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:
$$ y(x) = \begin{cases}
\sin\left(\frac{1}{x}\right) & \mbox{if $0\lt x \lt 1$,}\\\
0 & \mbox{if $x=0$,}\end{cases}$$
Best Answer
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 $B$.