The Topologist’s Sine Curve As A Connected Closure

The Oscillating Graph and Its Added Segment

Consider the graph of \(y=\sin(1/x)\) for \(0<x\le1\), viewed as a subset of the Euclidean plane. As \(x\) approaches \(0\), the graph oscillates ever more rapidly. Its values remain between \(-1\) and \(1\), and, in fact, every value in that interval occurs arbitrarily close to \(x=0\). The closure therefore adds a whole vertical segment, not just a single endpoint.

All subsets of \(\mathbb{R}^2\) below use the standard Euclidean topology, and subsets such as the graph inherit the subspace topology. Define the graph \(G\), the vertical segment \(V\), and their union \(T\) by

We will show that \(\overline{G}=T\). To use sequences in this computation, recall the metric-space closure test: a point lies in the closure of a set if and only if it is the limit of a sequence from that set. Here is the reason. If a point \(p\) lies in the closure, each ball of radius \(1/n\) about \(p\) meets the set, so choose a point of the set in each ball; the resulting sequence converges to \(p\). Conversely, if points of the set converge to \(p\), every open neighborhood of \(p\) contains all sufficiently late terms and so meets the set.

Identifying the Closure

Proof. First take \((0,y)\in V\), so \(-1\le y\le1\). Let \(\arcsin y\in[-\pi/2,\pi/2]\) be the angle whose sine is \(y\). For \(n\in\mathbb{N}\), set \(x_n=1/(2\pi n+\arcsin y)\). Since \(n\ge1\), the denominator is at least \(3\pi/2>1\), so \(0<x_n\le1\). Also \(x_n\to0\), and \(\sin(1/x_n)=\sin(2\pi n+\arcsin y)=y\). Thus \((x_n,\sin(1/x_n))\in G\) and these points converge to \((0,y)\). Every point of \(G\) already lies in \(\overline{G}\), so \(T\subseteq\overline{G}\).

For the reverse inclusion, let \((x,y)\in\overline{G}\). By the metric-space closure test, there is a sequence \((x_n,\sin(1/x_n))\) in \(G\) converging to \((x,y)\). Each \(x_n\) lies in \((0,1]\), so \(x\in[0,1]\); also \(-1\le\sin(1/x_n)\le1\) for every \(n\), so the limit satisfies \(-1\le y\le1\). If \(x=0\), these facts give \((x,y)\in V\). If \(x>0\), continuity of \(t\mapsto\sin(1/t)\) at \(x\) gives \(y=\sin(1/x)\), so \((x,y)\in G\). Hence \(\overline{G}\subseteq T\), proving the equality. \(\square\)

Compactness

A space is compact if every open cover has a finite subcover. The Heine–Borel theorem is the standard result that a subset of Euclidean space is compact if and only if it is closed and bounded. We use this theorem here; its proof is part of the standard earlier material on compactness in Euclidean space.

Proof. The closure of any set is closed: by definition it is an intersection of closed sets, and an intersection of closed sets is closed. The closure calculation gives \(T=\overline{G}\), so \(T\) is closed in \(\mathbb{R}^2\). Every \((x,y)\in T\) has \(0\le x\le1\) and \(-1\le y\le1\). Consequently \(T\subseteq[0,1]\times[-1,1]\), which is bounded in the Euclidean metric. By the Heine–Borel theorem, \(T\) is compact. \(\square\)

Compactness is a property of the topology, not just of a drawing or of the fact that the oscillating values stay in a bounded interval. The closure calculation supplies the closedness required for the compactness argument.

Connectedness

A space is connected if it cannot be written as the union of two disjoint, nonempty open subsets. A standard interval-connectedness theorem says that every interval in \(\mathbb{R}\) with its usual topology is connected. In particular, \((0,1]\) is connected. We will use that theorem and prove the other connectedness facts needed here.

Proof. Suppose instead that \(\overline{A}=P\cup Q\), where \(P\) and \(Q\) are disjoint, nonempty sets open in the subspace topology on \(\overline{A}\). Since \(P\) is relatively open, there is an open set \(U\subseteq X\) with \(P=U\cap\overline{A}\). Choose \(p\in P\). Because \(p\in\overline{A}\), the open neighborhood \(U\) of \(p\) meets \(A\). As \(A\subseteq\overline{A}\), every point of \(U\cap A\) lies in \(P\). Therefore \(A\cap P\ne\varnothing\). The same argument, starting with a point of \(Q\), gives \(A\cap Q\ne\varnothing\). Moreover, \(A\cap P\) and \(A\cap Q\) are disjoint, cover \(A\), and are open in the subspace topology on \(A\), since \(P\) and \(Q\) are open in \(\overline{A}\). They would separate \(A\), contradicting its connectedness. Hence \(\overline{A}\) is connected. \(\square\)

Proof. Define \(f:(0,1]\to\mathbb{R}^2\) by \(f(t)=(t,\sin(1/t))\). The map is continuous because both coordinate functions are continuous on \((0,1]\), and its image is \(G\). A continuous image of a connected space is connected: if the image had a separation into disjoint, nonempty relatively open sets, their preimages would be disjoint, nonempty open sets covering the domain, contradicting connectedness of the domain. The interval-connectedness theorem therefore implies that \(G\) is connected. By the closure calculation, \(T=\overline{G}\); the lemma now implies that \(T\) is connected. \(\square\)

A Common Pitfall: Bounded Does Not Mean Compact

A tempting but incorrect argument is: “The graph is bounded because \(0<x\le1\) and \(-1\le\sin(1/x)\le1\), so it is compact.” Boundedness alone does not imply compactness in \(\mathbb{R}^2\). In this case the graph is not even closed: the points \((1/(2\pi n),0)\) lie in \(G\) and converge to \((0,0)\), which is not in \(G\). The correction is to compute the closure, include the missing segment, and then apply Heine–Borel to the closed and bounded set \(T\).

The distinction also clarifies why the vertical segment is essential. It records all possible limiting sine values as the first coordinate tends to \(0\). The graph alone omits those boundary limits; its closure includes them, and that closure is the compact connected set proved above.