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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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marked as duplicate by Martin Sleziak, Adriano, Grumpy Parsnip, Potato, Amzoti Jun 27 '13 at 4:33

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this seems relevant: en.wikipedia.org/wiki/Topologist's_sine_curve –  Holdsworth88 Dec 11 '12 at 13:28
    
@Holdsworth88 It doesn't really prove that it is not path connected and i wanna ask if my argument presented above is correct? –  Mathematics Dec 11 '12 at 13:31

2 Answers 2

Indeed, your proof is correct, the graph $X$ of the function $x \longmapsto \sin(1/x)$ is clearly path connected, but if you add $(0,0)$, it is no longer true. Besides, considering $\bar{X}$ the closure of $X$ (we actually have $\bar{X}=X \cup I$, where $I=\{ (0,y), y \in [-1,1]\}$), $\bar{X}$ is a nice example of a set which is connected but not path connected.

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$X$ itself is connected but not path connected; there is no need to pass to $\overline{X}$. –  Rhys Dec 11 '12 at 13:42
    
Sorry if this was not clear but in my notations $(0,0)$ is not in $X$. Besides, I find it more natural to work with $\overline{X}$, which is obviously connected, but this may be a matter of personal taste. –  Corentin Dec 11 '12 at 14:39
    
Sorry, I missed how you defined $X$. My previous comment applies to the set defined in the OP. –  Rhys Dec 11 '12 at 14:58

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.

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