# Topologist's sine curve is not path-connected

Is there a (preferably elementary) proof that the graph of the (discontinuous) function $$y$$ defined on $$[0,1)$$ by $$y(x) =\begin{cases} \sin\left(\dfrac{1}{x}\right) & \mbox{if 0\lt x \lt 1,}\\\ 0 & \mbox{if x=0,}\end{cases}$$ is not path connected?

If $S=\{(0,0)\}\cup\{(x,\sin(1/x)):0<x<1\}$ and $f=(f_1,f_2):[0,1]\to S$ is a path with $f(0)=(0,0)$, then $f(t)=(0,0)$ for all $t$.

To see this by contradiction, suppose that $f(t)$ is not always $(0,0)$. Removing an initial part of the interval and then rescaling if necessary, assume that $0=\sup\{t:f([0,t])=\{(0,0)\}\}$. By continuity of $f_2$, there is a $\delta>0$ such that $|f_2(t)|<1$ for all $t<\delta$. Take $t_0$ with $0<t_0<\delta$ and $f_1(t_0)>0$. By continuity of $f_1$ and the intermediate value theorem, $[0,f_1(t_0)]$ is in the image of $f_1$ restricted to $[0,t_0]$. Since $f_2(t)=\sin(1/f_1(t))$ for all $t$ with $f_1(t)\neq0$ and $\sin(1/x)$ maps $]0,\varepsilon[$ onto $[-1,1]$ for all $\varepsilon>0$, it follows that $[-1,1]$ is in the image of $f_2$ restricted to $[0,t_0]$. This contradicts $t_0<\delta$.

• May I ask what exactly $f_1,f_2$ are here? Do you mean $f_1=x,f_2=\sin{(1/x)}$ or? – Cancan Apr 17 '13 at 18:55
• Cancan: $f:[0,1]\to S$ is a function whose codomain is a subset of $\mathbb R^2$, so for each $t\in [0,1]$, $f(t)$ is a point in $\mathbb R^2$, and that point is named $(f_1(t),f_2(t))$. Thus $f_1$ and $f_2$ are the component functions. If you want explicit formulas for $f_1(t)$ and $f_2(t)$, they are $f_1(t)\equiv 0$, and $f_2(t)\equiv 0$. However, that is not supposed at the beginning. $f$ is an arbitrary path in $S$ starting at $(0,0)$, and the goal is to prove that $f(t)=(0,0)$ for all $t$. – Jonas Meyer Apr 18 '13 at 0:20
• (continued) @Cancan: Because $f(t)$ is in $S$ for all $t$, if $f(t)$ is not equal to zero for some particular $t$, then $f_1(t)$ and $f_2(t)$ satisfy the equation $f_2(t)=\sin(1/f_1(t))$ for that $t$; that is by definition of $S$. – Jonas Meyer Apr 18 '13 at 0:21
• @Johans, I get it now, Thanks! Sorry, I confused myself. – Cancan Apr 18 '13 at 0:35
• The path can leave $(0,0)$ and then return to it some time later. So the assumption $0=\sup \{t: f(t)=(0,0)\}$ is in fact restrictive, and cannot be guaranteed by cutting and rescaling. Nonetheless the argument is essentially working. – Behnam Esmayli Nov 15 '18 at 22:16

Assume, to get a contradiction, that the graph $$G \subset \mathbb R \times \mathbb R$$ of the function

$$y(x) =\begin{cases} \sin\left(\dfrac{1}{x}\right) & \mbox{if 0\lt x \lt 1}\\\ 0 & \mbox{if x=0}\end{cases}$$

is path-connected.

Let $$\gamma$$ be a path in $$G$$ connecting $$(\frac{1}{2},sin(\frac{1}{2}))$$ to $$(0,0)$$. We know that the image $$K$$ of $$\gamma$$ is also compact.

The sets defined by

$$\tag 1 L_n = K \cap \{(x,y) \in \mathbb R \times \mathbb R \, | \, x \le \frac{1}{n} \text{ and } y = 1\} \text{ for } n \ge 2$$

define a decreasing chain of closed subsets of $$K$$.

