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.

  • $\begingroup$ Actually you are using the same notation $f$ to denote $\sin (1/x)$ and the continuous function $[0,1]\to S$, which is quite confusing. $\endgroup$ – user99914 May 6 '18 at 5:08
  • $\begingroup$ I had done changes accordingly .But Could you suggest something I had been with this problem For Yesterday?Still Not getting Sir $\endgroup$ – idon'tknow May 6 '18 at 5:14
  • 1
    $\begingroup$ Thanks for Helping Me .Major confusion happen due to fact that I assumed $g_2(t)=sin(1/t)$ Now I understand that argument . $\endgroup$ – idon'tknow May 6 '18 at 6:02

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.

  • $\begingroup$ Sir , I thought $f_2(x)$=$sin(1/x)$ and that function takes value 1 for many values in above setup.Where is I am missing? $\endgroup$ – idon'tknow May 6 '18 at 4:05
  • $\begingroup$ @idon'tknow. There is no f$_2$. $\endgroup$ – William Elliot Nov 29 '20 at 8:36

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.

  • $\begingroup$ Intuition is right But How to Prove that using $\epsilon -\delta$ definition that is my problem $\endgroup$ – idon'tknow May 6 '18 at 5:00
  • $\begingroup$ If $t_0 \in [0,1]$ is such that $g_2(t_0)=0$ then, for continuity, $|g_2(t)|<1$ for $t$ sufficiently close to $t_0$. That only can happen if $g_2$ equals $0$ around $t_0$. Take the supremmum of the set of $t \in [0,1]$ such that $g_2(s)=0$ for all $s \in [0,t]$. The argument above shows that this supremmum equals $1$. So, as you say, the only continous path starting at $(0,0)$ must be constant. $\endgroup$ – 97DL May 6 '18 at 5:24

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.

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.