Connectedness of the Topologist's Sine Curve

The Graph and Its Limiting Segment

The graph of \(y=\sin(1/x)\) becomes more and more oscillatory as \(x\) approaches \(0\). Although the formula is not defined at \(x=0\), the graph accumulates at every point of the vertical segment from height \(-1\) to height \(1\). The key to proving connectedness is to separate two tasks: prove that the graph itself is connected, then show that adjoining all its limiting points preserves connectedness.

All subsets of \(\mathbb{R}^2\) below have the subspace topology from the standard Euclidean topology. The interval \((0,1]\) has the subspace topology from \(\mathbb{R}\). Define the graph \(G\), the vertical segment \(V\), and the topologist’s sine curve \(S\) by

For a subset \(A\) of a space \(X\), its closure \(\overline{A}\) is the intersection of all closed subsets of \(X\) containing \(A\). We use the neighborhood test: \(x\in\overline{A}\) if and only if every open neighborhood of \(x\) meets \(A\). Indeed, if an open neighborhood \(U\) misses \(A\), then the closed set \(X\setminus U\) contains \(A\), so it contains \(\overline{A}\) and excludes \(x\). Conversely, if \(x\notin\overline{A}\), the open set \(X\setminus\overline{A}\) is a neighborhood of \(x\) disjoint from \(A\).

Why the Oscillating Graph Is Connected

The map \(f:(0,1]\to\mathbb{R}^2\) defined by \(f(x)=(x,\sin(1/x))\) is continuous: \(x\mapsto 1/x\) is continuous for \(x>0\), sine is continuous, and the two coordinates together define a continuous map into the Euclidean plane. Its image is exactly \(G\). We justify both facts about connectedness needed to conclude that \(G\) is connected.

Proof. It is enough to consider a nonempty interval \(J\). Suppose, for contradiction, that \(J=U\cup W\) is a separation. Choose one point from each set; after exchanging the names if necessary, there are \(u\in U\) and \(v\in W\) with \(u<v\). The set \(U\cap[u,v]\) is nonempty and bounded above, so let \(c=\sup(U\cap[u,v])\). Since \(U\) is relatively open and contains \(u\), it contains points of \(J\) strictly greater than \(u\) and sufficiently close to \(u\); hence \(c>u\). Since \(W\) is relatively open and contains \(v\), there is an interval immediately to the left of \(v\) containing no points of \(U\), so \(c<v\). Thus \(c\in J\), and it belongs to either \(U\) or \(W\). If \(c\in U\), relative openness gives points of \(J\cap(u,v)\) strictly greater than \(c\) that belong to \(U\), contradicting that \(c\) is an upper bound. If \(c\in W\), relative openness gives an interval around \(c\) whose points in \(J\) belong to \(W\). But the definition of supremum gives points of \(U\cap[u,v]\) arbitrarily close to \(c\) from below, a contradiction. Thus \(J\) has no separation. The empty interval is connected by the definition. \(\square\)

Proof. Let \(X\) be connected, let \(f:X\to Y\) be continuous, and give \(f(X)\) the subspace topology from \(Y\). Suppose \(f(X)=P\cup Q\) were a separation. Then \(f^{-1}(P)\) and \(f^{-1}(Q)\) are disjoint open subsets of \(X\): they are open by continuity, because \(P\) and \(Q\) are relatively open in \(f(X)\). They cover \(X\), and both are nonempty because every point of \(f(X)\) has a preimage. This contradicts connectedness of \(X\). Therefore \(f(X)\) is connected. \(\square\)

Every Height on the Segment Is a Limit

To show that the whole vertical segment belongs to the closure of \(G\), fix any \(y\in[-1,1]\). Let \(\arcsin(y)\) denote the value in \([-\pi/2,\pi/2]\) whose sine is \(y\). For \(n\in\mathbb{N}=\{1,2,\ldots\}\), set

Since \(t_n\ge2\pi-\pi/2=3\pi/2>1\), each \(x_n\) satisfies \(0<x_n<1\). Also, \(t_n\to\infty\), so \(x_n\to0\). Periodicity of sine gives \(\sin(1/x_n)=\sin(t_n)=y\). Thus \((x_n,y)\in G\) for every \(n\), and \((x_n,y)\to(0,y)\). Every open neighborhood of \((0,y)\) therefore meets \(G\), so the neighborhood test gives \((0,y)\in\overline{G}\).

