([a,a)\) would instead describe the empty set and is not included among the proposed basis elements.
The second condition is local: it asks for a basis element around each point of an intersection, not necessarily for the whole intersection to be one basis element. For half-open intervals we can prove something stronger: the intersection of any two basis elements is either empty or another basis element.
Verifying the Basis Criterion
Proof. First we verify coverage. Let \(x\in\mathbb{R}\). The interval \([x,x+1)\) belongs to \(\mathcal{B}\), since \(x<x+1\), and it contains \(x\), since \(x\le x<x+1\).
Now let \([a,b),[c,d)\in\mathcal{B}\). A real number \(x\) belongs to their intersection exactly when \(x\ge a\), \(x\ge c\), \(x<b\), and \(x<d\). Equivalently, \(x\ge\max(a,c)\) and \(x<\min(b,d)\). Therefore $$ [a,b)\cap[c,d)= \begin{cases} [\max(a,c),\min(b,d)),&\text{if }\max(a,c)<\min(b,d),\\ \varnothing,&\text{if }\max(a,c)\ge\min(b,d). \end{cases} $$ If \(x\in[a,b)\cap[c,d)\), then \(\max(a,c)\le x<\min(b,d)\). In particular, \(\max(a,c)<\min(b,d)\), so the intersection is itself a member of \(\mathcal{B}\). Taking that intersection as \(B_3\) verifies the local intersection condition at \(x\). Thus \(\mathcal{B}\) satisfies both basis conditions. \(\square\)
For completeness, the basis conditions do give a topology here. Take the collection of all unions of members of \(\mathcal{B}\). It contains \(\varnothing\), as the empty union, and it contains \(\mathbb{R}\), because coverage says that \(\mathbb{R}\) is the union of all members of \(\mathcal{B}\). An arbitrary union of unions of basis elements is again a union of basis elements.
To check finite intersections, let \(U\) and \(V\) be unions of basis elements. If \(x\in U\cap V\), there are \(B_1,B_2\in\mathcal{B}\) with \(x\in B_1\subseteq U\) and \(x\in B_2\subseteq V\). The intersection formula above gives \(B_1\cap B_2\in\mathcal{B}\), and \(x\in B_1\cap B_2\subseteq U\cap V\). Consequently every point of \(U\cap V\) belongs to a basis element contained in \(U\cap V\), so \(U\cap V\) is a union of basis elements. If \(U\cap V=\varnothing\), it is already the empty union. This proves closure under intersections of two sets, and repeated application proves closure under every finite intersection. Hence these unions form a topology. \(\square\)
The topology is specified by the chosen basis, not just by the underlying set \(\mathbb{R}\). The notation \(\mathbb{R}_{\ell}\) signals that these half-open intervals, rather than some other collection of intervals, are the designated building blocks.
Worked Intersections and Local Choices
Worked Example: Two Overlapping Intervals
Compute \([-2,3)\cap[1,5)\). The larger left endpoint is \(\max(-2,1)=1\), and the smaller right endpoint is \(\min(3,5)=3\). Since \(1<3\), the intersection formula gives $$ [-2,3)\cap[1,5)=[1,3). $$
Check: The point \(1\) belongs to both intervals: \(-2\le1<3\) and \(1\le1<5\). The point \(3\) does not belong to the first interval, because its right endpoint is excluded. Every point \(x\) with \(1\le x<3\) satisfies both sets of membership inequalities, and no point outside that range does. This confirms the computed intersection.
Worked Example: Finding a Basis Element Around a Point
Let \(B_1=[0,4)\), \(B_2=[1,3)\), and \(x=2.5\). First, \(x\in B_1\) because \(0\le2.5<4\), and \(x\in B_2\) because \(1\le2.5<3\). Thus \(x\in B_1\cap B_2\).
The intersection formula gives \(\max(0,1)=1\) and \(\min(4,3)=3\), so \(B_1\cap B_2=[1,3)\). This intersection itself is a basis element containing \(x\), and it is contained in \(B_1\cap B_2\), as required by the criterion.
Check: The inequalities \(1\le2.5<3\) confirm membership in the proposed basis element. Its endpoints are also correct: \(1\) belongs to both original intervals, while \(3\) does not belong to \(B_2\).
Worked Example: Intervals That Meet at an Endpoint
Compute \([-1,2)\cap[2,6)\). The larger left endpoint is \(\max(-1,2)=2\), and the smaller right endpoint is \(\min(2,6)=2\). Since \(2\not<2\), the intersection formula gives \(\varnothing\).
Check: Any point in the second interval must be at least \(2\), while any point in the first interval must be strictly less than \(2\). No real number satisfies both conditions. In particular, \(2\) is not in the first interval, so the intervals do not share that endpoint.
Endpoint Care and the Role of the Criterion
Common pitfall. A frequent incorrect calculation is \([0,2)\cap[2,4)=\{2\}\), on the mistaken reasoning that the intervals meet at \(2\). But membership in the first interval requires \(x<2\), whereas membership in the second requires \(x\ge2\). No \(x\) can satisfy both inequalities, so the correct answer is $$ [0,2)\cap[2,4)=\varnothing. $$ The general formula gives the same result: the larger left endpoint and smaller right endpoint are both \(2\), and the strict condition \(2<2\) fails.
This endpoint check is not an incidental detail. The basis condition concerns points that actually lie in the intersection. If two intervals only touch at an endpoint excluded by one of them, there is no such point, and the local intersection condition has nothing to check there. If the intersection is nonempty, its left endpoint is included and its right endpoint is excluded, exactly as the formula states.
Once the basis criterion has been verified, it supplies a consistent way to build open sets: take unions of the half-open intervals. The coverage condition ensures that every real number is available as a point of an interval. The intersection condition ensures that when two such unions overlap, points in the overlap still have basis elements contained inside it. These are the structural reasons the intervals generate a topology, rather than merely a convenient collection of subsets.
When checking a proposed basis of this form, keep the two tasks separate. First show that every point is covered. Then take two arbitrary proposed basis elements and calculate their intersection, paying attention to strict and non-strict inequalities at the endpoints. For the lower limit topology, the calculation shows that every nonempty intersection is itself a proposed basis element, which immediately supplies the local set required by the criterion.
Check Your Understanding
Use the basis criterion and the endpoint inequalities to answer each question.
- Compute \([-3,1)\cap[-1,2)\), and identify which endpoints belong to the intersection.
- For \(x=0\), give a member of \(\mathcal{B}\) that contains \(x\). Verify both endpoint inequalities.
- Determine whether \([1,4)\cap[4,7)\) is empty, and justify the answer using the intersection formula.
- Explain why the intervals \([a,b)\) cover \(\mathbb{R}\) when \(a<b\) ranges over all real endpoints.
- State the two conditions in the basis criterion and identify where each was checked for \(\mathcal{B}\).