Topologist's sine curve is not path-connected Here I encounter Proof Of Topologist Sine curve is not path connected .But I had doubts in understanding that .
$$ f(x) =\begin{cases}
\sin\left(\dfrac{1}{x}\right) & \mbox{if $0\lt x \lt 1$,}\\\
0 & \mbox{if $x=0$,}\end{cases}$$
S is Range of f(x).I wanted to show that there exist no continous function such that for any a,b$\in$S,$f(0)$=$a$,$f(1)$=$b$.It is apperenernt that a and b are 2 tupple elements .
If $S=\{(0,0)\}\cup\{(x,\sin(1/x)):0<x<1\}$ and $g=(g_1,g_2):[0,1]\to S$ is a path with $g(0)=(0,0)$, then $g(t)=(0,0)$ for all $t$. So we have to show that for any function there is no curve leaving (0,0) point .Any continuous function taking value $(0,0)$ must be single point constant function [Is this my interpretation is right ?].
So On contrary Assume there exist continuous function which take value $(0,0)$ but also take other value .
Here Jonas Meyer Sir claim that By continuity of $g_2$ function There exist $\delta$>0 such that $g_2(t)<1$ whenever $t<\delta $But How this is possible Because $sin(\frac {1}{x}$) takes value [-1,1] for that condition .He also Obtained contradiction at the end using same argument .
Where is I am Missing ? Any Help will be appreciated.
[Math] Proof of Topologist Sine curve is not path connected .
connectednessgeneral-topologypath-connected
Best Answer
Recall an old fashion definition of continuity.
For all $\epsilon $ > 0, there is some $ \delta$ >0 with for all x,
in the domain of f,
$d(x,a)$ < $ \delta$ implies $ d(f(x),f(a)) $< $\epsilon $
Set $a $= 0, $\epsilon $ = 1 and get the claimed results.
In this case, the assumption of continuity leads to a
contradiction if the domain is a non-trivial interval.