Path Components of the Topologist's Sine Curve

The Curve and Its Two Pieces

The topologist’s sine curve combines a vertical segment with a graph that oscillates infinitely often as it approaches that segment. The two pieces lie close together, and points on the graph accumulate on the vertical segment. Yet proximity does not guarantee that a continuous path can travel from one piece to the other. We will construct paths within each piece and prove that a path cannot cross between them.

We use the standard topology on $\mathbb{R}^2$ and its subspace topology on the set $T$ defined below. A path in $T$ is a continuous map $\gamma:[0,1]\to T$, where $[0,1]$ has its standard topology. A subset is path-connected if each pair of its points can be joined by a path whose image lies in that subset. The path component of a point consists of all points that can be joined to it by paths. Constant paths, reversed paths, and concatenated paths show that being joined by a path is an equivalence relation.

Definition: The topologist’s sine curve considered here is $$ T=V\cup G,\qquad V=\{(0,y):-1\le y\le 1\},\qquad G=\{(x,\sin(1/x)):0<x\le 1\}. $$ The set $V$ is the vertical segment, and $G$ is the oscillating graph. They are disjoint because the first coordinate is zero on $V$ and positive on $G$.

The subspace topology matters: continuity of a path into $T$ is continuity into the plane with all its values lying in $T$. In particular, if $\gamma(t)=(u(t),v(t))$ is a path in $T$, then its coordinate functions $u$ and $v$ are continuous real-valued functions, since the coordinate projections from $\mathbb{R}^2$ are continuous.

Paths Within Each Piece

Proposition: The vertical segment $V$ and the graph $G$ are each path-connected.

Proof. Take points $(0,a),(0,b)\in V$, so $-1\le a,b\le1$. The map $$ \gamma(t)=(0,(1-t)a+tb),\qquad 0\le t\le1, $$ is continuous, starts at $(0,a)$, and ends at $(0,b)$. Since $0\le t\le1$, the second coordinate is a weighted average of $a$ and $b$, so it lies between them and therefore belongs to $[-1,1]$. Thus $\gamma(t)\in V$ for every $t$.

Now take $(x,\sin(1/x)),(y,\sin(1/y))\in G$, where $0<x,y\le1$. Define $$ \eta(t)=\big((1-t)x+ty,\ \sin(1/((1-t)x+ty))\big). $$ The first coordinate lies between the positive numbers $x$ and $y$, so it remains positive and at most $1$. The reciprocal and sine are continuous wherever the first coordinate is positive, making $\eta$ continuous. Substituting $t=0$ and $t=1$ gives the required endpoints, and the formula puts every value of $\eta$ in $G$. Hence both pieces are path-connected. $\square$

Worked Example: A Path Along the Vertical Segment

Join $(0,-1)$ to $(0,1)$ in $V$. Substituting $a=-1$ and $b=1$ into the linear path gives $$ \gamma(t)=(0,(1-t)(-1)+t(1))=(0,-1+2t). $$ At $t=0$, this is $(0,-1+2(0))=(0,-1)$. At $t=1$, it is $(0,-1+2(1))=(0,1)$. At $t=1/4$, it is $(0,-1+2(1/4))=(0,-1/2)$, which lies in $V$. For every $t\in[0,1]$, $-1\le -1+2t\le1$, so the whole path stays in the segment.

Check: the endpoint at $t=1/2$ is $(0,-1+2(1/2))=(0,0)$, also in $V$; the formula gives the stated endpoints exactly.

Worked Example: A Path on the Oscillating Graph

Join the graph points with first coordinates $1$ and $1/2$. The corresponding points are $(1,\sin 1)$ and $(1/2,\sin 2)$. Linear interpolation of the first coordinates gives $$ x(t)=(1-t)(1)+t(1/2)=1-t/2. $$ Thus the graph path is $\eta(t)=(1-t/2,\sin(1/(1-t/2)))$. At $t=0$, it equals $(1,\sin 1)$; at $t=1$, it equals $(1/2,\sin 2)$. At $t=1/2$, the first coordinate is $1-(1/2)/2=3/4$, and the second is $\sin(1/(3/4))=\sin(4/3)$. Since $1/2\le1-t/2\le1$ for $0\le t\le1$, this path remains in $G$.

Check: at the midpoint, the graph equation gives $\sin(1/(3/4))=\sin(4/3)$, as in the path formula. No decimal approximation is needed to verify that the point lies on $G$.

Why No Path Crosses Between the Pieces

Theorem: No path in $T$ joins a point of $V$ to a point of $G$.

Proof. Suppose, to obtain a contradiction, that a path $\gamma(t)=(u(t),v(t))$ starts at a point of $V$ and ends at a point of $G$. Then $u(0)=0$ and $u(1)>0$. Consider the set of times at which the first coordinate is positive. The interval of such times that contains $1$ has the form $(a,1]$ for some $0\le a<1$; throughout this interval $u(t)>0$. Its left endpoint satisfies $u(a)=0$. If $a=0$, this follows from $u(0)=0$. If $a>0$ and $u(a)>0$, continuity would make $u$ positive on an interval around $a$, extending the interval of positive times to the left, contrary to its choice.

