Closed Topologist's Sine Curve is not Path-Connected

Theorem
Let $$G$$ be the graph of the function $$y = \sin \left({\frac 1 x}\right)$$ for $$x > 0$$ is connected, but not path-connected.

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

Then $$G \cup J$$ is connected.

Proof

 * First we need to show that $$G \cup J$$ is connected.

Since the open interval $$\left({0 \, . \, . \, \infty}\right)$$ is connected, then so is $$G$$ by Continuous Image of Connected Space is Connected.

It is enough, from Subset of Closure of Connected Subspace, to show that $$J \subseteq \operatorname{cl}\left({G}\right)$$.

Let $$p \in J$$, say, $$\left({0, y}\right)$$ where $$-1 \le y \le 1$$.

We need to show that $$\forall \epsilon > 0: N_\epsilon \left({p}\right) \cap G \ne \varnothing$$, where $$N_\epsilon \left({p}\right)$$ is the $\epsilon$-neighborhood of $$p$$.

Let us choose $$n \in \N: \frac 1 {2 n \pi} < \epsilon$$.

From Nature of Sine Function, $$\sin \left({\frac {\left({4n + 1}\right) \pi} 2}\right) = 1$$ and $$\sin \left({\frac {\left({4n + 3}\right) \pi} 2}\right) = -1$$.

So by the Intermediate Value Theorem, $$\sin \left({\frac 1 x}\right)$$ takes every value between $$-1$$ and $$1$$ in the closed interval $$\left[{\frac {2}{\left({4n + 3}\right) \pi} \,. \, . \, \frac {2}{\left({4n + 1}\right) \pi}}\right]$$.

In particular, $$\sin \left({\frac 1 {x_0}}\right) = y$$ for some $$x_0$$ in this interval.

The distance between the points $$\left({0, y}\right)$$ and $$\left({x_0, \sin \left({\frac 1 {x_0}}\right)}\right) = \left({x_0, y}\right)$$ is $$x_0 < \epsilon$$.

So $$\left({x_0, \sin \left({\frac 1 {x_0}}\right)}\right) \in N_\epsilon \left({p}\right) \cap G$$, as required.


 * Now we show that $$G$$ is not path-connected.

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

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

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 $$pr_1 \circ i \circ f$$ and $$pr_2 \circ i \circ f$$, where $$i$$ is the inclusion mapping and $$pr_1, pr_2$$ 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 Continuous Mapping from Compact Metric Space is Uniformly Continuous, $$f_2$$ 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 $$\forall i = 0, 1, \ldots, n: f_1 \left({t_i}\right) > \frac {2} {\left({2i + 3}\right) \pi}$$.

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

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

Suppose also, to get a contradiction, that $$f_1 \left({t_{i+1}}\right) \ge \frac {2} {\left({2i + 5}\right) \pi}$$.

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

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

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

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

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