# Topologist's sine curve

Is there a (preferably elementary) proof that the graph of the function $y$ defined on $[0,1)$ by $$y(x) = \left\{\begin{array}{ll} \sin\left(\frac{1}{x}\right) & \mbox{if 0\lt x \lt 1,}\\ 0 & \mbox{if x=0,}\end{array}\right.$$ is not path connected?

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Somehow, it looks as if the choice of notations for an open interval and a point's coordinates seems unfortunate when used in the same sentence. – J. M. Apr 25 '11 at 18:30
@J.M. Perhaps my fault. I'm trying to figure out how to phrase it. – Arturo Magidin Apr 25 '11 at 18:30

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 $t_0<\delta$.
May I ask what exactly $f_1,f_2$ are here? Do you mean $f_1=x,f_2=\sin{(1/x)}$ or? – Cancan Apr 17 '13 at 18:55
Cancan: $f:[0,1]\to S$ is a function whose codomain is a subset of $\mathbb R^2$, so for each $t\in [0,1]$, $f(t)$ is a point in $\mathbb R^2$, and that point is named $(f_1(t),f_2(t))$. Thus $f_1$ and $f_2$ are the component functions. If you want explicit formulas for $f_1(t)$ and $f_2(t)$, they are $f_1(t)\equiv 0$, and $f_2(t)\equiv 0$. However, that is not supposed at the beginning. $f$ is an arbitrary path in $S$ starting at $(0,0)$, and the goal is to prove that $f(t)=(0,0)$ for all $t$. – Jonas Meyer Apr 18 '13 at 0:20
(continued) @Cancan: Because $f(t)$ is in $S$ for all $t$, if $f(t)$ is not equal to zero for some particular $t$, then $f_1(t)$ and $f_2(t)$ satisfy the equation $f_2(t)=\sin(1/f_1(t))$ for that $t$; that is by definition of $S$. – Jonas Meyer Apr 18 '13 at 0:21