Take the tour ×  I just came across the example of the
topologist's sine curve that is connected but not path-
connected. The rigorous proof of the 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: $$ y(x) = \begin{cases}
\sin\left(\frac{1}{x}\right) & \mbox{if $0\lt x \lt 1$,}\\\ 0 &
\mbox{if $x=0$,}\end{cases}$$  share|improve this question Draw
a picture. The set $ \{(x,\sin \frac{1}{x}) | x \in (0,1) \}$ is
connected