Compactness of the Circle as a Continuous Image

From a Compact Interval to the Circle

The unit circle can be described by one real parameter that runs through a full turn. This gives a direct route to its compactness: start with the compact interval \(I=[0,1]\), map it continuously onto the circle, and transfer compactness to the image. The map need not be one-to-one; what matters is that it is continuous and reaches every point.

Give \(I=[0,1]\) its usual subspace topology from \(\mathbb{R}\), and give \(\mathbb{R}^2\) its Euclidean topology. The unit circle is the subset \(S^1=\{(x,y)\in\mathbb{R}^2:x^2+y^2=1\}\), with the subspace topology inherited from \(\mathbb{R}^2\). Compactness here means that every open cover has a finite subcover.

The formula does take values in \(S^1\), because \(\cos^2(2\pi t)+\sin^2(2\pi t)=1\) for every \(t\in I\). The main points to check are that \(\gamma\) is continuous and that it reaches every point of \(S^1\). We will also establish the compactness facts needed for the conclusion.

Compactness of the Interval

Proof. Let \(\mathcal{U}\) be an open cover of \(I\). Define \(E\) to be the set of all \(x\in I\) for which \([0,x]\) is covered by finitely many members of \(\mathcal{U}\). Since some member of the cover contains \(0\), we have \(0\in E\). Thus \(E\) is nonempty and bounded above by \(1\). Let \(c=\sup E\).

Choose a member \(U\in\mathcal{U}\) containing \(c\). Since \(U\) is open in \(I\), there is a \(\delta>0\) such that \((c-\delta,c+\delta)\cap I\subseteq U\). By the definition of supremum, there is \(s\in E\) with \(s>c-\delta/2\); if \(c=0\), take \(s=0\), which also satisfies this inequality. A finite collection from \(\mathcal{U}\) covers \([0,s]\). The set \(U\) covers every point from \(s\) to \(c\), because \(s>c-\delta\) and \([s,c]\subseteq(c-\delta,c+\delta)\cap I\). Hence \([0,c]\) has a finite subcover, so \(c\in E\).

If \(c<1\), choose \(\eta>0\) with \(\eta<\delta\) and \(c+\eta\leq1\). The same member \(U\) covers \([c,c+\eta]\), while a finite subcollection covers \([0,c]\). Their union is a finite cover of \([0,c+\eta]\), so \(c+\eta\in E\), contradicting that \(c\) is an upper bound for \(E\). Therefore \(c=1\). Since \(c\in E\), a finite subcollection covers \([0,1]\). This proves that every open cover of \(I\) has a finite subcover. \(\square\)

Why Continuous Images Preserve Compactness

Proof. Let \(\mathcal{V}\) be an open cover of \(f(K)\), where each member is open in the subspace topology. For each \(V\in\mathcal{V}\), the set \(f^{-1}(V)\) is open in \(K\): by continuity, it is the preimage of an open set in \(Y\) whose intersection with \(f(K)\) is \(V\). These preimages cover \(K\), since every \(k\in K\) has \(f(k)\in f(K)\) and some \(V\in\mathcal{V}\) contains \(f(k)\). Compactness of \(K\) gives finitely many members \(V_1,\ldots,V_n\) of \(\mathcal{V}\) whose preimages cover \(K\). For any \(y\in f(K)\), write \(y=f(k)\) for some \(k\in K\). Since the preimages cover \(K\), \(k\in f^{-1}(V_i)\) for some \(i\), and therefore \(y\in V_i\). Thus \(V_1,\ldots,V_n\) cover \(f(K)\), proving compactness. \(\square\)

This argument requires continuity so that the preimages of the open cover are open, and compactness of the domain so that those preimages have a finite subcover. It does not require \(f\) to be injective or to have an inverse.

Checking the Circle Parametrization

Proof. The functions \(t\mapsto\cos(2\pi t)\) and \(t\mapsto\sin(2\pi t)\) are continuous on \(I\), so the coordinate map \(t\mapsto(\cos(2\pi t),\sin(2\pi t))\) is continuous as a map into \(\mathbb{R}^2\). Its values lie in \(S^1\). If \(O\) is open in \(S^1\), then \(O=S^1\cap W\) for some open \(W\subseteq\mathbb{R}^2\). Since \(\gamma(t)\in S^1\), the preimage of \(O\) under \(\gamma\) equals the preimage of \(W\) under the coordinate map. That preimage is open in \(I\), so \(\gamma\) is continuous as a map into \(S^1\).

To prove surjectivity, take any \((x,y)\in S^1\), so \(x^2+y^2=1\). Let \(\alpha=\arccos(x)\), where \(\alpha\in[0,\pi]\). Then \(\cos\alpha=x\) and \(\sin\alpha=\sqrt{1-x^2}=|y|\). If \(y\geq0\), take \(t=\alpha/(2\pi)\in I\). Then \(\cos(2\pi t)=x\) and \(\sin(2\pi t)=y\). If \(y<0\), take \(t=1-\alpha/(2\pi)\in I\). The identities \(\cos(2\pi-\alpha)=\cos\alpha\) and \(\sin(2\pi-\alpha)=-\sin\alpha\) give \(\cos(2\pi t)=x\) and \(\sin(2\pi t)=y\). In either case \(\gamma(t)=(x,y)\), so every point of \(S^1\) is attained. \(\square\)

The Compactness Conclusion

Proof. The interval \(I\) is compact by the open-cover argument above. The map \(\gamma:I\to S^1\) is continuous and surjective, so \(\gamma(I)=S^1\). The continuous-image theorem therefore implies that \(S^1\) is compact. In open-cover terms, an open cover of \(S^1\) pulls back to an open cover of \(I\); a finite collection of those preimages covers \(I\), and the corresponding finite collection of original sets covers \(S^1\) because every point of \(S^1\) is \(\gamma(t)\) for some \(t\in I\). \(\square\)

Common pitfall. One might assume that \(\gamma\) must be a continuous bijection to transfer compactness. It is not injective: direct substitution gives \(\gamma(0)=(\cos 0,\sin 0)=(1,0)\) and \(\gamma(1)=(\cos 2\pi,\sin 2\pi)=(1,0)\), although \(0\ne1\). This causes no problem: the continuous-image theorem requires a continuous map, not a bijection. Another error is to omit the factor \(2\pi\) and use \((\cos t,\sin t)\) for \(t\in[0,1]\); that only traces angles from \(0\) to \(1\) radian, not a full turn. The \(2\pi\) factor is what makes the stated map surjective.

The argument illustrates a useful role of parametrizations in topology. A space can be shown compact by realizing it as the continuous image of a familiar compact space, even when the map identifies some points. Here the interval traverses the circle and identifies its two endpoints, while still reaching the entire circle.