Intervals Inherit Openness by Intersection
An interval can have open sets that are not open in the real line. The reason is that openness is always relative to a specified space: when the space is an interval, we test sets by intersecting them with open sets of \(\mathbb{R}\). This gives a direct way to calculate the topology on an interval, including when the interval itself is closed or unbounded.
Throughout, \(\mathbb{R}\) has its standard topology, whose open sets are unions of open intervals. If \(X\subseteq\mathbb{R}\), we give \(X\) the subspace topology.
The set \(U\) in this definition need not be contained in \(X\); only its intersection with \(X\) matters. Since every open set of \(\mathbb{R}\) is a union of open intervals, we can calculate relative open sets by taking unions of intersections \(X\cap(a,b)\). The interval \((a,b)\) has \(a<b\), and either endpoint may lie outside \(X\).
Proof. Suppose first that \(A=X\cap U\) for an open set \(U\subseteq\mathbb{R}\), and take \(x\in A\). Then \(x\in U\). Since \(U\) is open in \(\mathbb{R}\), there is an open interval \((a,b)\) with \(x\in(a,b)\subseteq U\). Intersecting with \(X\) gives \(x\in X\cap(a,b)\subseteq X\cap U=A\).
Conversely, suppose that for every \(x\in A\) there is an interval \((a_x,b_x)\) with \(x\in X\cap(a_x,b_x)\subseteq A\). Let \(U=\bigcup_{x\in A}(a_x,b_x)\). This is open in \(\mathbb{R}\), because it is a union of open intervals. Every \(x\in A\) belongs to \(X\cap U\), so \(A\subseteq X\cap U\). If \(y\in X\cap U\), then \(y\in(a_x,b_x)\) for some \(x\in A\). Since \(y\in X\), we have \(y\in X\cap(a_x,b_x)\subseteq A\). Thus \(X\cap U\subseteq A\), proving \(A=X\cap U\). If \(A=\varnothing\), take \(U=\varnothing\), and the same conclusion holds. \(\square\)
This test says that around each point of a relatively open set, some ambient open interval cuts out a piece lying entirely in the set. The interval itself may extend past the ends of \(X\); the intersection is what must lie inside \(A\).
Open Intervals as Subspaces
Worked Example: Cutting an Open Interval
Let \(X=(1,5)\) with its subspace topology, and let \(U=(-2,3)\cup(4,8)\), open in \(\mathbb{R}\). Compute \(X\cap U\), and decide whether \(A=(1,3]\) is relatively open in \(X\).
Intersect each component of \(U\) with \(X\). First, \((1,5)\cap(-2,3)=(1,3)\), because points must be greater than \(1\) and less than \(3\). Second, \((1,5)\cap(4,8)=(4,5)\), because points must be greater than \(4\) and less than \(5\). Therefore \(X\cap U=(1,3)\cup(4,5)\).
Check: \(1\) and \(5\) are excluded because neither belongs to \(X\); \(3\) and \(4\) are excluded because the two intervals making up \(U\) have strict endpoints there. Thus the computed intersection has exactly the stated endpoints.
Now consider \(A=(1,3]\). The point \(3\) belongs to both \(A\) and \(X\). For any open interval \((a,b)\) containing \(3\), we have \(b>3\). Choose \(y=(3+\min\{b,5\})/2\). Since \(b>3\) and \(5>3\), this gives \(3<y<\min\{b,5\}\), so \(y\in X\cap(a,b)\) but \(y\notin A\). No interval around \(3\) can therefore pass the interval-basis test. Hence \(A\) is not relatively open in \(X\).
For an open interval \(X=(c,d)\), the relative topology is still formed by intersections \(X\cap(a,b)\). Some such intersections equal ordinary open intervals, while others have \(c\) or \(d\) as an endpoint. Whether an endpoint appears in the intersection depends on whether that point belongs to \(X\) and to the ambient interval being intersected.
Closed Intervals as Subspaces
Worked Example: A Closed Interval and a Union
Let \(X=[-2,3]\) with its subspace topology, and take the ambient open set \(U=(-3,1)\cup(2,4)\). Calculate \(X\cap U\) and explain why it is relatively open in \(X\).
For the first component, \(X\cap(-3,1)=[-2,1)\): the point \(-2\) is included because \(-2\in X\) and \(-3<-2<1\), while \(1\) is excluded by the strict upper bound. For the second, \(X\cap(2,4)=(2,3]\): \(2\) is excluded by the strict lower bound, while \(3\) is included because \(3\in X\) and \(2<3<4\). The two pieces are disjoint, so \(X\cap U=[-2,1)\cup(2,3]\).
