TITLE: Topologist's sine curve is connected
QUESTION [31 upvotes]: 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:
$$ y(x) = \begin{cases}
\sin\left(\frac{1}{x}\right) & \mbox{if $0\lt x \lt 1$,}\\\
0 & \mbox{if $x=0$,}\end{cases}$$

REPLY [2 votes]: 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!