The Topologist's Sine Curve as a Subspace

The Graph and Its Closure

The function $x\mapsto\sin(1/x)$ oscillates more and more rapidly as $x$ approaches $0$ from the right. Its graph has no points with first coordinate $0$, but points on the graph can approach many different heights above $0$. The topologist’s sine curve records precisely what is added when we take the closure of this graph in the plane.

Throughout, $\mathbb{R}^2$ has its Euclidean topology, determined by the metric $d((x,y),(u,v))=\sqrt{(x-u)^2+(y-v)^2}$. A subset of $\mathbb{R}^2$, including the topologist’s sine curve, is given the subspace topology: its open sets are intersections of the subset with open sets in $\mathbb{R}^2$.

Definition: Let $G=\{(x,\sin(1/x)):0<x\leq1\}\subseteq\mathbb{R}^2$, the graph of $\sin(1/x)$ on the indicated domain. The topologist’s sine curve is the subspace $S=\overline{G}$ of $\mathbb{R}^2$, where $\overline{G}$ denotes the closure of $G$ in the Euclidean plane.

There are two distinct sets to keep track of: $G$, which consists only of the graph points with positive first coordinate, and $S$, which may also contain limit points not in $G$. To find $S$, we will identify exactly which points of the plane can be limits of sequences from $G$.

Testing Whether a Point Lies in the Closure

For a set $A$ in a topological space $X$, its closure $\overline{A}$ is the intersection of all closed subsets of $X$ that contain $A$. In a metric space, this definition gives a useful local test.

Theorem (Neighborhood test): A point $p$ belongs to $\overline{A}$ if and only if every open neighborhood of $p$ intersects $A$.

Proof. Suppose $p\in\overline{A}$. If an open neighborhood $U$ of $p$ did not intersect $A$, then $A\subseteq X\setminus U$. The set $X\setminus U$ is closed, so the definition of closure would give $\overline{A}\subseteq X\setminus U$, contradicting $p\in U$. Conversely, suppose $p\notin\overline{A}$. By the definition of closure as an intersection, some closed set $C$ contains $A$ but does not contain $p$. Then $X\setminus C$ is an open neighborhood of $p$ disjoint from $A$. Thus the neighborhood condition fails. $\square$

Theorem (Sequence test for metric spaces): If $(X,d)$ is a metric space and $A\subseteq X$, then $p\in\overline{A}$ if and only if there is a sequence $(a_n)$ in $A$ such that $a_n\to p$.

Proof. Suppose $p\in\overline{A}$. For each $n\in\mathbb{N}$, the open ball $B(p,1/n)$ intersects $A$ by the neighborhood test. Choose $a_n\in A\cap B(p,1/n)$. Then $d(a_n,p)<1/n$, so for every $\varepsilon>0$, choosing $N>1/\varepsilon$ gives $d(a_n,p)<\varepsilon$ whenever $n\geq N$. Hence $a_n\to p$. Conversely, suppose $a_n\in A$ and $a_n\to p$. Every open neighborhood $U$ of $p$ contains a ball $B(p,r)$ for some $r>0$. Convergence gives an $N$ with $d(a_n,p)<r$ for all $n\geq N$, so $a_N\in U\cap A$. Every open neighborhood of $p$ therefore intersects $A$, and the neighborhood test gives $p\in\overline{A}$. $\square$

Identifying the Added Segment

Theorem: In the Euclidean plane, $$ \overline{G}=G\cup\bigl(\{0\}\times[-1,1]\bigr). $$

Proof. First take any $(x,y)\in\overline{G}$. By the sequence test, there is a sequence $(x_n,\sin(1/x_n))$ in $G$ converging to $(x,y)$. Since $0<x_n\leq1$ for every $n$, taking limits gives $0\leq x\leq1$. Also, $-1\leq\sin(1/x_n)\leq1$ for every $n$, so taking limits gives $-1\leq y\leq1$.

If $x>0$, then $x_n\to x$ and the function $t\mapsto\sin(1/t)$ is continuous at $x$. Consequently, $y=\lim_{n\to\infty}\sin(1/x_n)=\sin(1/x)$, so $(x,y)\in G$. If $x=0$, the bounds on $y$ show that $(x,y)\in\{0\}\times[-1,1]$. This proves that $\overline{G}$ is contained in the stated union.

For the reverse inclusion, every point of $G$ belongs to $\overline{G}$, since every set is contained in its closure. Now fix any $y\in[-1,1]$, and let $\theta=\arcsin(y)$, so $-\pi/2\leq\theta\leq\pi/2$ and $\sin\theta=y$. For $n\in\mathbb{N}$, set $x_n=1/(2\pi n+\theta)$. The denominator is positive: $2\pi n+\theta\geq2\pi-\pi/2=3\pi/2>0$. It is also greater than $1$, so $0<x_n\leq1$. Thus $(x_n,\sin(1/x_n))\in G$, and $$ \sin(1/x_n)=\sin(2\pi n+\theta)=\sin\theta=y. $$ As $n\to\infty$, $x_n\to0$, and hence $(x_n,\sin(1/x_n))\to(0,y)$. By the sequence test, $(0,y)\in\overline{G}$. This holds for every $y\in[-1,1]$, proving the reverse inclusion and the theorem. $\square$

The formula has two parts. For $x>0$, the closure contains exactly the original graph points. At $x=0$, it contains every height between $-1$ and $1$, inclusive. There are no points at heights outside that interval, because the sine function always takes values in $[-1,1]$.

