The Set and the Path Question
Some connected spaces contain points that cannot be joined by paths. The topologist’s sine curve is a standard example: its graph oscillates more and more rapidly as it approaches a vertical segment. The essential issue is not merely that the graph oscillates as its first coordinate tends to zero. A path might try to approach the segment in some other way. We will show that continuity of the first coordinate forces the path to pass through values at which the second coordinate alternates between \(1\) and \(-1\).
We use the Euclidean topology on \(\mathbb{R}^2\), the standard topology on \(I=[0,1]\), and the subspace topology on the following subset \(S\) of \(\mathbb{R}^2\):
Here \(G\) is the oscillating graph and \(V\) is the limiting vertical segment. A path in \(S\) is a continuous map \(\gamma:I\to S\). We will prove that no path has one endpoint in \(V\) and the other in \(G\).
The Oscillation Forced by a Positive Coordinate
The key fact is that a continuous function that starts at zero and stays positive must take every sufficiently small positive value. At those values, the sine term can be forced to equal \(1\) or \(-1\).
Proof. For \(n\in\mathbb{N}\), define \(r_n=1/(\pi/2+2\pi n)\) and \(q_n=1/(3\pi/2+2\pi n)\). Their denominators tend to infinity, so \(r_n\to0\) and \(q_n\to0\). Also, \(\sin(1/r_n)=\sin(\pi/2+2\pi n)=1\) and \(\sin(1/q_n)=\sin(3\pi/2+2\pi n)=-1\).
For all sufficiently large \(n\), both \(r_n\) and \(q_n\) are less than \(u(c)\). The intermediate value theorem applied to \(u\) on \([a,c]\), where \(u(a)=0\), therefore gives points \(s_n,t_n\in(a,c)\) such that \(u(s_n)=r_n\) and \(u(t_n)=q_n\).
We check that \(s_n\to a\) and \(t_n\to a\). Fix any \(\delta\) with \(0<\delta<c-a\). The function \(u\) is continuous and strictly positive on the compact interval \([a+\delta,c]\). It therefore has a minimum \(m_\delta>0\) there. For all sufficiently large \(n\), \(r_n<m_\delta\), so \(s_n\) cannot lie in \([a+\delta,c]\). Thus \(a<s_n<a+\delta\). The same argument using \(q_n<m_\delta\) gives \(a<t_n<a+\delta\) for all sufficiently large \(n\). Since this holds for every such \(\delta\), both sequences tend to \(a\).
But \(\sin(1/u(s_n))=1\) and \(\sin(1/u(t_n))=-1\) for every sufficiently large \(n\). If \(\sin(1/u(t))\) had a limit as \(t\) approached \(a\), its values along both sequences would have to tend to that same limit. One sequence is constantly \(1\) and the other constantly \(-1\), which is impossible. \(\square\)
No Path Can Join the Two Parts
Proof. Suppose first that a path \(\gamma:I\to S\) satisfies \(\gamma(0)\in V\) and \(\gamma(1)\in G\). Write \(\gamma(t)=(u(t),v(t))\). Then \(u(0)=0\) and \(u(1)>0\).
We identify a final interval on which \(u\) stays positive. Let \(E=\{r\in I:u(s)>0\text{ for every }s\in[r,1]\}\). Continuity of \(u\) and \(u(1)>0\) imply that \(u\) is positive on some interval ending at \(1\), so \(E\) is nonempty. Set \(a=\inf E\). Then \(a<1\). If \(t>a\), the definition of infimum gives some \(r\in E\) with \(r<t\), and hence \(u(t)>0\). Also \(u(a)=0\): continuity and positivity immediately to the right imply \(u(a)\geq0\); if \(u(a)>0\), continuity would make \(u\) positive on an interval to the left of \(a\) as well, contradicting that \(a\) is the infimum of \(E\).
Choose any \(c\) with \(a<c\leq1\). On \((a,c]\), \(u(t)>0\). Since \(\gamma(t)\in S\), every point of \(S\) with positive first coordinate lies on \(G\). Consequently, \(v(t)=\sin(1/u(t))\) for every \(t\in(a,c]\). The lemma says this expression has no limit as \(t\) approaches \(a\) from above. However, \(v\) is continuous at \(a\), so \(v(t)\to v(a)\). This contradiction rules out a path oriented from \(V\) to \(G\).
If a path were instead oriented from \(G\) to \(V\), reversing its parameter would produce a path oriented from \(V\) to \(G\): if \(\eta:I\to S\) were that path, then \(t\mapsto\eta(1-t)\) would be continuous and have the opposite endpoints. The case already proved excludes it as well. \(\square\)
Worked Example: A Path Along the Vertical Segment
Consider \(\gamma:I\to S\) given by \(\gamma(t)=(0,2t-1)\). We check directly that this is a path from \((0,-1)\) to \((0,1)\).
For \(0\leq t\leq1\), multiplying by \(2\) gives \(0\leq2t\leq2\), and subtracting \(1\) gives \(-1\leq2t-1\leq1\). Thus \(\gamma(t)\in\{0\}\times[-1,1]=V\subseteq S\) throughout. Its endpoint calculations are \(\gamma(0)=(0,2(0)-1)=(0,-1)\) and \(\gamma(1)=(0,2(1)-1)=(0,1)\). Both coordinate functions are continuous, so \(\gamma\) is continuous. This path stays entirely on the vertical segment; it does not join that segment to the graph.
