Topologist's Sine Curve

$(0,\infty)$ is connected (it is an interval — the same $\sup$ argument as for $\mathbb{R}$ applies verbatim). The map $f(x) = (x, \sin(1/x))$ is continuous since both coordinate functions are, and its image is exactly $S$.

The continuous image of a connected set is connected: if $f(X) = A \cup B$ were a separation into disjoint nonempty relatively open sets, then $X = f^{-1}(A) \cup f^{-1}(B)$ would be a separation of $X$ into disjoint nonempty relatively open sets ($A, B$ nonempty implies their preimages are nonempty), contradicting the connectedness of $X$.

Hence $S = f((0,\infty))$ is connected. The closure of a connected set is connected, so $T = \overline{S}$ is connected as well — problem 2 below computes this closure.

$\supseteq$: $S \subseteq \overline{S}$ is trivial. For $(0,y)$ with $|y| \le 1$, put $x_n = \frac{1}{\arcsin y + 2\pi n}$ ($n \ge 1$); then $x_n \to 0^+$ and $\sin(1/x_n) = \sin(\arcsin y + 2\pi n) = y$.

Hence $(x_n, y) \in S$ converges to $(0,y)$, so every point of $\{0\} \times [-1,1]$ is a limit point of $S$.

$\subseteq$: let $p \in \overline{S}$. If $p = (x,y)$ with $x < 0$, the ball $B(p, |x|/2)$ contains no point of $S$ (whose first coordinates are all positive) — impossible. If $x > 0$ and $p \notin S$, then $K = f([x/2,\, 3x/2])$ ($f$ from the problem above) is compact as a continuous image of a compact set, hence closed.

Now let $r > 0$ be arbitrary: points of $S$ meeting $B(p, \min(r, x/2))$ have first coordinate in $(x/2,\, 3x/2)$, so they lie in $K$. Since $p \in \overline{S}$, this ball meets $S$, hence meets $K$, so $\operatorname{dist}(p, K) < r$. As $r > 0$ was arbitrary, $p \in \overline{K} = K \subseteq S$ — a contradiction. Thus $x > 0$ implies $p \in S$.

Finally, if $p = (0,y)$: for $(u,v) \in S$, $|p - (u,v)| \ge |y - v| \ge |y| - 1$ (since $|v| = |\sin| \le 1$). If $|y| > 1$, the ball $B(p, (|y|-1)/2)$ misses $S$, contradicting $p \in \overline{S}$.

Hence $|y| \le 1$, i.e. $p \in \{0\} \times [-1,1]$. The three cases together give $\overline{S} = S \cup (\{0\} \times [-1,1])$.

Assume a continuous $\gamma: [0,1] \to T$ with $\gamma(0) \in \{0\} \times [-1,1]$ and $\gamma(1) \in S$, aiming for a contradiction. Write $\gamma(t) = (x(t), y(t))$ and put $c = \sup\{t \in [0,1] : x(t) = 0\}$ (nonempty, containing $0$). By continuity $x(c) = 0$, and $c < 1$ since $x(1) > 0$.

Moreover $T \subseteq \{x \ge 0\}$ by the definition of $T = \overline{S}$ (points of $S$ have positive first coordinate and the limit segment has first coordinate $0$), so $x(t) \ge 0$ for all $t$. For $t \in (c,1]$ we also have $x(t) \ne 0$ (otherwise the supremum would be larger), hence $x(t) > 0$ on $(c,1]$.

For $t \in (c,1]$, $\gamma(t) \in T$ has positive first coordinate, so $\gamma(t) \in S$ and $y(t) = \sin(1/x(t))$. Since $v_n = \frac{1}{2\pi n + \pi/2} \to 0$, we have $v_n < x(1)$ for large $n$. On $[c,1]$, $x(c) = 0 < v_n < x(1)$, so the intermediate value theorem gives $t_n \in (c,1]$ with $x(t_n) = v_n$.

Then $y(t_n) = \sin(2\pi n + \pi/2) = 1$. Every subsequential limit $t^* \in [c,1]$ of $(t_n)$ satisfies $x(t^*) = \lim v_n = 0$, forcing $t^* = c$ (as $x > 0$ on $(c,1]$). Hence $t_n \to c$, and by continuity $y(c) = \lim y(t_n) = 1$.

Likewise, with $w_n = \frac{1}{2\pi n + 3\pi/2}$, pick $s_n \in (c,1]$ with $x(s_n) = w_n$; then $y(s_n) = \sin(2\pi n + 3\pi/2) = -1$, and the same argument gives $s_n \to c$, so $y(c) = -1$. This contradicts $y(c) = 1$, so the assumed path cannot exist.

$T$ is connected but not path-connected.