# Shrinking Topologist's Sine Curve

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?

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

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

• Indeed, as $\lim_{x\to 0} x\sin{(\frac{1}{x})} = 0 = f(0)$ – Rustyn Mar 17 '13 at 5:09
• @Rustyn: Exactly. (But I figured that that was pretty well known.) – Brian M. Scott Mar 17 '13 at 5:10
• +1 nice answer@ Brian sir – user525416 May 19 '18 at 1:13

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.