TITLE: Topologist's sine curve is not path-connected
QUESTION [36 upvotes]: Is there a (preferably elementary) proof that the graph of the (discontinuous) function $y$ defined on $[0,1)$ by
$$ y(x) =\begin{cases}
\sin\left(\dfrac{1}{x}\right) & \mbox{if $0\lt x \lt 1$,}\\\
0 & \mbox{if $x=0$,}\end{cases}$$
is not path connected?

REPLY [4 votes]: An attempt at rewriting Jonas Meyer's answer. Also I prove that closure of sine curve, which contains a whole segment on $y$-axis is not path-connected:
Let $f=(f_1,f_2):[a,b]\to S$ be any continuous map such that $f(a)=(0,0)$. Since projections $(x,y) \to x$ and $(x,y) \to y$ are continuous maps $\mathbb{R}^2 \to \mathbb{R}$, we see that both $f_1:[a,b] \to \mathbb{R}$ and $f_2$ are continuous.
By continuity at $a$ of $f_2$, there exists a $\delta > 0$ such that
$$ (*) \ \ \ \ |f_2(x) - 0| < \frac{1}{2} \ ,$$ for all $a\leq x \leq a+\delta$.
Claim: For all $a\leq x \leq  a+\delta$, $f_1(x)=0$.
Proof of the claim: Assume $f(x^*) = \tau >0$ for some $a < x^* \leq a+\delta$, $f_1(x)=0$, then since $f_1$ is continuous, by Intermediate Value Theorem $f_1([a,x^*]) \supset [0,\tau) \ $  .
Therefore there exists some $t=\frac{1}{2n\pi+\pi/2}$ in the range of $f_1([0,\delta)) \ $, i.e there exists $\tilde{x} \in [a,a+\delta]$ such that $f_1(\tilde{x})=t$ We see that $$f_2(\tilde{x}) = \sin(1/t) =1, $$
which contradicts (*) above.
This proves the claim. The claim means that $f$ cannot leave the $y$-axis for some while ($\delta$ here.)
But now applying the same argument to the continuous map $[a+\delta , b] \to S$, we see that $f$ will NEVER be able to leave the $y$-axis.$^1$ In other words, we proved that any continuous path into $S$ starting on $y$-axis remains on $y$-axis. This refutes the possibility of paths between $y$-axis points and other points outside it.
$^1$ To make this precise we define $sup$ of such delta, and then reach a contradiction if this supremum is smaller than $b$. Similar construction done by Jonas in his answer.