Choose $c_n=a+(1-a)/(n+1)$ for each $n\in\mathbb{N}$. Then $a<c_n<1$, $c_n\to a$, and $u(c_n)>0$. For each $n$, choose positive integers $k_n$ and $l_n$ large enough that $$ r_n=\frac{1}{\pi/2+2\pi k_n}<u(c_n), \qquad s_n=\frac{1}{3\pi/2+2\pi l_n}<u(c_n). $$ Such choices are possible because both displayed reciprocal sequences tend to zero as their integer indices increase. Since $u(a)=0$, continuity and the intermediate value theorem on $[a,c_n]$ give times $p_n,q_n\in(a,c_n)$ with $u(p_n)=r_n$ and $u(q_n)=s_n$. As $a<p_n,q_n<c_n$ and $c_n\to a$, both time sequences converge to $a$.

For every $t\in(a,1]$, the point $\gamma(t)$ lies in $T$ and has positive first coordinate, so it lies in $G$. Consequently $v(t)=\sin(1/u(t))$ there. Our choices give $$ v(p_n)=\sin(\pi/2+2\pi k_n)=1,\qquad v(q_n)=\sin(3\pi/2+2\pi l_n)=-1. $$ But continuity of $v$ at $a$, together with $p_n\to a$ and $q_n\to a$, would require both sequences of values to converge to the same number $v(a)$. One is constantly $1$ and the other constantly $-1$, which is impossible. This contradiction rules out a path from $V$ to $G$. If a path were given in the opposite direction, reversing its parameter by $t\mapsto1-t$ would produce a path from $V$ to $G$, also impossible. $\square$

The argument relies on more than the graph’s oscillation in a picture. Continuity of the first coordinate forces the path to pass through arbitrarily small positive first-coordinate values near the time it leaves $V$. Those values can be chosen so that the graph’s second coordinate alternates between $1$ and $-1$, contradicting continuity of that coordinate.

Worked Example: A Point’s Path Component

Consider $p=(0,0)$. Since $p\in V$, every point of $V$ can be joined to $p$ by the vertical linear path. In particular, for a point $(0,y)\in V$, the path is $\gamma(t)=(0,ty)$; its endpoints are $(0,0)$ and $(0,y)$, and $-1\le ty\le1$.

No point of $G$ can be joined to $p$, by the theorem. For a concrete graph point, take $q=(1,\sin 1)$, which belongs to $G$ because its first coordinate is $1$ and its second coordinate is $\sin(1/1)=\sin 1$. The theorem rules out a path from $p$ to $q$. Thus the path component of $p$ is exactly $V$: it contains all of $V$, and none of $G$.

Check: every point of $T$ lies in $V$ or $G$ by the definition $T=V\cup G$, and the sets are disjoint. Therefore there is no third piece of $T$ that could belong to the path component of $p$.

The Two Path Components

Theorem: The path components of $T$ are exactly $V$ and $G$.

Proof. The proposition shows that any two points of $V$ can be joined by a path in $V$, so all points of $V$ belong to one path component. It also shows that all points of $G$ belong to one path component. The no-crossing theorem says that no point of $V$ can be joined to a point of $G$, so these two path components are distinct. Finally, $T=V\cup G$, so every point of $T$ is already in one of them. There are exactly two path components. $\square$

Common pitfall. A sequence of graph points can approach a point on the vertical segment, but that does not construct a path to the point. For $n\in\mathbb{N}$, set $x_n=1/(\pi/2+2\pi n)$. The denominators exceed $1$, so $0<x_n\le1$; also $x_n\to0$, and $$ \sin(1/x_n)=\sin(\pi/2+2\pi n)=1. $$ Hence $(x_n,\sin(1/x_n))=(x_n,1)$ approaches $(0,1)\in V$. It would be incorrect to infer that the graph can be joined to the vertical segment by a path: the theorem rules this out. Approaching a point by a sequence and reaching it by a continuous path are different claims.

Another tempting but invalid formula is to set $\gamma(0)=(0,1)$ and $\gamma(t)=(t,\sin(1/t))$ for $t>0$. Although the first coordinate tends to zero, the second does not tend to $1$ (or to any other number). For example, $t_n=1/(\pi/2+2\pi n)$ gives $\sin(1/t_n)=1$, while $s_n=1/(3\pi/2+2\pi n)$ gives $\sin(1/s_n)=-1$; both sequences tend to zero. Thus this proposed map is not continuous at $0$.

Check Your Understanding

Use the path formulas and the no-crossing argument to answer the following.

Write a path in $V$ from $(0,1)$ to $(0,-1/2)$, and verify that its second coordinate stays in $[-1,1]$.
Write the graph path joining the points of $G$ with first coordinates $1/3$ and $1$. What is its first coordinate at $t=1/2$?
Why does a path from a graph point to a vertical-segment point also contradict the no-crossing theorem?
For $n\in\mathbb{N}$, evaluate $\sin(\pi/2+2\pi n)$ and $\sin(3\pi/2+2\pi n)$. Explain how these values enter the proof.
Does the existence of graph points converging to $(0,1)$ imply that $(0,1)$ is in the same path component as the graph? Give the distinction that resolves the question.

Path Components of the Topologist's Sine Curve

What You'll Learn