Theorem 19.17.
The topologist’s sine curve is connected but not path connected.
Section Path Connectedness and Connectedness in Infinite Topological Spaces
Given that connectedness and path connectedness are equivalent in finite topological spaces, a reasonable question now is whether the converse of Theorem 19.10 is true in arbitrary topological spaces. As we will see, the answer is no. To find a counterexample, we need to look in an infinite topological space. There are many examples, but a standard example to consider is the topologist’s sine curve. This curve \(S\) is defined as the union of the sets
\begin{equation*}
S_1 = \{(0,y) \mid -1 \leq y \leq 1\} \ \text{ and } \ S_2 = \left\{ \left(x,\sin\left(\frac{1}{x}\right)\right) \Big| \ 0 \lt x \leq 1\right\}\text{.}
\end{equation*}
A picture of \(S\) is shown in Figure 19.16.
To understand if \(S\) is connected, let us consider the relationship between \(S\) and \(S_2\text{.}\) Figure 19.16 seems to indicate that \(S = \overline{S_2}\text{.}\) To see if this is true, let \(q=(0,y) \in S_1\text{,}\) and let \(N\) be a neighborhood of \(q\text{.}\) Then there is an \(\epsilon \gt 0\) such that \(B = B(q, \epsilon) \subseteq N\text{.}\) Choose \(K \in \Z^+\) such that \(\frac{1}{\arcsin(y)+2 \pi K} \lt \epsilon\text{,}\) and let \(z = \frac{1}{\arcsin(y)+2 \pi K}\text{.}\) Then
\begin{align*}
d_E\left(q,\left(z, \sin\left(\frac{1}{z}\right)\right)\right) \amp = d_E\left((0,y),(z, \sin(\arcsin(y)+2 \pi K)\right)\\
\amp = d_E((0,y), (z,\sin(\arcsin(y))))\\
\amp = d_E((0,y), (z,y))\\
\amp = \la z \ra\\
\amp \lt \epsilon\text{,}
\end{align*}
and so \(\left(z, \arcsin(z)\right) \in B(q, \epsilon)\) and every neighborhood of \(q\) contains a point in \(S_2\text{.}\) Therefore, \(S_1 \subseteq S_2' \subseteq \overline{S_2}\) and \(\overline{S_2} = S\) in \(S\text{.}\) The fact that \(S\) is connected follows from Theorem 18.8.
Now that we know that \(S\) is connected, the following theorem demonstrates that \(S\) is a connected space that is not path connected.
Proof.
We know that \(S\) is connected, so it remains to show that \(S\) is not path connected. The sets \(S_1\) and \(S_2\) are connected (as continuous images of the interval \([0,1]\) and \((0,1]\text{,}\) respectively). We will prove that there is no path \(p\) in \(S\) from \(p(0) = (0,0)\) to \(p(1) = b\) for any point \(b \in S_2\) by contradiction. Assume the existence of such a path \(p\text{.}\) Let \(U = p^{-1}(S_1)\) and \(V = p^{-1}(S_2)\text{.}\) Then
\begin{equation}
[0,1] = p^{-1}(S) = p^{-1}(S_1 \cup S_2) = p^{-1}(S_1) \cup p^{-1}(S_2) = U \cup V\text{.}\tag{19.2}
\end{equation}
Note that \(S_2\) is an open subset of \(S\text{,}\) since \(S_2 = \left( \bigcup_{z = (x,y) \in S_2} B\left(z, \frac{x}{2}\right)\right) \cap S\text{.}\) So the continuity of \(p\) implies that \(V\) is an open subset of \([0,1]\text{.}\) Also, the fact that \(p(0) \in S_1\) means that \(U \neq \emptyset\text{,}\) and the fact that \(p(1) \in S_2\) means that \(V \neq \emptyset\text{.}\) If we demonstrate that \(U\) is an open subset of \([0,1]\text{,}\) then Equation (19.2) will imply that \([0,1]\) is not connected, a contradiction. So we proceed to prove that \(U\) is open in \([0,1]\text{.}\)
Let \(x \in U\text{,}\) and so \(p(x)\) in \(S_1\text{.}\) The set \(O = B_S\left(p(x), \frac{1}{2}\right) \cap S\) is open in \(S\text{.}\) The continuity of \(p\) then tells us that \(p^{-1}(O)\) is open in \([0,1]\text{.}\) So there is a \(\delta \gt 0\) such that the open ball \(B=B_{[0,1]}(x, \delta)\) is a subset of \(p^{-1}(O)\text{.}\) We will prove that \(p(B) \subseteq S_1\text{.}\) This will imply that \(B \subseteq U\) and so \(U\) is a neighborhood of each of its points, and \(U\) is therefore an open set.
