proving a set is not path-connected Let 
$$A=\{(x,\sin(1/x)) \mid x > 0 \} \cup \{(0,y) \mid y\in[-1,1]\} \subset\mathbb{R}^2$$
how to prove that this set is not path-connected(or just not connected)?
Thank you
 A: Let $C$ be that sine curve part, and $S$ be the vertical segment. Suppose there exists a path $\omega: [0, 1] \to A$ that connects a point $(0, 1) \in S$ with $(\frac{1}{\pi}, 0) \in C$. Consider the set $U = \{t \in [0, 1]: t \in C \} = \omega^{-1}(C)$. As sine curve is open in $A$, $U$ is an open subset of $[0, 1]$. Let $t_0 = \inf U$. As $U$ is open and $0 \not \in U$ (because $\omega(0) = (0, 1) \in S$, $\omega(t_0) \in S$, but for some $t_1 > t_0$, $\omega((t_0, t_1)) \subset C$. Consider a small open neighbourhood $V \subset A$ of $t_0$ that has infinitely many disconnected components. By continuity of $\omega$, there's some interval $(t_0 - \epsilon, t_0 + \epsilon)$ such that $\omega((t_0 - \epsilon, t_0 + \epsilon)) \subset V$. For some point in $t \in (t_0, t_0 + \epsilon)$, $\omega(t)$ is in a different component of $V$ than $t_0$, which is a contradiction, because $\omega((t_0 - \epsilon, t_0 + \epsilon))$ is connected, as a continuous image of a connected set.
A: Suppose this set is path connected. Consider the point $(0,0)$ and the point $(1,\sin(1))$. The only way to link these two points is along the path $(x,\sin(\frac{1}{x}))$. However $\lim_{x \rightarrow 0}\sin(\frac{1}{x})=$DNE. Hence there is no path connecting $(0,0)$ to $(1,\sin(1))$, a contradiction. Therefore the set is not path connected.
