Find $\operatorname{Cl }A$ in topological space $\mathbb{R^2}$ with dictionary order topology. Let $ A = \left\{  (x,y) \in \mathbb{R^2} \mid  y= \sin ( \frac{1}{x}) , \  0 < x \leq 1 \right\}$ . Find $\operatorname{Cl} A$ in topological space $\mathbb{R^2}$ with dictionary order topology.
I guess $ \operatorname{Cl} A = A $? 

You're quite right. To prove it, I would take an arbitrary point not in $A,$ and find an open interval around it containing no points of $A.$ This shows that the complement of $A$ is open, so that $A$ is closed.