Worked Example: Approaching the Top and Bottom of the Segment

We construct graph sequences converging to the two endpoint heights of the segment above $x=0$.

For the height $1$, take $x_n=1/(\pi/2+2\pi n)$, where $n\in\mathbb{N}$. The denominator is at least $\pi/2+2\pi=5\pi/2>1$, so $0<x_n\leq1$. Also, $$ \sin(1/x_n)=\sin(\pi/2+2\pi n)=1, $$ and $x_n\to0$. Thus the graph points $(x_n,1)$ converge to $(0,1)$.

For the height $-1$, take $u_n=1/(3\pi/2+2\pi n)$. The denominator is at least $3\pi/2+2\pi=7\pi/2>1$, so $0<u_n\leq1$. Now $$ \sin(1/u_n)=\sin(3\pi/2+2\pi n)=-1, $$ and $u_n\to0$. Hence $(u_n,-1)\to(0,-1)$.

Check: the first sequence has second coordinate $1$ at every index, and the second has second coordinate $-1$ at every index. Their first coordinates tend to $0$, so the limits are exactly the two claimed endpoints.

Worked Example: A Point at an Interior Height

To approach $(0,1/2)$, use $\arcsin(1/2)=\pi/6$ and define $x_n=1/(2\pi n+\pi/6)$. For $n\geq1$, the denominator is at least $2\pi+\pi/6=13\pi/6>1$, so $0<x_n\leq1$. At each index, $$ \sin(1/x_n)=\sin(2\pi n+\pi/6)=1/2. $$ The denominator $2\pi n+\pi/6$ tends to infinity, so $x_n\to0$, and therefore $(x_n,\sin(1/x_n))\to(0,1/2)$. This verifies directly that $(0,1/2)$ belongs to the closure, even though it is not a point of $G$.

Check: substituting the limit coordinates into the closure formula gives first coordinate $0$ and second coordinate $1/2\in[-1,1]$, in agreement with the sequence calculation.

Worked Example: Checking Several Points

Consider $(1,\sin 1)$, $(1,0)$, $(0,2)$, $(-1/4,0)$, and $(0,-1/2)$.

The point $(1,\sin 1)$ belongs to $G$, because $0<1\leq1$ and its second coordinate is $\sin(1/1)=\sin1$. The point $(1,0)$ is not in the closure: the closure formula would require its second coordinate to equal $\sin1$, and $\sin1>0$ since $0<1<\pi$. The point $(0,2)$ is excluded because $2\notin[-1,1]$. The point $(-1/4,0)$ is excluded because its first coordinate is negative, whereas every point in the closure has first coordinate in $[0,1]$. Finally, $(0,-1/2)$ belongs to the closure. For a sequence witnessing this, take $v_n=1/(2\pi n-\pi/6)$. Its denominator is at least $2\pi-\pi/6=11\pi/6>1$, so $0<v_n\leq1$; moreover $\sin(1/v_n)=\sin(2\pi n-\pi/6)=-1/2$ and $v_n\to0$.

Check: the closure formula classifies these points, in order, as included, excluded, excluded, excluded, and included. The sequence for the final point verifies its inclusion independently.

Common Pitfall: Confusing the Graph with Its Closure

A common mistake is to reason that because $x=0$ is not in the domain, the closure cannot contain any point whose first coordinate is $0$. That confuses membership in the graph with membership in its closure. The explicit sequences above show that graph points can converge to points absent from the graph itself.

The opposite overreach is to add the entire vertical line $\{0\}\times\mathbb{R}$. The sine values never leave $[-1,1]$, and the closure theorem proves that heights outside this interval are excluded. For instance, the open ball of radius $1/2$ about $(0,2)$ misses $G$: every point $(x,\sin(1/x))\in G$ has vertical coordinate at most $1$, so its Euclidean distance from $(0,2)$ is at least $|2-\sin(1/x)|\geq1$. Thus $(0,2)\notin\overline{G}$.

The defining formula is useful because it converts an informal description of rapid oscillation into a precise membership test. A point with positive first coordinate belongs to the topologist’s sine curve exactly when it lies on the graph; a point with first coordinate $0$ belongs exactly when its height is between $-1$ and $1$. This description concerns the subspace as a set and its inherited topology; it does not require any conclusion about connectedness or paths.

Check Your Understanding

Use the closure formula and, where requested, construct or analyze a sequence in $G$.

Does $(0,1)$ belong to $G$, to $\overline{G}$, to both, or to neither?
Construct a sequence of points in $G$ converging to $(0,-1/2)$.
Which points of the form $(0,y)$ belong to $\overline{G}$? Include the endpoint cases.
Explain why $(0,3/2)$ is not in $\overline{G}$, using either the closure formula or an open ball.
For $0<x\leq1$, give the exact condition on $y$ for $(x,y)\in\overline{G}$.

The Topologist's Sine Curve as a Subspace

What You'll Learn