Topologist's sine curve is not path-connected

If $S=\{(0,0)\}\cup\{(x,\sin(1/x)):0<x<1\}$ and $f=(f_1,f_2):[0,1]\to S$ is a path with $f(0)=(0,0)$, then $f(t)=(0,0)$ for all $t$.

To see this by contradiction, suppose that $f(t)$ is not always $(0,0)$. Removing an initial part of the interval and then rescaling if necessary, assume that $0=\sup\{t:f([0,t])=\{(0,0)\}\}$. By continuity of $f_2$, there is a $\delta>0$ such that $|f_2(t)|<1$ for all $t<\delta$. Take $t_0$ with $0<t_0<\delta$ and $f_1(t_0)>0$. By continuity of $f_1$ and the intermediate value theorem, $[0,f_1(t_0)]$ is in the image of $f_1$ restricted to $[0,t_0]$. Since $f_2(t)=\sin(1/f_1(t))$ for all $t$ with $f_1(t)\neq0$ and $\sin(1/x)$ maps $]0,\varepsilon[$ onto $[-1,1]$ for all $\varepsilon>0$, it follows that $[-1,1]$ is in the image of $f_2$ restricted to $[0,t_0]$. This contradicts $t_0<\delta$.

Assume, to get a contradiction, that the graph $G \subset \mathbb R \times \mathbb R$ of the function

$$ y(x) =\begin{cases} \sin\left(\dfrac{1}{x}\right) & \mbox{if $0\lt x \lt 1$}\\\ 0 & \mbox{if $x=0$}\end{cases}$$

is path-connected.

Let $\gamma$ be a path in $G$ connecting $(\frac{1}{2},sin(\frac{1}{2}))$ to $(0,0)$. We know that the image $K$ of $\gamma$ is also compact.

The sets defined by

$\tag 1 L_n = K \cap \{(x,y) \in \mathbb R \times \mathbb R \, | \, x \le \frac{1}{n} \text{ and } y = 1\} \text{ for } n \ge 2$

define a decreasing chain of closed subsets of $K$.

Now for any $0\lt\alpha\lt\frac{1}{2}$, the path $\gamma$ must pass through $(\alpha, sin(\frac{1}{\alpha})$. This follows since the image of $\gamma$ is connected and can't be disconnected by the two open half planes defined by $x = \alpha$. Also note that for every $k \ge 1$, the point $\left(sin(\frac{\pi}{2} + 2 \pi k)^{-1}), 1\right)$ belongs to $G$. So each $L_n$ must be a non-empty closed subset $K$.

Recall the following general theory for a compact topological space $X$:

Any collection of closed subsets of X with the finite intersection property has nonempty intersection.

We know that our chain $L_2 \supset L_3 \supset L_4\supset \dots$ of closed sets in $K$ satisfies the finite intersection property, so the intersection must be nonempty.

A simple argument shows that the intersection of the $L_n$ must exclude all of $G$ except perhaps the singleton set containing $(0,0)$. But this point does not lie on the line $y = 1$, so the intersection is indeed empty. But this is absurd.

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Let $\beta \in \mathbb R$. Using the argument above, we can also show that the graph of the function

$$ y(x) =\begin{cases} \sin\left(\dfrac{1}{x}\right) & \mbox{if $0\lt x \lt 1$}\\\ \beta & \mbox{if $x=0$}\end{cases}$$

can't be path-connected.

Using this fact, one can show that the Topologist's sine curve as defined by Munkres is also not path-connected; see this stackexchange answer.