The Circle's Relative Topology in the Plane

From Plane Disks to Circle Neighborhoods

The unit circle inherits its topology from the plane: a set is open in the circle when it is the intersection of the circle with an open set in \(\mathbb{R}^2\). This means that an ordinary open disk in the plane cuts out a relatively open part of the circle. The shape of that intersection can be calculated exactly, and the calculation helps distinguish open arcs from closed and half-open arcs.

Throughout, \(\mathbb{R}^2\) has its Euclidean topology and Euclidean distance, and \(S^1=\{(x,y)\in\mathbb{R}^2:x^2+y^2=1\}\) has the subspace topology. Write \(B(c,r)=\{z\in\mathbb{R}^2:\|z-c\|<r\}\) for the open disk with center \(c\) and radius \(r>0\). A relative open neighborhood of \(p\in S^1\) is a relatively open subset of \(S^1\) containing \(p\). A neighborhood, without the word “open,” may be any set that contains a relative open neighborhood of \(p\).

Open disks suffice for testing relative openness locally. Indeed, if \(O\) is open in the plane and \(p\in O\), the definition of the Euclidean topology gives some \(r>0\) with \(B(p,r)\subseteq O\). Thus \(S^1\cap B(p,r)\subseteq S^1\cap O\). In particular, if \(V\) is a neighborhood of \(p\) (not necessarily relatively open), first choose a relative open set \(W\) with \(p\in W\subseteq V\). Write \(W=S^1\cap O\), then choose \(r>0\) with \(B(p,r)\subseteq O\). It follows that \(p\in S^1\cap B(p,r)\subseteq V\). This is the precise sense in which disk intersections provide small relative neighborhoods.

What a Disk Centered on the Circle Cuts Out

We can calculate the intersection when the disk is centered at a point \(p\) of the circle. For two points on \(S^1\), let their angular distance be the smaller of the two distances around the circle, a number in \([0,\pi]\). If this angular distance is \(\delta\), their Euclidean distance is \(2\sin(\delta/2)\). To verify the formula, rotate coordinates so that one point has angle \(0\) and the other angle \(\delta\). Their squared distance is \((\cos\delta-1)^2+\sin^2\delta=2-2\cos\delta=4\sin^2(\delta/2)\), so their distance is \(2\sin(\delta/2)\).

Proof. Let \(q\in S^1\), and let \(\delta\in[0,\pi]\) be its angular distance from \(p\). The distance formula gives \(\|q-p\|=2\sin(\delta/2)\). Since \(0\leq\delta/2\leq\pi/2\), the sine function is strictly increasing on this interval. If \(0<r<2\), then \(q\in B(p,r)\) exactly when \(2\sin(\delta/2)<r\), equivalently \(\delta<2\arcsin(r/2)\). If \(r=2\), the inequality \(2\sin(\delta/2)<2\) holds exactly when \(\delta<\pi\), so it excludes just the antipodal point \(-p\). If \(r>2\), every point satisfies \(\|q-p\|\leq2<r\), so the intersection is the whole circle. \(\square\)

For \(0<r<2\), the theorem describes an open arc centered at \(p\), with its two endpoints excluded because the disk is open. The strict inequality matters: a point at distance exactly \(r\) is on the boundary of the disk and is not in it.

Open and Closed Arcs in the Relative Topology

For real numbers \(\alpha<\beta<\alpha+2\pi\), define the open arc and closed arc with these angular endpoints by

The restriction \(\beta-\alpha<2\pi\) ensures that the arc is not the entire circle. In the next result, “open” and “closed” mean open and closed as subsets of \(S^1\), not as subsets of the plane.

Proof. First consider \(p=(\cos t,\sin t)\in A(\alpha,\beta)\). Choose \(\eta>0\) smaller than each of \(t-\alpha\), \(\beta-t\), and \(\pi\). The disk \(B(p,2\sin(\eta/2))\) intersects \(S^1\) in points at angular distance less than \(\eta\) from \(p\), by the disk-intersection formula. Every such point has an angle, modulo \(2\pi\), within \(\eta\) of \(t\); because \(\eta<t-\alpha\) and \(\eta<\beta-t\), that angle lies strictly between \(\alpha\) and \(\beta\). Thus this disk intersection is contained in \(A(\alpha,\beta)\). Each point of the arc has a relative open neighborhood inside it, so the arc is relatively open.

The complement of \(C(\alpha,\beta)\) in \(S^1\) is the open arc traversed from angle \(\beta\) to angle \(\alpha+2\pi\), with both endpoints excluded. It is relatively open by the argument just given, so \(C(\alpha,\beta)\) is relatively closed.

For a proper open arc, the endpoint with angle \(\alpha\) is not in the arc. Every disk centered at that endpoint has an intersection with \(S^1\) containing points with angles slightly greater than \(\alpha\), which lie in the arc. Hence every relative open neighborhood of that endpoint meets the arc, so the endpoint belongs to its relative closure; the open arc is not closed. For the closed arc, consider its included endpoint with angle \(\alpha\). Every disk centered there also contains circle points with angles slightly less than \(\alpha\), outside the closed arc. No relative open neighborhood of that endpoint is contained in the closed arc, so the closed arc is not open.

Finally, the half-open arc contains its endpoint at \(\alpha\), but every relative open neighborhood of that point contains circle points with angles slightly less than \(\alpha\), which are not in the half-open arc. It is not open. Its omitted endpoint at \(\beta\) lies in its relative closure, because every disk centered there contains circle points with angles slightly less than \(\beta\), which belong to the half-open arc. It is not closed. \(\square\)

Reading Arc Status Correctly

The words “open arc” and “closed arc” describe which endpoints are included in a particular parametrization; the proposition explains their topological status in \(S^1\). A proper open arc is relatively open but not relatively closed. A proper closed arc is relatively closed but not relatively open. A proper half-open arc is neither. The empty set and the whole circle are both relatively open and relatively closed, so the properness condition in the proposition matters.

A common mistake is to replace the strict disk condition \(\|q-p\|<r\) with \(\|q-p\|\leq r\). For the disk \(B((1,0),1)\), that incorrect calculation would include the points at angles \(\pm\pi/3\), since their distance is \(1\). But an open disk contains only points whose distance is strictly less than \(1\); those boundary points are excluded. Replacing an open disk by a closed disk changes the intersection and its endpoint status.

There is also a useful distinction between ambient and relative openness. The set \(B((1,0),1)\cap S^1\) is open in \(S^1\) because it is the intersection of \(S^1\) with an open plane disk. That does not assert that the arc itself is open in \(\mathbb{R}^2\). For relative questions, the ambient intersection is the relevant test.