Is A = $\{(0,0)\}\cup \{(x,\sin(1/x))|0 < x \le 1\}$ $\subseteq \mathbb{R}^2$ under the usual metric on $\mathbb{R}$ is compact?

Question

Is A = $\{(0,0)\}\cup \{(x,\sin(1/x))|0 < x \le 1\}$ $\subseteq \mathbb{R}^2$ under the usual metric on $\mathbb{R}$ is compact ?

I thinks yes , because A is closed and bounded

Is its True??

Answer 1

The set is not closed. $(0,1)$ is a point in the closure but it is not in this set. [Note that $(\frac 1 {(4n+1)\pi /2}, \sin ((4n+1)\pi /2)) \to (0,1)$].