The Closure of the Graph Is the Whole Curve

We have proved \(V\subseteq\overline{G}\), and \(G\subseteq\overline{G}\) by the definition of closure. This gives \(S\subseteq\overline{G}\). To establish equality, we also need to check that \(S\) is closed in \(\mathbb{R}^2\).

Proof. We show that every point \((a,b)\) outside \(S\) has an open neighborhood disjoint from \(S\). If \(a<0\) or \(a>1\), choose a sufficiently small open ball around \((a,b)\) whose first coordinates remain outside \([0,1]\); it misses both \(G\) and \(V\). If \(a\in[0,1]\) and \(b\notin[-1,1]\), choose a sufficiently small open ball whose second coordinates remain outside \([-1,1]\). This also misses \(G\) and \(V\), because the sine values and the heights in \(V\) all lie in \([-1,1]\). The only remaining possibility for a point outside \(S\) is \(0<a\le1\), \(b\in[-1,1]\), and \(b\ne\sin(1/a)\). On the open set \((0,\infty)\times\mathbb{R}\), the function \(h(x,y)=y-\sin(1/x)\) is continuous. Since \(h(a,b)\ne0\), the inverse image of \(\mathbb{R}\setminus\{0\}\) gives an open neighborhood of \((a,b)\) within that open set on which \(h\ne0\). This neighborhood misses \(G\), and it misses \(V\) because its first coordinates are positive. Thus the complement of \(S\) is open, so \(S\) is closed. \(\square\)

Since \(S\) is closed and contains \(G\), the definition of closure gives \(\overline{G}\subseteq S\). Combined with \(S\subseteq\overline{G}\), this proves \(\overline{G}=S\).

Proof. If \(A\) is empty, then \(\overline{A}\) is empty and is connected. Otherwise, suppose \(\overline{A}=P\cup Q\) were a separation. Each of \(P\) and \(Q\) is nonempty and relatively open in \(\overline{A}\), so each has the form \(O\cap\overline{A}\) for some open set \(O\) in \(X\). Any point \(p\in P\) lies in \(\overline{A}\), so the open set \(O\) defining \(P\) meets \(A\); because \(A\subseteq\overline{A}\), this gives \(P\cap A\ne\varnothing\). The same reasoning gives \(Q\cap A\ne\varnothing\). The sets \(P\cap A\) and \(Q\cap A\) are disjoint, relatively open in \(A\), nonempty, and cover \(A\). They would separate \(A\), contradicting its connectedness. Therefore \(\overline{A}\) is connected. \(\square\)

Apply Theorem 3 in \(\mathbb{R}^2\) to \(A=G\). The graph \(G\) is connected, and its closure is \(S\). Therefore the topologist’s sine curve \(S\) is connected.

A Common Mistake: Treating Oscillation as Membership

The fact that graph points approach \((0,1)\) does not mean \((0,1)\) belongs to \(G\). The definition of \(G\) requires \(x>0\); its formula does not assign a value at \(x=0\). For example, the sequence \(x_n=1/(2\pi n+\pi/2)\) has \(\sin(1/x_n)=1\) and \(x_n\to0\), so it proves that \((0,1)\) is in the closure. It does not make \((0,1)\) a point of the graph.

A related incomplete argument is to use just that sequence and conclude that the whole vertical segment is in the closure. That calculation establishes only the upper endpoint. For a general height \(y\in[-1,1]\), the corrected calculation uses \(x_n=1/(2\pi n+\arcsin(y))\); it gives \(\sin(1/x_n)=y\) for every \(n\) and \(x_n\to0\). The parameter \(y\) is arbitrary, so it proves the assertion for every point of \(V\).

The proof separates the roles of the two pieces: continuous images establish connectedness of the oscillating graph, while the closure calculation accounts for the missing limiting segment. Because taking the closure of a connected set preserves connectedness, no additional geometric intuition about the oscillations is needed.