What are the interior and boundary of $\Bbb R^2 - \{(x,sin(1/x))|x>0\}$ in $\Bbb R^2$?

Question

In $\Bbb R^2$, is the interior of $\Bbb R^2 - \{(x,sin(1/x))|x>0\}$ itself? The curve is continuous on $(0,\infty)$ so its graph is closed, and hence the set is open. Moreover, its boundary is $\{(x,sin(1/x))|x>0\}$.

Am I correct?

Answer 1

The boundary of $$\Bbb R^2 - \{(x,sin(1/x))|x>0\}$$ is $$B=\{(x,sin(1/x))|x>0\}\cup (\{0\}\times [-1,1])$$

The interior is$$\Bbb R^2 - B$$