# Understanding the following proof about path-disconnectedness of topologist sine curve.

Given the function $$f$$ from $$\Bbb R^+_0$$ to $$\Bbb R$$ defined by the equation $$f(x):=\begin{cases}\sin\Big(\frac 1 x\Big),\,\text{if }x\in\Bbb R^+\\0,\,\text{if }x=0\end{cases}$$ for any $$x\in\Bbb R^+_0$$ then we call Topologist sine curve the graph of $$f$$ that is the set $$\mathcal G_f:=\biggl\{x\in\Bbb R^2:x_1\ge 0\wedge x_2=\sin\Big(\frac 1 {x_1}\Big)\biggl\}\cup\big\{\mathbf 0\big\}$$

Now it is a well know result that $$\cal G_f$$ is connected but not path-connected and thus I would like to discuss this about the path-disconnectedness of $$\cal G_f$$: so I do not really understand why it is possible I have to remove a part of $$[0,1]$$ and thus rescaling; then I do not understand why the path-disconnectedness of $$\cal G_f$$ follows by showing that $$\sup\Big\{t\in[0,1]:f\big([0,t]\big)=\mathbf 0\Big\}=0$$ finally I would like to see a formal proof about the subjectivity of $$\sin\Big(\frac 1 x\Big)$$ on $$(0,\varepsilon)$$ onto $$[-1,1]$$.

Please do not closed this question: unfortunately the users who wrote the linked answer and question are inactive by many time so that I cannot use comments to ask clarifications but the linked answer seems very simple and unfortuately I did not understand other proofs about the path-disconnectedness of $$\cal G_f$$ so that I really like to understand the linked proof: so could somoene help me, please?

• For surjectivity of $\sin \frac 1x$ on $(0,\epsilon)$ onto $[-1,1]$, you just need to find two points whose images are $1$ and $-1$ (use appropriate values of $x$ for which the sine equals $\pm 1$) and apply the intermediate value theorem. Jun 20, 2022 at 11:14
• @SarveshRavichandranIyer Okay, so I know that $\sin\Big(\frac 1 x\Big)=1$ if and only if $\frac 1 x=\frac \pi 2+2k\pi$ for any $k\in\Bbb Z$ . Now the last identity show that $x=\frac 1{\frac \pi 2+2k\pi}$ for any $k\in\Bbb Z$ but the sequence $$x_n:=\frac 1{\frac \pi 2+2n\pi}$$ converges to $0$ so that for any $\epsilon$ there exsist $n_0\in\Bbb N$ such that $$x_n\in (0,\epsilon)$$ which is such that $$\sin\Big(\frac 1{x_n}\Big)=1$$ Then by analogous argumens we can find $x\in (0,\epsilon)$ such that $\sin\Big(\frac 1 x\Big)=-1$, right? Jun 20, 2022 at 11:22
• You're right, Antonio. (Some of the MathJax didn't render in your comment, but I read off the obvious expressions). Now, apply the intermediate value theorem as follows : imagine you got $z \in (0,\epsilon)$ such that $\sin \frac{1}{z} = 1$ and $z' \in (0,\epsilon)$ such that $\sin \frac{1}{z'} = -1$. Then, in the interval $[z,z']$ (or $[z',z]$ depending on which is bigger), for every $-1 \leq c \leq 1$ there must be a $y \in [z,z']$ such that $\sin \frac{1}{y} = c$. So what does this tell you about the range of $\sin \frac{1}{x}$ as a function on $(0,\epsilon)$? Jun 20, 2022 at 11:27
• Another proof of this property of the topologist’s sine curve, with two simpler examples presented first, is at kconrad.math.uconn.edu/blurbs/topology/connnotpathconn.pdf and you might prefer the argument there.
– KCd
Jun 20, 2022 at 13:33

I do not understand why the path-disconnectedness of $$\mathcal{G}_f$$ follows by showing that: $$\sup_{t\in[0,1]}\{f([0,t])=\{0,0\}\}=0$$

This is a misinterpretation of the linked answer. They don’t show that, indeed they suppose that by a “without loss of generality” argument.

