Open Caps and Neighborhoods on the Sphere

The Sphere with Its Relative Topology

On a sphere, a neighborhood can be described using ordinary Euclidean balls in the surrounding three-dimensional space. Intersecting such a ball with the sphere gives a neighborhood on the sphere, even though the intersection is generally a curved patch rather than a three-dimensional open set. This description lets us test relative openness and calculate the shape of many spherical caps by comparing distances.

Let \(\mathbb{R}^3\) have its Euclidean topology and norm \(\|x\|=\sqrt{x_1^2+x_2^2+x_3^2}\). The unit sphere is \(S^2=\{x\in\mathbb{R}^3:\|x\|=1\}\), equipped with the subspace topology inherited from \(\mathbb{R}^3\). Thus a set \(U\subseteq S^2\) is relatively open if there is an open set \(O\subseteq\mathbb{R}^3\) such that \(U=S^2\cap O\). For \(p\in\mathbb{R}^3\) and \(r>0\), write \(B(p,r)=\{x\in\mathbb{R}^3:\|x-p\|<r\}\).

Ambient Balls Give a Neighborhood Basis

Proof. The Euclidean ball \(B(p,r)\) is open in \(\mathbb{R}^3\). Its intersection with \(S^2\) is therefore relatively open by the definition of the subspace topology, and it contains \(p\) because \(\|p-p\|=0<r\). Now let \(U\) be any relatively open neighborhood of \(p\). There is an open set \(O\subseteq\mathbb{R}^3\) with \(U=S^2\cap O\). Since \(p\in U\), we have \(p\in O\). By Euclidean openness, some \(r>0\) satisfies \(B(p,r)\subseteq O\). Intersecting with \(S^2\) gives \(S^2\cap B(p,r)\subseteq U\). This proves both claims. \(\square\)

The theorem gives a local test for relative neighborhoods: it is enough to consider intersections with balls centered at the point in question. The radius can be made smaller to fit inside any given relatively open neighborhood. It is important that the ball is open and that the intersection is taken with the sphere. The ambient ball itself is not a subset of \(S^2\).

Caps as Intersections with Balls

For \(u,x\in S^2\), the distance between them is related to their dot product by expanding the square of the norm:

Consequently, for \(c\leq 1\), the cap \(C(u,c)\) is exactly the portion of the sphere inside a ball centered at \(u\). Indeed, \(x\cdot u>c\) is equivalent to \(\|x-u\|^2<2-2c\), and both sides compare nonnegative quantities. In particular, \(C(u,c)=S^2\cap B(u,\sqrt{2-2c})\). This distance formula also proves relative openness directly.

Proof. If \(c\leq1\), the distance identity shows that \(x\in C(u,c)\) if and only if \(x\in S^2\) and \(\|x-u\|<\sqrt{2-2c}\). The square root is defined because \(2-2c\geq0\), so the ball description follows; when \(c=1\), the radius is zero and the open ball is empty. The intersection of an open ball with \(S^2\) is relatively open. If \(c>1\), then \(x\cdot u\leq\|x\|\|u\|=1\) for every \(x\in S^2\), by the Cauchy–Schwarz inequality. Thus no \(x\) can satisfy \(x\cdot u>c\), so \(C(u,c)=\varnothing\), which is relatively open. \(\square\)

For \(-1<c<1\), the cap is nonempty and proper: it contains \(u\), since \(u\cdot u=1>c\), and it excludes \(-u\), since \((-u)\cdot u=-1<c\). At the extreme thresholds, \(c=-1\) gives the sphere with the antipodal point \(-u\) removed, while \(c<-1\) gives all of \(S^2\). These edge cases follow from the same dot-product bounds and use the strict inequality in the cap definition.

What an Ambient Ball Cuts Out

The center of an ambient ball need not lie on the sphere. If \(a\in\mathbb{R}^3\), \(r>0\), and \(x\in S^2\), expansion gives \(\|x-a\|^2=1-2x\cdot a+\|a\|^2\). Hence membership in \(B(a,r)\) is equivalent to \(x\cdot a>(1+\|a\|^2-r^2)/2\). When \(a\ne0\), this is a cap condition along the unit direction \(a/\|a\|\), after dividing the inequality by the positive number \(\|a\|\). If \(a=0\), the intersection is all of \(S^2\) when \(r>1\), and empty when \(r\leq1\), since every point of the sphere has distance exactly \(1\) from the origin.

Finding a Smaller Neighborhood Inside a Cap

The basis theorem applies at any point of a cap, not just at its axis. The next example carries out the containment calculation explicitly. It illustrates how an ambient ball centered at a point of the sphere can be chosen small enough to stay inside a larger relative neighborhood.

A Common Mistake: Replacing “Greater Than” by “Greater Than or Equal To”

The strict inequality in the cap definition matters. For example, the set \(H=\{x\in S^2:x_3\geq0\}\) is a closed hemisphere, not an open cap. A tempting but incorrect calculation is to identify \(H\) with \(S^2\cap B(e_3,\sqrt{2})\): the ball condition gives \(\|x-e_3\|^2<2\), which is equivalent to \(x_3>0\), not \(x_3\geq0\). Points on the equator have distance exactly \(\sqrt{2}\) from \(e_3\), so the open ball excludes them.

In fact, \(H\) is not relatively open. Let \(q=(1,0,0)\), which belongs to \(H\). For any \(r>0\), choose \(t\) with \(0<t<\min(r,\pi/2)\), and set \(y=(\cos t,0,-\sin t)\). Then \(\|y\|^2=\cos^2t+\sin^2t=1\), so \(y\in S^2\), while \(y_3=-\sin t<0\), so \(y\notin H\). Moreover \(\|y-q\|=\sqrt{2-2\cos t}=2\sin(t/2)<t<r\). Thus every ambient ball centered at \(q\) contains a point of the sphere outside \(H\); by the neighborhood-basis theorem, no relatively open neighborhood of \(q\) is contained in \(H\). The strict cap \(C(e_3,0)=\{x\in S^2:x_3>0\}\), by contrast, is relatively open.

These calculations show why ambient balls are useful on the sphere: they give a concrete neighborhood basis and convert many geometric-looking conditions into exact inequalities. When checking a cap or a ball section, keep track of whether the boundary is included. An open ball produces a strict inequality, while replacing it by a non-strict inequality can change a relatively open set into a set that is not open.