A Continuous Bijection from a Half-Open Interval to the Circle

The Interval and the Circle

Angles provide a natural way to describe the unit circle: as an angle increases from \(0\) to \(2\pi\), the corresponding point travels once around the circle. At the end of this trip the point is back where it started. Restricting the angle to be strictly less than \(2\pi\) removes that duplicate, producing a bijection. The resulting map is continuous, but its inverse has a discontinuity at the point where the trip began and ended.

Give \([0,1)\) the subspace topology inherited from \(\mathbb{R}\) with its standard topology. Give \(\mathbb{R}^2\) its Euclidean topology, and give the unit circle the subspace topology from \(\mathbb{R}^2\). Thus \(S^1=\{(x,y)\in\mathbb{R}^2:x^2+y^2=1\}\). Consider the map

The trigonometric identity \(\cos^2 u+\sin^2 u=1\) shows that \(f(t)\in S^1\) for every \(t\in[0,1)\), so the formula indeed defines a map into the circle. A homeomorphism is a bijection that is continuous and whose inverse is also continuous. We will show that \(f\) has the first two properties but not the third.

Continuity of the Parametrization

Proof. First consider the same formula as a map \(F:[0,1)\to\mathbb{R}^2\). For \(s,t\in[0,1)\), the mean value theorem and the bounds \(|\sin u|\leq 1\), \(|\cos u|\leq 1\) give \(|\cos(2\pi t)-\cos(2\pi s)|\leq 2\pi|t-s|\) and \(|\sin(2\pi t)-\sin(2\pi s)|\leq 2\pi|t-s|\). Therefore

so \(\|F(t)-F(s)\|\leq 2\pi\sqrt{2}|t-s|\). Given \(\varepsilon>0\), choosing \(\delta=\varepsilon/(2\pi\sqrt{2})\) ensures that \(|t-s|<\delta\) implies \(\|F(t)-F(s)\|<\varepsilon\). Thus \(F\) is continuous. To regard \(F\) as a map into \(S^1\), let \(O\) be open in \(S^1\). By the subspace topology, \(O=S^1\cap W\) for some open \(W\subseteq\mathbb{R}^2\). Since \(F(t)\in S^1\), \(f^{-1}(O)=F^{-1}(W)\), which is open in \([0,1)\) by continuity of \(F\). This proves that \(f\) is continuous. \(\square\)

Why the Map Is a Bijection

Proof. For injectivity, suppose \(f(s)=f(t)\), where \(s,t\in[0,1)\). Write \(\alpha=2\pi s\) and \(\beta=2\pi t\). Equality of the two coordinates, together with the angle-difference identities, gives \(\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta=1\) and \(\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta=0\). The angles with cosine \(1\) and sine \(0\) are exactly the integer multiples of \(2\pi\); hence \(\alpha-\beta=2\pi k\) for some integer \(k\). Dividing by \(2\pi\) gives \(s-t=k\). Since \(s,t\in[0,1)\), we have \(-1<s-t<1\). The only integer in that interval is \(0\), so \(s=t\).

For surjectivity, take any \((x,y)\in S^1\), so \(x^2+y^2=1\). If \(y=0\), then \(x=1\) or \(x=-1\); the point is \(f(0)=(1,0)\) in the first case and \(f(1/2)=(-1,0)\) in the second. If \(y>0\), let \(\theta=\arccos x\), where \(\arccos x\in[0,\pi]\). Then \(\cos\theta=x\), and \(\sin\theta=\sqrt{1-x^2}=y\), using \(x^2+y^2=1\) and \(y>0\). If \(y<0\), let \(\theta=2\pi-\arccos x\). Then \(\cos\theta=x\) and \(\sin\theta=-\sqrt{1-x^2}=y\). In both cases with \(y\ne0\), \(0<\theta<2\pi\), so \(t=\theta/(2\pi)\) belongs to \([0,1)\) and \(f(t)=(x,y)\). The two cases \(y=0\) were already covered, so every point of \(S^1\) is in the image. Thus \(f\) is both injective and surjective. \(\square\)

The Inverse Is Not Continuous

Since \(f\) is a bijection, its inverse \(f^{-1}:S^1\to[0,1)\) exists as a function. It assigns to each point its unique parameter in the half-open interval. At \((1,0)\), that parameter is \(0\). But points on the circle just before the parameter reaches \(1\) are close to \((1,0)\), while their parameters are close to \(1\), not to \(0\). The following open-set test makes this failure precise.

Proof. Let \(V=[0,1/2)\). This is open in \([0,1)\), since \(V=(-1,1/2)\cap[0,1)\), where \((-1,1/2)\) is open in \(\mathbb{R}\). Because \(f\) is bijective, \((f^{-1})^{-1}(V)=f(V)\). The point \((1,0)=f(0)\) belongs to \(f(V)\).

For each integer \(n\geq3\), set \(p_n=f(1-1/n)\). The parameter \(1-1/n\) belongs to \([0,1)\) and is strictly greater than \(1/2\), so injectivity implies \(p_n\notin f(V)\). On the other hand, \(p_n\to(1,0)\). Indeed, using the trigonometric identities for angles differing from \(2\pi\),

Thus \(\|p_n-(1,0)\|=2\sin(\pi/n)\), which tends to \(0\) as \(n\to\infty\). Every open neighborhood of \((1,0)\) in \(S^1\) contains \(p_n\) for all sufficiently large \(n\): it contains the intersection of \(S^1\) with some Euclidean open ball centered at \((1,0)\), and the distances just computed eventually fall below that ball's radius. Since all these \(p_n\) lie outside \(f(V)\), no open neighborhood of \((1,0)\) is contained in \(f(V)\). Therefore \(f(V)\) is not open in \(S^1\). Continuity of \(f^{-1}\) would require the preimage of every open set, including \(V\), to be open. This contradiction proves that \(f^{-1}\) is not continuous at \((1,0)\). \(\square\)

What the Half-Open Endpoint Accomplishes

The choice of domain determines whether the parametrization is one-to-one. If the formula were instead applied to the closed interval \([0,1]\), then \(f(0)=(1,0)\) and \(f(1)=(\cos 2\pi,\sin 2\pi)=(1,0)\), so the map would not be injective. On \([0,1)\), the input \(1\) is excluded, and the injectivity proof shows that no two allowed parameters give the same point.

Common pitfall. A tempting but incorrect calculation is to say, “The map is not injective because \(f(0)=f(1)\).” The calculation of the two formula values is correct, but \(1\notin[0,1)\), so it does not compare two elements of the stated domain. The half-open interval is precisely what removes this duplicate. A different mistake is to infer that the inverse must be continuous just because the map is a continuous bijection. The neighborhood test above disproves that inference: the open set \([0,1/2)\) has a preimage under \(f^{-1}\) that is not open in the circle.

This example isolates the role of each condition in a homeomorphism. The formula gives a continuous map; the half-open domain makes it bijective; and the behavior near the omitted endpoint prevents the inverse from being continuous. The circle is not missing a point: rather, the parameter values approaching \(1\) approach the same circle point assigned parameter \(0\). The two spaces therefore have a continuous bijection between them, but this particular bijection does not identify their topologies by a homeomorphism.