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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?

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    this seems relevant: http://en.wikipedia.org/wiki/Topologist's_sine_curve2012-12-11
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    @Holdsworth88 It doesn't really prove that it is not path connected and i wanna ask if my argument presented above is correct?2012-12-11

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