By definition this is relatively open in \(X\), since it is the intersection of \(X\) with the open set \(U\) of \(\mathbb{R}\). The displayed brackets at \(-2\) and \(3\) do not change that: those endpoints are included in \(X\), and the ambient intervals used here contain them.
Check: The intersection contains \(-2\), every point strictly between \(-2\) and \(1\), every point strictly between \(2\) and \(3\), and \(3\). It contains neither \(1\) nor \(2\). These inclusions and exclusions agree with both \(X\) and \(U\).
This example illustrates a useful distinction: the notation \([ -2,1)\) describes a set of real numbers, but it does not say whether that set is open in \(\mathbb{R}\). Here it is open in the subspace \([-2,3]\) because it is obtained by intersecting that space with an ambient open set. Relative openness is always judged in the stated space.
Unbounded Intervals as Subspaces
Worked Example: Intersecting an Unbounded Interval
Let \(X=(-\infty,2]\) with its subspace topology, and let \(V=(-4,0)\cup(1,5)\), open in \(\mathbb{R}\). Find \(X\cap V\).
For the first component, every point of \((-4,0)\) is less than \(2\), so \((-\infty,2]\cap(-4,0)=(-4,0)\). For the second, points must be greater than \(1\), less than \(5\), and at most \(2\). These conditions give \((-\infty,2]\cap(1,5)=(1,2]\). Consequently \(X\cap V=(-4,0)\cup(1,2]\).
Check: The first interval lies entirely below \(2\), and neither \(-4\) nor \(0\) is included. In the second interval \(1\) is excluded by \(V\), while \(2\) is included by both \(X\) and \(V\). Points greater than \(2\) are excluded by \(X\), so no part of \((2,5)\) appears.
The calculation does not require \(X\) to be bounded. It uses the same intersection rule as for bounded intervals: retain precisely those points that satisfy the membership conditions for both sets.
Proof. If \(U\) is open in \(\mathbb{R}\), then \(X\cap U\) belongs to the subspace topology by its definition, so it is relatively open. Conversely, the definition of relative openness states that a relatively open subset of \(X\) is \(X\cap U\) for some ambient open set \(U\). \(\square\)
When \(U\) is a union of open intervals, distributing intersection over the union gives \(X\cap\bigl(\bigcup_i(a_i,b_i)\bigr)=\bigcup_i\bigl(X\cap(a_i,b_i)\bigr)\). Indeed, a point lies in the left side exactly when it lies in \(X\) and in at least one interval \((a_i,b_i)\); that is exactly the condition for belonging to the right side. This lets us calculate complicated relative open sets one interval at a time.
A Common Mistake: Dropping an Included Endpoint
Take \(X=[0,4]\) and the ambient open interval \(U=(-1,2)\). A mistaken calculation may say \(X\cap U=(0,2)\), on the grounds that the intersection should have open endpoints. But \(0\in X\) and \(-1<0<2\), so \(0\in U\) as well. Therefore \(0\) must be included, and the correct answer is \(X\cap U=[0,2)\). The upper endpoint \(2\) is excluded because \(2\notin U\).
The correction is to check membership in both sets separately, rather than assume that an intersection must have two strict endpoints. In general, an endpoint is retained exactly when it belongs to each set being intersected. This pointwise check works for open, closed, and unbounded intervals alike.
| Subspace \(X\) | Ambient open set \(U\) | Relative open set \(X\cap U\) |
|---|---|---|
| \((1,5)\) | \((-2,3)\cup(4,8)\) | \((1,3)\cup(4,5)\) |
| \([-2,3]\) | \((-3,1)\cup(2,4)\) | \([-2,1)\cup(2,3]\) |
| \((-\infty,2]\) | \((-4,0)\cup(1,5)\) | \((-4,0)\cup(1,2]\) |
For any interval \(X\), the reliable procedure is to state \(X\), state the ambient open set, intersect each interval component separately, and check every endpoint against both sets. The result is relatively open by construction; the interval-basis test also lets us reject proposed sets that fail to contain a full relative interval around one of their points.
Check Your Understanding
For each question, specify the subspace and check endpoint membership in both sets.
- In \(X=(0,6)\), calculate \(X\cap\bigl((-2,2)\cup(4,8)\bigr)\).
- In \(X=[-1,2]\), express \([-1,1)\) as \(X\cap U\) for an open set \(U\subseteq\mathbb{R}\).
- In \(X=(-\infty,3]\), calculate \(X\cap(1,5)\).
- Is \((0,2]\) relatively open in \(X=(0,4)\)? Use the interval-basis test.
- Give an ambient open interval \(U\) for which \(X\cap U=[0,3)\), where \(X=[0,5]\).