Show that $\{(x,y): 0 < x \leq 1; y = \sin{1/x}\} \cup \{z: x = 0; -1 \leq y \leq 1\}$ is not path connected

Question

Show that $\{(x,y): 0 < x \leq 1; y = \sin{1/x}\} \cup \{z: x = 0; -1 \leq y \leq 1\}$ is not path connected. The book gives this set as an example of a closed, connected but not path connected set. I don't understand how it is possible though. The first component is continuous everywhere except $(0,0)$ and the second component contains $(0,0)$ too and is continuous on every point. Clearly they are path connected individually. Which are those 2 points one on the first component, the other on second component, that can't be connected with a curve from the given set?

Answer 1

Suppose there is a path $f(t)=(a(t), b(t))$, $t\in [0, 1]$ and $f(0)$ is a point in the upright segment and $f(1)$ is a point on the given curve. Then if $t_0$ such that the path $f$ leaves the upright segment then the limit $\lim_{t\to t_0^{+}} f(t)$ should exist with also $b(t)=\sin (1/t)$ for $t\in (t_0, 1]$. But arbitary near any point of the upright segment $b(t)$ takes values from $1 $ to $-1$.

EDIT

Consider a circle of arbritary small radius and center the point $A=f(t_0)$ then from the semi plane $x>0$ the topologist's sine curve goes in and out this circle infinity many times. That means that the path cannot be continuous near the point $A$, which is absurd.

Please also note that the continuous path $f(t)$ lying in the semi plane $x>0$ may not have the parametrization of $f(t)=(t,\sin(1/t))$, (although I consider as such for simplicity) but in all cases should have the same track of the curve $(t,\sin(1/t)), t\in (0, 1]$.