Every element in \(B\) is mapped into \(O\) by the path \(p\text{.}\) The set \(O\) is complicated, consisting of infinitely many sub-curves of the curve \(S_2\text{,}\) along with points in \(S_1\text{,}\) as illustrated in Figure 19.18. To simplify our analysis, let us consider the projection onto the \(x\)-axis. The function \(P_x : \R^2 \to \R\) defined by \(P_x(x,y) = x\) is a continuous function. Let \(I = P_x(p(B))\text{.}\) Since \(p(B) \subseteq O\text{,}\) we know that \(I \subseteq P_x(O)\text{.}\) Let \(Z = P_x(O)\text{.}\) So \(I \subseteq Z\text{.}\) Since \(B\) is a connected set (\(B\) is an interval), we know that \(p(B)\) is a connected set. The fact that \(P_x\) is continuous means that \(I = P_x(p(B))\) is connected as well. Now \(I\) is a bounded subset of \(\R\text{,}\) so \(I\) must be a bounded interval. Recall that \(x \in B\) and so \(p(x) \in p(B)\text{.}\) The fact that \(p(x) \in S_1\) tells us that \(0 = P_x(p(x)) \in P_x(p(B)) = I\text{.}\) So \(I \neq \emptyset\text{.}\) There are two possibilities for \(I\text{:}\) either \(I = \{0\}\text{,}\) or \(I\) is an interval of positive length. We consider the cases.
Suppose \(I = \{0\}\text{.}\) Then the projection of \(p(B)\) onto the \(x\)-axis is the single point \(0\) and \(p(B) \subseteq S_1\) as desired. Suppose that \(I\) is an interval of the form \([0,d]\) or \([0,d)\) for some positive number \(d\text{.}\) The structure of \(O\) would indicate that there must be some gaps in the set \(Z\text{,}\) the projection of \(O\) onto the \(x\)-axis. This implies that \(I\) cannot be a connected interval. We proceed to show this. In other words, we will prove that \(I \setminus Z \neq \emptyset\) (which is impossible since \(I \subseteq Z\)). Remember that \(p(x) \in S_1\text{,}\) so let \(p(x) = (0, q)\text{.}\) We consider what happens if \(q \lt \frac{1}{2}\) and when \(q \geq \frac{1}{2}\text{.}\)
Suppose \(q \lt \frac{1}{2}\text{.}\) Then the ball \(B_S\left(p(x), \frac{1}{2}\right)\) contains only points with \(y\) value less than 1. Let \(N \in \Z^+\) so that \(t=\frac{1}{\pi/2+2N\pi} \lt d\text{.}\) Then \(t \in I\text{.}\) But \(\sin\left(\frac{1}{t}\right) = \sin(\pi/2 + 2N\pi) = \sin(\pi/2) = 1\text{,}\) and so \(\left(t,\sin\left(\frac{1}{t}\right)\right)\) is not in \(O\text{.}\) Thus, \(t \notin Z\text{.}\) Thus we have found a point in \(I \setminus Z\text{.}\)
Finally, suppose \(q \geq \frac{1}{2}\text{.}\) Then the ball \(B_S\left(p(x), \frac{1}{2}\right)\) contains only points with \(y\) value greater than \(-1\text{.}\) Let \(N \in \Z^+\) so that \(t=\frac{1}{3\pi/2+2N\pi} \lt d\text{.}\) Then \(t \in I\text{.}\) But \(\sin\left(\frac{1}{t}\right) = \sin(3\pi/2 + 2N\pi) = \sin(3\pi/2) = -1\text{,}\) and so \(t \notin Z\text{.}\) Thus we have found a point in \(I \setminus Z\text{.}\)
We conclude that there can be no path in \(S\) from \((0,0)\) to any point in \(S_2\text{,}\) completing our proof that \(S\) is not path connected. (In fact, the argument given shows that there is no path in \(S\) from any point in \(S_1\) to any point in \(S_2\text{.}\)
