# Topologist's sine curve is connected

I just came across the example of the topologist's sine curve that is connected but not path-connected. The rigorous proof of the non-path-connectedness can be found here.

But how can I prove that the curve is connected? To be honest, even intuitively I am not being able to see that the curve is connected. I am thinking if it is proved that the limit point of $\sin(1/x)$ as $x \to 0=0$, then it would be proved. But, why is this true? IMO, this limit doesn't exist. Intuitively also, it seems that the graph would behave crazily and not approach a particular value as a tends to $0.$

EDIT (Brett Frankel): There are a few different working definitions of the topologist's since curve. For the sake of clarity/consistency, I have copied below the definition used in the linked post: $$y(x) = \begin{cases} \sin\left(\frac{1}{x}\right) & \mbox{if 0\lt x \lt 1,}\\\ 0 & \mbox{if x=0,}\end{cases}$$

• Draw a picture. The set $\{(x,\sin \frac{1}{x}) | x \in (0,1) \}$ is connected simply because $\sin$ is continuous. So the only issue is $(0,0)$. Any open set that contains $(0,0)$ must intersect with the other part of the curve, so $(0,0)$ cannot lie in a disconnected component. – copper.hat Feb 28 '13 at 19:59
• @copper.hat Could you explain with more detail the connected issue in $(0,0)?$ Are you saying that $(0,0)$ lie in the connected set $\{(x,\sin \frac{1}{x}) | x \in (0,1) \}$? – user441848 Jan 24 '18 at 20:45
• @Anneliset.: No, the point $(0,0)$ does not lie in the connected set $C=\{(x,\sin \frac{1}{x}) | x \in (0,1) \}$. However, every open set that contains $(0,0)$ must also contain a point of $C$. Hence the curve is connected. – copper.hat Jan 25 '18 at 4:07
• Why is the set containing only (0,0) not an open set? Is it because the Euclidean topology doesn't allow balls of radius 0? – Alexander Kartun-Giles 1 hour ago

Call the topologist's sine curve $T$, and let $A = \{(x,\sin 1/x)\in\mathbb{R}^2\mid x\in\mathbb{R}^+\}$, $B = \{(x,\sin 1/x)\in\mathbb{R}^2\mid x\in\mathbb{R}^-\}$. Then $T \subseteq \overline{A\cup B} = \overline{A}\cup\overline{B}$. It isn't difficult to show that $A$ and $B$ are connected (even path connected!) and then you just need two lemmas to show that $T$ is connected:

Lemma 1: If $A\subseteq X$ is a connected subset of a metric space $X$ and $A\subseteq B\subseteq \overline{A}$, then $B$ is also connected. Edit: Stefan H. reminds us that this result also holds when $X$ is a general topological space, not just a metric space.

Lemma 2: If $A$ and $B$ are connected, and $A\cap B\neq\emptyset$, then $A\cup B$ is also connected.

And neither of these should be too hard to show (Hint: use the fact that if $X$ is connected and $f : X\to\{0,1\}$ is continuous, then $f$ is constant).

• I would just like to add that Lemma 1 works in arbitrary topological spaces. – Stefan Hamcke Feb 28 '13 at 20:31
• is it true that $T=\overline{A \cup B}$? – creepyrodent Dec 3 '18 at 0:35
• No, $\overline{A\cup B}$ contains all points $(0,y)$ for $y \in [-1,1]$, but $T$ only contains $(0,0)$ by definition for points of the form $(0,y)$. – James Yang Oct 24 at 3:21

If the graph $X$ of the topologist's sine curve were not connected, then there would be disjoint non-empty open sets $A,B$ covering $X$. Let's assume a point $(x,\ \sin(1/x))\in B$ for some $x>0$. Then the whole graph for positive $x$ is contained in $B$, only leaving the point $(0,0)$ for the set $A$. But any open set about $(0,0)$ would contain $(1/n\pi,\ \sin(n\pi))$ for large enough $n\in\mathbb N$, thus $A$ would intersect $B$.

• Doesn't disconnected mean that there are two disjoint open sets $A$ and $B$ such that the entire set $S$ is equal to $A\cup B$? But when you take your definitions of $A$ and $B$, the union of them is more than what is needed, no? – AlanH Jun 17 '13 at 22:48
• @AlanH: "$A$ and $B$ cover a set" can be interpreted in two ways: 1.) There union contains $S$, or 2.) There union equals $S$. But if $A$ and $B$ are sets in $\Bbb R^2$ covering $S$, then $A\cap S$ and $B\cap S$ are sets in $S$ whose union is $S$, and if they are open, then their intersections with $S$ are open relative to $S$. Furthermore, we want them to be disjoint in $S$, but this doesn't requite $A$ and $B$ to be disjoint in $\Bbb R^2$. – Stefan Hamcke Jun 18 '13 at 14:32
• You can just stop at "leaving only $(0,0)$ for the set $A$" because $A=\{0,0\}$ is not open. – user309475 Apr 26 '17 at 14:44

It's clear that two of the "pieces" of this set are connected--path connected, in fact. So what we need to argue is that there is no separation between them. That is, it is not possible to find a pair of disjoint open sets such that the "$\sin$" part of the curve is contained in one and the origin is in the other.

So it will suffice to show that any open set that contains the origin will also intersect the other piece. What happens if you consider a small ball centered at $0$?

It's connected because it is the closure of a path-connected (hence connected) set.

If $S=\{x\times \sin(1/x): 0<x\le 1\}$ then $S$ is the image of the connected set $(0,1]$ under a continuous map, hence $S$ is (path) connected. And the topologist's sine curve is $(0\times[-1,1])\cup S=\bar S$ and we know that the closure of a connected set is connected.

you can define the topologist's sine curve as $S\subset \mathbb{R}^2$ with $S=((x,sin(1/x), x\in (0,1]))$. Note that a continuous function with $f((0,1])\subset S$ is the continuous image of a connected set, and hence connected. Just define $f:(0,1]\rightarrow \mathbb{R}^2$, and note that the restriction of a continuous map is continuous $f:[0,1)\rightarrow S$

You can actually replace $(0,1]$ with any connected subset $C\subset \mathbb{R}$ which does not contain $0$. Properties of the space $S$ are not required to show connectedness, just construct a continuous function from a connected interval without $0$ into $S$ and you're done!