Proof of Topologist Sine curve is not path connected .

Question

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.

Answer 1

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.

Answer 2

Actually, you want to show that there are two points in $S$ that cannot be connected by a continous function $f \colon [0,1] \to S$. Just observe that if $x=0$ and $y \in (0,1)$ the existence of a continous path from $(y,f(y))$ to $(x,f(x))$ should implie that $f$ has a limit on $x=0$, which from calculus you should know is not true. The essence of the statement is proving what I'm telling you.

Answer 3

In your question you stated that $S$ is the range of $f$. You meant to say let $S$ be the graph of the function $f$. It is best to simply think of $S$ as a well defined set and to think of the function $f$ as an artifact.

The following is not difficult to prove (c.f. Cantor's intersection theorem):

Proposition 1: Let $\gamma: [0,1] \to \mathbb R \times \mathbb R$ be any path and $\alpha_n$ a sequence of positive numbers that converge to $0$. For each $n$ set $A_n = \gamma(\,\left(0,\alpha_n\right]\,)$. Then the intersection (direct limit) of the closures of the $A_n$ is a singleton,

$\tag 1 \bigcap \overline{A_n} = \{\,\gamma(0)\,\}$

To show that $S$ is not path connected, it suffices to show that there is no path connecting $(0,0) \in S$ to a different point in $S$.

In what follows we make use of $\pi_x$, the projection mapping onto the $x\text{-axis}$.

To get a contradiction, if such a path exists, we can construct another path connecting the same endpoints satisfying

$\tag 2 \gamma(0) = (0,0) \text{ and for } t \gt 0, \;(\pi_x \circ \gamma)\,(t) \gt 0$

(see Jonas Meyer's argument).

Exercise 1: Show that for any $0 \lt \delta \le 1$, the range of $\pi_x \circ \gamma$ applied to the interval $(0,\delta]$ contains the interval $(0, \delta^{'}]$ where $\delta^{'} = \pi_x \circ \gamma(\delta)$.
Hint: The range of any path is connected.

Exercise 2: Show that the positive numbers $\alpha_n = (\pi_x \circ \gamma)\,(\frac{1}{n})$ converge to zero.

Exercise 3: Show that the LHS of (1) contains the two points $(0,0)$ and $(0,1)$.

But upon completing exercise 3, you find that $\text{(1)}$ is false, a contradiction.

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The above is essentially the same solution that Jonas Meyer provides, but with a more elaborate and step-by-step exposition.