Now for any $$0\lt\alpha\lt\frac{1}{2}$$, the path $$\gamma$$ must pass through $$(\alpha, sin(\frac{1}{\alpha})$$. This follows since the image of $$\gamma$$ is connected and can't be disconnected by the two open half planes defined by $$x = \alpha$$. Also note that for every $$k \ge 1$$, the point $$\left(sin(\frac{\pi}{2} + 2 \pi k)^{-1}), 1\right)$$ belongs to $$G$$. So each $$L_n$$ must be a non-empty closed subset $$K$$.

Recall the following general theory for a compact topological space $$X$$:

Any collection of closed subsets of X with the finite intersection property has nonempty intersection.

We know that our chain $$L_2 \supset L_3 \supset L_4\supset \dots$$ of closed sets in $$K$$ satisfies the finite intersection property, so the intersection must be nonempty.

A simple argument shows that the intersection of the $$L_n$$ must exclude all of $$G$$ except perhaps the singleton set containing $$(0,0)$$. But this point does not lie on the line $$y = 1$$, so the intersection is indeed empty. But this is absurd.

Let $$\beta \in \mathbb R$$. Using the argument above, we can also show that the graph of the function

$$y(x) =\begin{cases} \sin\left(\dfrac{1}{x}\right) & \mbox{if 0\lt x \lt 1}\\\ \beta & \mbox{if x=0}\end{cases}$$

can't be path-connected.

Using this fact, one can show that the Topologist's sine curve as defined by Munkres is also not path-connected; see this stackexchange answer.

• You answer this too late but still helpful! – Cloud JR Sep 23 '18 at 17:33
• I think there is a small typo in this point $\left(\frac{\pi}{2} + 2 \pi k)^{-1}, 1\right)$ ? $\ ($ is missing. – Bellatrix Sep 27 '18 at 23:31

An attempt at rewriting Jonas Meyer's answer. Also I prove that closure of sine curve, which contains a whole segment on $$y$$-axis is not path-connected:

Let $$f=(f_1,f_2):[a,b]\to S$$ be any continuous map such that $$f(a)=(0,0)$$. Since projections $$(x,y) \to x$$ and $$(x,y) \to y$$ are continuous maps $$\mathbb{R}^2 \to \mathbb{R}$$, we see that both $$f_1:[a,b] \to \mathbb{R}$$ and $$f_2$$ are continuous.

By continuity at $$a$$ of $$f_2$$, there exists a $$\delta > 0$$ such that $$(*) \ \ \ \ |f_2(x) - 0| < \frac{1}{2} \ ,$$ for all $$a\leq x \leq a+\delta$$.

Claim: For all $$a\leq x \leq a+\delta$$, $$f_1(x)=0$$.

Proof of the claim: Assume $$f(x^*) = \tau >0$$ for some $$a < x^* \leq a+\delta$$, $$f_1(x)=0$$, then since $$f_1$$ is continuous, by Intermediate Value Theorem $$f_1([a,x^*]) \supset [0,\tau) \$$ .

Therefore there exists some $$t=\frac{1}{2n\pi+\pi/2}$$ in the range of $$f([0,\delta)) \$$, i.e there exists $$\tilde{x} \in [a,a+\delta]$$ such that $$f_1(\tilde{x})=t$$ We see that $$f_2(\tilde{x}) = \sin(1/t) =1,$$ which contradicts (*) above.

This proves the claim. The claim means that $$f$$ cannot leave the $$y$$-axis for some while ($$\delta$$ here.)

But now applying the same argument to the continuous map $$[a+\delta , b] \to S$$, we see that $$f$$ will NEVER be able to leave the $$y$$-axis.$$^1$$ In other words, we proved that any continuous path into $$S$$ starting on $$y$$-axis remains on $$y$$-axis. This refutes the possibility of paths between $$y$$-axis points and other points outside it.

$$^1$$ To make this precise we define $$sup$$ of such delta, and then reach a contradiction if this supremum is smaller than $$b$$. Similar construction done by Jonas in his answer.