Check: at \(t=1/2\), \(\gamma(1/2)=(0,2(1/2)-1)=(0,0)\), which is in \(V\). The second coordinate remains between \(-1\) and \(1\) at every parameter value, as required.
Worked Example: A Path Along the Graph
We construct a path between two points of \(G\). Set \(\gamma(t)=(x(t),\sin(1/x(t)))\), where \(x(t)=(1-t)/\pi+t/(2\pi)=(2-t)/(2\pi)\).
For \(0\leq t\leq1\), \(1\leq2-t\leq2\). Hence \(x(t)>0\), and \(1/(2\pi)\leq x(t)\leq1/\pi\leq1\). It follows from the definition of \(G\) that \(\gamma(t)\in G\) for every \(t\). The reciprocal used in the second coordinate is \(1/x(t)=2\pi/(2-t)\), which is continuous on \(I\) because \(2-t\geq1>0\). Thus both coordinates, and therefore \(\gamma\), are continuous.
At the endpoints, \(x(0)=2/(2\pi)=1/\pi\), so \(\gamma(0)=(1/\pi,\sin\pi)=(1/\pi,0)\). Also \(x(1)=1/(2\pi)\), so \(\gamma(1)=(1/(2\pi),\sin(2\pi))=(1/(2\pi),0)\). Both points belong to \(G\), and the path stays in \(G\) throughout.
Check: substituting \(t=1/2\) gives \(x(1/2)=(3/2)/(2\pi)=3/(4\pi)>0\), and the second coordinate is \(\sin(4\pi/3)=-\sqrt{3}/2\). Thus this intermediate point is exactly \((3/(4\pi),-\sqrt{3}/2)\), a point of \(G\) by its defining equation.
Worked Example: Testing a Proposed Approach from the Segment
A natural attempt to join \((0,0)\in V\) to \((1,\sin1)\in G\) is to set \(\gamma(t)=(t,\sin(1/t))\) for \(0<t\leq1\), and to define \(\gamma(0)=(0,0)\). For positive \(t\), this formula lies in \(G\), and at \(t=1\) it gives \((1,\sin1)\). The question is whether it is continuous at \(0\).
For \(n\in\mathbb{N}\), let \(t_n=1/(\pi/2+2\pi n)\) and \(w_n=1/(3\pi/2+2\pi n)\). Both sequences are positive and tend to \(0\), since their denominators tend to infinity. Direct substitution gives \(1/t_n=\pi/2+2\pi n\) and \(1/w_n=3\pi/2+2\pi n\). Therefore \(\gamma(t_n)=(t_n,1)\) and \(\gamma(w_n)=(w_n,-1)\). These image sequences tend to \((0,1)\) and \((0,-1)\), respectively. They do not both tend to the proposed value \((0,0)\), so the proposed map is not continuous at \(0\).
Check: the first coordinates \(t_n\) and \(w_n\) do tend to \(0\), but the second coordinates are constantly \(1\) and \(-1\) along the two sequences. Continuity in \(\mathbb{R}^2\) requires both coordinates to converge to the coordinates of \(\gamma(0)\); the first-coordinate calculation alone cannot establish continuity.
Common Pitfall: Checking Only the First Coordinate
A tempting but incorrect argument is: “As the graph point approaches the vertical segment, its first coordinate tends to zero; therefore the point tends to \((0,0)\).” The example above gives the concrete error. Along \(t_n=1/(\pi/2+2\pi n)\), the points are \((t_n,1)\), which approach \((0,1)\); along \(w_n=1/(3\pi/2+2\pi n)\), the points are \((w_n,-1)\), which approach \((0,-1)\). Neither sequence approaches \((0,0)\).
For a general path, it is not enough simply to substitute a sequence of small first coordinates into \(\sin(1/x)\): one must establish that the path actually takes those first-coordinate values. That is why the proof uses the intermediate value theorem. Once the first coordinate starts at zero and remains positive on an interval, every sufficiently small positive value is attained there. The oscillation then contradicts continuity of the second coordinate.
Check Your Understanding
Use the coordinate argument and the oscillation lemma to answer the following.
- Why must the second coordinate of a path in \(S\) satisfy \(v(t)=\sin(1/u(t))\) whenever \(u(t)>0\)?
- In the lemma, why can the points with \(u(s_n)=r_n\) be chosen inside \((a,c)\), rather than only at an endpoint?
- Compute \(\sin(1/r_n)\) and \(\sin(1/q_n)\) for \(r_n=1/(\pi/2+2\pi n)\) and \(q_n=1/(3\pi/2+2\pi n)\).
- Why does continuity of \(v\) at \(a\) require both \(v(s_n)\) and \(v(t_n)\) to converge to \(v(a)\) when \(s_n,t_n\to a\)?
- What change to the parameter produces a path with reversed endpoints, and why does that show the theorem covers either endpoint ordering?