Topologist's Sine Curve | Real Analysis

The closure of

\{(x, \sin(1/x)) : x > 0\}

adds the segment

\{0\}\times[-1,1]

on the

y

-axis. Connected as the closure of a connected set, yet no path runs from the segment into the curve — the living example of why connected is weaker than path-connected.

Examples & Exercises

Prove completely that $S = \{(x, \sin(1/x)) : x > 0\} \subseteq \mathbb{R}^2$ is connected.

S = f((0,\infty)), \qquad f(x) = (x,\, \sin(1/x))

x_{\min} = 0.05

: the curve

S

alone oscillates infinitely near 0, yet it is one unbroken continuous curve.

Show proof

$(0,\infty)$ 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.

Prove completely that $\overline{S} = S \cup \big(\{0\} \times [-1,1]\big)$ : the closure of $S$ is $S$ together with the vertical segment $\{0\} \times [-1,1]$ .

\overline{S} = S \cup \big(\{0\} \times [-1,1]\big)

x_{\min} = 0.01

: zooming closer to 0 shows the sequences

(x_n, 1)

(x_n, -1)

converging onto the endpoints of the red segment.

Show proof

$\supseteq$ : $S \subseteq \overline{S}$ 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. Points of $S$ meeting $B(p, x/2)$ have first coordinate in $(x/2,\, 3x/2)$ , so they lie in $K$ — giving $p \in \overline{K} = K \subseteq S$ , 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 $\overline{S} = S \cup (\{0\} \times [-1,1])$ .

Prove completely that $T = \overline{S}$ is not path-connected: no path in $T$ leads from a point of the segment $\{0\} \times [-1,1]$ to a point of $S$ .

\gamma(0) \in \{0\} \times [-1,1],\ \gamma(1) \in S \implies \text{no continuous } \gamma: [0,1] \to T

Try "walking" from the red segment into the curve — the infinite oscillation near 0 blocks every path. The proof captures this obstruction with sequences.

Show proof

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 $x(t) > 0$ for $t \in (c,1]$ (otherwise the supremum would be larger).
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.