Is the usual topology on the upper half plane same as that induced by Riemannian metric?

Question

The upper half plane is a Riemannian manifold, with the Riemannian metric given by $(ds)^2 = (dx^2+dy^2)/y^2$ and thus has a metric topology induced by this metric. Is this topology same as the topology inherited as a subspace of $\mathbb{R}^2$ (or $\mathbb{C}$)?

Answer 1

Yes, it is a general fact that the topology induced by the metric that is induced by the Riemannian metric is the same as the original topology of the manifold. In this case you can check it by hand by drawing $\epsilon$ balls of the standard metric, and the metric induced by your Riemannian metric.