Endpoint Neighborhoods in Half-Open and Closed Intervals

Why Endpoints Need a Relative Test

An interval can have open sets that are not open in the real line. The reason is that openness depends on which space supplies the neighborhoods. Near an included endpoint, an interval contains points on only one side, so a neighborhood in the interval is a clipped, one-sided version of an open interval in \(\mathbb{R}\). Near an endpoint that is not in the space, there is no point at which to impose an endpoint condition.

Throughout, \(a<b\) are real numbers and the intervals have the subspace topology inherited from \(\mathbb{R}\) with its standard topology. Thus, for an interval \(X\subseteq\mathbb{R}\), a set \(U\subseteq X\) is open in \(X\) when \(U=G\cap X\) for some open set \(G\) in \(\mathbb{R}\). We call such a set relatively open. An open neighborhood in \(X\) of \(x\in X\) is a relatively open set containing \(x\).

This test is local: it checks each point of \(U\) separately. The intersection with \(X\) is essential. It discards the parts of the real-line interval that are not in the space.

The Endpoint Neighborhoods

Proof. Suppose first that \(U\) is open in \(X\), so \(U=G\cap X\) for an open \(G\subseteq\mathbb{R}\). For any \(x\in U\), we have \(x\in G\). Since \(G\) is open in \(\mathbb{R}\), there is an \(\varepsilon>0\) with \((x-\varepsilon,x+\varepsilon)\subseteq G\). Intersecting with \(X\) gives \((x-\varepsilon,x+\varepsilon)\cap X\subseteq G\cap X=U\).

Conversely, suppose the stated condition holds at every \(x\in U\). For each such \(x\), choose an interval \((x-\varepsilon_x,x+\varepsilon_x)\) whose intersection with \(X\) is contained in \(U\). Then \[ U=\bigcup_{x\in U}\bigl((x-\varepsilon_x,x+\varepsilon_x)\cap X\bigr). \] The right side is a union of sets open in the subspace topology, so \(U\) is open in \(X\). If \(U=\varnothing\), the union is empty and is open as well. This proves both directions, including the empty-set case. \(\square\)

At an included endpoint, the test takes a particularly simple form. For \(0<\varepsilon\le b-a\), the relevant intersections are \[ (a-\varepsilon,a+\varepsilon)\cap[a,b]=[a,a+\varepsilon),\qquad (b-\varepsilon,b+\varepsilon)\cap[a,b]=(b-\varepsilon,b]. \] The same formulas apply at \(a\) in \([a,b)\) and at \(b\) in \((a,b]\). The left endpoint of \([a,b)\) therefore has a right-sided neighborhood, while its missing endpoint \(b\) is not a point of the space. In \((a,b]\), it is \(b\) that has a left-sided neighborhood, and \(a\) is absent.

More explicitly, a relatively open subset of \([a,b]\) that contains \(a\) must contain \([a,a+\varepsilon)\) for some \(\varepsilon>0\); if it contains \(b\), it must contain \((b-\varepsilon,b]\) for some \(\varepsilon>0\). For \([a,b)\), only the condition at an included endpoint \(a\) applies; for \((a,b]\), only the condition at \(b\) applies. At every other point, use the interval-intersection test in the theorem. These are requirements for some positive \(\varepsilon\), not for every positive \(\varepsilon\).

Relative Openness Versus Openness in the Real Line

Proof. If \(U\) is open in \(\mathbb{R}\), then it is relatively open in \(X\), since \(U=U\cap X\). It cannot contain an endpoint of \(X\) that is included in \(X\): every real-line open interval around that endpoint also contains points outside \(X\), and hence outside \(U\).

For the converse, suppose \(U\) is relatively open and excludes every included endpoint of \(X\). Every \(x\in U\) then lies strictly between \(a\) and \(b\). By the relative-neighborhood test, some \(\varepsilon>0\) satisfies \((x-\varepsilon,x+\varepsilon)\cap X\subseteq U\). Choose \[ \delta=\min\left(\frac{\varepsilon}{2},\frac{x-a}{2},\frac{b-x}{2}\right)>0. \] Then \((x-\delta,x+\delta)\subseteq(a,b)\subseteq X\), and also \((x-\delta,x+\delta)\subseteq U\). Thus every point of \(U\) has a real-line open interval contained in \(U\), so \(U\) is open in \(\mathbb{R}\). For \(U=\varnothing\), the conclusion holds directly. This proves the stated criteria for all three intervals. \(\square\)

The proposition makes the distinction precise. A relatively open set can reach an included endpoint, provided it contains a one-sided segment there. A set open in \(\mathbb{R}\) cannot reach an included endpoint of any of these bounded intervals. The excluded endpoint imposes no restriction because it is not a point of the subspace.

These statements assume \(a<b\). In the degenerate case \(a=b\), the interval \([a,a]\) consists of the single point \(a\), and its subspace topology has just the open sets \(\varnothing\) and \(\{a\}\). Both \([a,a)\) and \((a,a]\) are empty. In particular, \(\{a\}\) is relatively open in \([a,a]\) but is not open in \(\mathbb{R}\).

A Common Endpoint Mistake

Common pitfall. The claim “\([0,1)\) is not open because it contains \(0\)” uses the wrong space for the neighborhood test. It is correct that \([0,1)\) is not open in \(\mathbb{R}\), but it is open in \(X=[0,2]\). Indeed, \((-1,1)\) is open in \(\mathbb{R}\), and \[ (-1,1)\cap[0,2]=[0,1). \] At \(0\), the relative neighborhood \([0,1/2)\) lies inside the set. The correction is to name the space first, then intersect real-line neighborhoods with that space. For a relative test, points outside \(X\) do not need to be included.

One practical reason to keep this distinction is that the same subset can have different openness answers depending on its ambient space. Endpoint calculations make that dependence explicit instead of relying on a sketch or on the appearance of brackets in interval notation.