= Topologist's sine curve
{title2=$T=\overline G$}
{wiki}
Consider the <graph of a function> $x\mapsto\sin(1/x)$, using the <sine> of the <multiplicative inverse> of $x$, together with a vertical <closed interval>:
$$
G=\{(x,\sin(1/x)):0<x\le1\},\qquad L=\{0\}\times[-1,1],\qquad T=G\cup L.
$$
Give $T$ the <subspace topology> from the <Euclidean plane>. The symbol $\times$ denotes a <Cartesian product>. The right endpoint $(1,\sin1)\approx(1,0.841471)$ belongs to $G$. The <continuous map> $x\mapsto(x,\sin(1/x))$ sends the <connected> <real interval> $(0,1]$ onto $G$. Hence $G$ is <connected>, and its <closure> $\overline G=T$ is <connected> by <closure of a connected set>. Every height in $[-1,1]$ is approached as $x$ tends to zero; for example,
$$
p_n=\frac1{\pi/2+2\pi n},\qquad q_n=\frac1{3\pi/2+2\pi n},\qquad \sin(1/p_n)=1,\qquad\sin(1/q_n)=-1\quad(n=0,1,\ldots).
$$
For $n=1$, these inputs are $p_1=2/(5\pi)\approx0.127324$ and $q_1=2/(7\pi)\approx0.090946$. For any other height $y\in[-1,1]$, choose $\theta\ge1$ with $\sin\theta=y$; then the <sequence> $(1/(\theta+2\pi n),y)$ lies in $G$ and has <limit of a sequence> $(0,y)$.
Nevertheless, $T$ has exactly two <path components>, $G$ and $L$. Within $G$, join inputs $a,b\in(0,1]$ by the <continuous path> $t\mapsto((1-t)a+tb,\sin(1/((1-t)a+tb)))$. Within $L$, use $t\mapsto(0,(1-t)c+td)$ for heights $c,d\in[-1,1]$.
The obstruction to a <continuous path> between $L$ and $G$ is persistent oscillation, not a visible gap. Proof outline: suppose such a <path> exists, directed from $L$ to $G$, and take its last time in $L$. Immediately afterwards its first coordinate $x(t)>0$ approaches zero and its second coordinate is $\sin(1/x(t))$. The <intermediate value theorem> forces the first coordinate to pass through inputs of both displayed types arbitrarily close to that departure time. The second coordinate therefore takes both values $1$ and $-1$ arbitrarily close to that time, contradicting <continuity>. https://math.mit.edu/~hrm/palestine/munkres-topology.pdf[Munkres, Topology: A First Course, section 3-2, example 7 on p. 158]{ref}
\Image[/topologist-sine-curve-connection.png]
{title=The <topologist's sine curve> has one <connected component> and two <path components>; the <graph of a function> is drawn only for $x\ge0.005$, with further oscillations omitted near the vertical <closed interval>}
The sampled <graph of a function> illustrates the formulas; it does not prove <connectedness> or the failure of <path connectedness>. Taking a <closure> preserves <connectedness>, but this example shows that it can destroy <path connectedness>.
Back to article page
