# Closed Topologist's Sine Curve is not Path-Connected

## Theorem

Let $G$ be the graph of the function $y = \sin \left({\dfrac 1 x}\right)$ for $x > 0$.

Let $J$ be the line segment joining the points $\left({0, -1}\right)$ and $\left({0, 1}\right)$ in $\R^2$.

Then while $G \cup J$ is connected, it is not path-connected.

## Proof

From Closed Topologist's Sine Curve is Connected, $G \cup J$ is connected.

Let $I \subseteq \R$ be the closed real interval $\left[{0 \,.\,.\, 1}\right]$.

Let $A = \left({\dfrac 1 \pi, 0}\right) \in \R^2$.

This proof is based on the fact that a continuous path $f: I \to G \cup J$ beginning at $A$ will never actually arrive at $0 = \left({0, 0}\right) \in \R^2$ because $I$ is compact.

Suppose $f: I \to G \cup J$ is continuous and that $f \left({0}\right) = A$.

Let $f_1: I \to \R$ and $f_2: I \to \R$ denote ${\operatorname{pr}_1} \circ i \circ f$ and ${\operatorname{pr}_2} \circ i \circ f$, where $i$ is the inclusion mapping and $\operatorname{pr}_1, \operatorname{pr}_2$ the first and second projections on the $x$ and $y$ axes.

Thus $f_2$ describes the vertical movement of the graph, and $f_1$ the horizontal movement.

Now $f_2$ is continuous and $I$ compact.

So by the Heine-Cantor Theorem, $f_2$ is uniformly continuous on $I$.

Let $\delta > 0$ be such that $\left|{f_2 \left({t}\right) - f_2 \left({t'}\right)}\right| < 2$ for any $t, t' \in I$ such that $\left|{t - t'}\right| < \delta$.

Let $0 < t_0 < t_1 < \ldots < t_n = 1$ be such that $t_i - t_{i-1} < \delta$ for all $i = 1, 2, \ldots, n$.

Now: as $t$ goes from $t_0$ to $t_1$, $f \left({t}\right)$ (which starts at $A$), can not reach a point $C$ where $y = 1$ without passing through a point $B$ where $y = -1$ on the way.

But then $f \left({t}\right) = B$ and $f \left({t'}\right) = C$ for some $t, t'$ where $\left|{t - t'}\right| < \delta$.

And since $B$ and $C$ are $2$ apart, $\left|{f_2 \left({t}\right) - f_2 \left({t'}\right)}\right| = 2$ which contradicts the choice of $\delta$.

Similarly, when going from $t_1$ to $t_2$, $t$ can similarly not get past more than one hump.

So, as $t$ goes from $0$ to $1$, $f \left({t}\right)$ can not traverse more than $n$ humps.

We formalize this discussion by induction.

We will show that:

$\forall i = 0, 1, \ldots, n: f_1 \left({t_i}\right) > \dfrac 2 {\left({2i + 3}\right) \pi}$

Since $f \left({t_0}\right) = f \left({0}\right) = \left({\dfrac 1 \pi, 0}\right)$, we have $f_1 \left({t_0}\right) = \dfrac 1 \pi > \dfrac 2 {3 \pi}$.

Suppose that $f_1 \left({t_i}\right) > \dfrac 2 {\left({2 i + 3}\right) \pi}$ for some $i \ge 0$.

Aiming for a contradiction, suppose also that $f_1 \left({t_{i+1}}\right) \ge \dfrac 2 {\left({2 i + 5}\right) \pi}$.

Then since $f_1$ is continuous, by the I.V.P. there exists $t, t' \in \left[{t_1 \,.\,.\, t_{i + 1} }\right]$ such that:

$f_1 \left({t_i}\right) = \dfrac 2 {\left({2 i + 3}\right) \pi}, f_1 \left({t_{i + 1} }\right) = \dfrac 2 {\left({2 i + 5}\right) \pi}$

But the only point in $G \cup J$ whose first coordinate is $\dfrac 2 {\left({2 i + 3}\right) \pi}$ is $\left({\dfrac 2 {\left({2 i + 3}\right) \pi}, \sin \left({\dfrac {\left({2 i + 3}\right) \pi} 2}\right)}\right)$.

So:

$f_2 \left({t}\right) = \sin \left({\dfrac {\left({2 i + 3}\right) \pi} 2}\right)$

Similarly:

$f_2 \left({t'}\right) = \sin \left({\dfrac {\left({2 i + 5}\right) \pi} 2}\right)$

Hence $\left|{f_2 \left({t}\right) - f_2 \left({t'}\right)}\right| = 2$ while $\left|{t - t'}\right| < \delta$.

This contradicts the choice of $\delta$.

This proves the induction.

The result follows from the fact that:

$f_1 \left({1}\right) > \dfrac 2 {\left({2 n + 3}\right) \pi} > 0$, so $f_1 \left({1}\right) \ne 0$

$\blacksquare$