Not closed path-component

Question

Where can I read the proof Is there any path-component not closed? with more details (for example, why graph is a path-component)?

Answer 1

In the topologist's sine curve $X= \{0\} \times [-1,1] \cup \{(x, \sin(\frac{1}{x}: x \in (0,1]\}$ the two sets we are taking the union of (the $y$-axis part $Y$ and the graph-part $G$) are also its path-components, because the usual proof (found many times over on this site and others) shows that we cannot have a continuous path between a point of $Y$ and one of $G$. As $Y \simeq [0,1]$, its path-connected and $G$ is clearly the continous image of the path-connected set $(0,1]$, so also path-connected. The non-path fact implies that these sets are maximally path-connected, so the path-components.

And $G$ is not closed, but even dense in $X$ as all points of $Y$ lie in its closure.