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The Topologist's Sine Curve is well-known:

The set $S = \{(0,0)\} \cup \{(x, \sin(1/x))\ |\ x \in \left]0,1\right]\}$, as a subspace of $\mathbb{R}^2$, is connected but not path-connected.

An intuitive reason is that no path from $S - \{(0,0)\}$ can reach $(0,0)$ in a finite amount of time. However, what if I shrink the distances involved? That is, consider $$ S' = \{(0,0)\} \cup \{(x, x\sin(1/x))\ |\ x \in \left]0,1\right]\}. $$

Is $S'$ path-connected?

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Let $$f:[0,1]\to\Bbb R:x\mapsto\begin{cases}x\sin\frac1x,&\text{if }0<x\le 1\\\\0,&\text{if }x=0\;;\end{cases}$$

then $f$ itself is already a path connecting any two points of $S'$.

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  • $\begingroup$ Indeed, as $\lim_{x\to 0} x\sin{(\frac{1}{x})} = 0 = f(0)$ $\endgroup$ – Rustyn Mar 17 '13 at 5:09
  • $\begingroup$ @Rustyn: Exactly. (But I figured that that was pretty well known.) $\endgroup$ – Brian M. Scott Mar 17 '13 at 5:10
  • $\begingroup$ +1 nice answer@ Brian sir $\endgroup$ – user525416 May 19 '18 at 1:13
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I should point out some subtlety here. The problem is never about infinite length of these curves. Remember that we have space filling (continuous) curves into $R^2$. So, why is it possible to turn the shrinking sine cure into a path, but the same is not possible for the usual sine curve?

The real answer is the non-/existence of the limit at zero. In both cases we already have a curve defined from $(0,1]$ into your space. In one case we can assign a value at $0$, and extend the function's domain continuously to the closed $[0,1]$, while in the other we cannot, since there is no limit at zero.

When can we extend a continuous function from $(0,1]$ to all of $[0,1]$? Precisely, if and only if the (original) function is uniformly continuous.

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