[example: The Topologist's Sine Curve Is Not Path-Connected]
The canonical example demonstrating the gap between connectedness and path connectedness is the **topologist's sine curve**. Define $S \subset \mathbb{R}^2$ by:
\begin{align*}
S := \left\{\left(x, \sin\frac{1}{x}\right) : x \in (0, 1]\right\} \cup \left(\{0\} \times [-1, 1]\right).
\end{align*}
The graph $\Gamma := \{(x, \sin(1/x)) : x \in (0, 1]\}$ is the continuous image of the connected interval $(0, 1]$, hence connected. Every point of the vertical segment $I := \{0\} \times [-1, 1]$ is a limit of points of $\Gamma$: for any $y \in [-1, 1]$, the function $\sin$ takes the value $y$ infinitely often on $(0, \varepsilon)$ for every $\varepsilon > 0$, so there exist $x_k \to 0^+$ with $\sin(1/x_k) = y$. Therefore $S = \overline{\Gamma}$, and since the [closure of a connected set is connected](/page/Connectedness), $S$ is connected.
However, $S$ is **not** path-connected. Suppose for contradiction that $\gamma: [0, 1] \to S$ is a continuous path with $\gamma(0) = (0, 0) \in I$ and $\gamma(1) = (1, \sin 1) \in \Gamma$. Write $\gamma(t) = (\gamma_1(t), \gamma_2(t))$. Since $\gamma_1$ is continuous and $\gamma_1(0) = 0$, the set $\{t \in [0, 1] : \gamma_1(t) = 0\}$ is closed; let $t_0$ be its supremum. Then $\gamma_1(t_0) = 0$, and for $t > t_0$ we have $\gamma_1(t) > 0$, so $\gamma_2(t) = \sin(1/\gamma_1(t))$. As $t \to t_0^+$, we have $\gamma_1(t) \to 0^+$, so $\sin(1/\gamma_1(t))$ oscillates between $-1$ and $1$ without converging. But continuity of $\gamma_2$ requires $\gamma_2(t) \to \gamma_2(t_0)$, a contradiction.
The essential mechanism is that the graph oscillates with unbounded frequency near $x = 0$: any path approaching the vertical segment from the graph side must traverse infinitely many oscillations in finite "time," which is incompatible with continuity. This phenomenon cannot occur in an open subset of $\mathbb{R}^n$, where every point has a convex (hence path-connected) neighbourhood.
[/example]