Let me call the sine curve $$T$$ and the first and second components of a function into $$T$$ by subscript $$1,2$$ respectively. For such a function, $$f_2=\sin(1/f_1)$$ is forced (when $$f_1\neq0$$). Suppose $$f$$ is any path in $$T$$ joining $$\{0,0\}$$ to some other anonymous distinct point. It is impossible for the above supremum to equal $$1$$ (why?). Furthermore $$f(0)=\{0,0\}$$, and a point implicitly used later in their answer is that if $$f_1(t)=0$$ then $$f_2(t)=0$$ is forced. If $$t_0\gt0$$ is the supremum, by continuity $$f(t_0)=\{0,0\}$$. Then a rescaling of the function $$f$$ to the continuous path (with the same endpoints): $$g:[0,1]\to T, \,t\mapsto\begin{cases}f(t_0+t(1-t_0))&t\in(0,1]\\\{0,0\}&t=0\end{cases}$$Would set the supremum to $$0$$ and moreover not change the validity of the following argument.

The argument can be restated as follows. We take without loss of generality this modified path $$g$$ which has the supremum of $$0$$ and by continuity $$g(t)\neq\{0,0\}$$ for $$t\gt0$$. Continuously pullback the unit ball about the origin by $$g$$, yielding an interval $$[0,t’)$$ such that $$|g_2|\lt1$$ on this interval. This is the problematic step: try visualising this, it is impossible for any path touching $$\{0,0\}$$ to avoid $$1$$ continuously. Now by the intermediate value theorem $$g_1$$ attains an interval of nonzero values in $$g_1[0,t’)$$ (by the without-loss-of-generality argument and the italicised remark $$g_1[0,t’)$$ is a nondegenerate interval), so by the surjectivity of the reciprocal sine curve (consider $$\frac{2}{n\pi},n\in\Bbb Z$$, will eventually fall into any interval about $$0$$) it must be the case that $$g_2([0,t’))=[-1,1]$$, a contradiction (recall $$g_2=\sin(1/g_1)$$).

I hope that clarifies things, feel free to query further. Note that there is also nothing special about the number $$1$$ here; it is just aesthetic.

• So if $f(1)\neq \mathbf 0$ then surely $t_0\neq 1$, right?; then if $f_1(t)=0$ it cannot be $f_2(t)\neq 0$ because otherwise it would be $$f_2(t)=\sin\frac 1{f_1(t)}$$ and this is impossible if $f_1(t)$ is zero, right?; then if $t<t_0$ there must exists $t<t^*<t_0$ such that $f\big[[0,t^*]\big]=\mathbf 0$ so that $f(t)=\mathbf 0$ for any $t<t_0$ and thus if $(t_n)_{n\in\Bbb N}$ is a sequence in $[0,t_0)$ converging to $t_0$ then by continuity $$f(t_0)=\lim_{n\in\Bbb N}f(t_n)=\mathbf 0$$ but now I am not able to find a such sequence and this seem a problem to me, so what can you say about? Jun 20, 2022 at 13:55
• Now wait some minutes for other clarifications: I have to read carefully the last part of the answer. Jun 20, 2022 at 14:03
• $f^{-1}\{0,0\}$ is a closed set by continuity and contains its supremum. It cannot contain $1$ as $f(1)\neq\{0,0\}$ by definition of path, and thus $t_0\neq1$. For the same reason, $t_0\in f^{-1}\{0,0\}$, hence $f(t_0)=\{0,0\}$. @AntonioMariaDiMauro It is often helpful to go for pure topological argument rather than using sequences, this clarifies the situation (context dependent - of course sequences are very common and useful in topology generally) Jun 20, 2022 at 14:08
• I should add that by definition of the topologist’s sine curve, $f_1(t)=0$ forces $f_2(t)=0$ (as already remarked, and as used later on in the post) Jun 20, 2022 at 14:09
• The “same reason” is: $\{0,0\}$ is closed and $f$ is continuous, hence $f^{-1}\{0,0\}$ is closed and contains its limit points, in particular its supremum. As $t_0=\sup f^{-1}\{0,0\}$, $t_0\in\{0,0\}$ and $f(t_0)=\{0,0\}$. We have $t_0\neq1$ immediately from this since $f$ must connect to another point distinct from the origin. Jun 20, 2022 at 14:14