TITLE: Is $\{(0,0)\}\cup\{(x,\sin{1\over x}):x\in\mathbb{R},x>0\}$ path connected?
QUESTION [5 upvotes]: Is $\{(0,0)\cup\{(x,\sin{1\over x}):x\in\mathbb{R},x>0\}$ path connected? 
I think it is path connected if we neglect the point$(0,0)$ it is as we can define a continuous function easily from $[0,1]$ but if we included the point $(0,0)$, then any continuous functions seems would disconnected at $(0,0)$ as the $\sin{1\over x}$ vibrate very quick between the value $0,1$for $x$ close to$0$ so for any $\delta$-ball at the origin, then choose $\epsilon={1\over2}$, then there must exist some $x\in B_{\delta}((0,0))$ where $\sin{1\over x}>\epsilon$. Is it correct? if not, is there any way can show it more clearly? Or how to present it in a better way?

REPLY [5 votes]: Suppose that there is a continuous path $\phi:[0,1]\rightarrow\mathbb{R}^2$ joining $(0,0)=\phi(0)$ and any point in the graph of $\sin(\frac{1}{x})$. 
Now take any sequence $x_n>0$ such that $x_n\rightarrow 0$. Because $\displaystyle\sin(\frac{1}{x_n})=\phi(y_n)$ for some sequence $y_n>0$ such that $y_n\rightarrow 0$, you can conclude that $\sin(\frac{1}{x})\rightarrow 0$, which is na absurd.