Showing that the inclusion $i : \mathbb{C} \to \mathbb{C}_\infty$ is continuous

Question

I need to show that the inclusion map from the complex plane to the extended complex plane is continuous, where the metric on $\mathbb{C}$ is the usual one and on $\mathbb{C}_\infty$, it's:

$$d(z,w) = \frac{2|z-w|}{\sqrt{1+|z|^2}\sqrt{1+|w|^2}}$$

Since $\infty$ isn't really involved here, this basically boils down to showing that the Euclidean metric and this metric (considered as metrics on $\mathbb{C}$) are equivalent, right?

I'm not sure how to do this, since I can't see any way to bound one by the other, really. Any helpful tricks would be appreciated.

Answer 1

Note that the metrics $d$ and $|\cdot - \cdot|$ are not equivalent on $\mathbf C$ (for example because $(n)$ is a Cauchy-sequence in $(\mathbf C, d)$, but not with respect to the Euclidean metric). But this is not needed for $i$ being continuous: We have that $$ d(z,w) \le 2|z-w| $$ as the denominator is $\ge 1$. Therefore $z_n \to z$ in $\mathbf C$ with the Euclidean metric implies $$ d(z_n, z) \le 2|z_n -z| \to 0 $$ that is $i(z_n) \to i(z)$.

Answer 2

You can use the definition of continuity using sequences. Denote by $d_{\operatorname{euc}}(z,w) := |z - w|$ the standard metric on $\mathbb{C}$ and note that for $z,w \in \mathbb{C}$ we have

$$ d(z,w) = \frac{2d_{\operatorname{euc}}(z,w)}{\sqrt{1 + |z|^2}\sqrt{1 + |w|^2}} \leq 2 d_{\operatorname{euc}}(z,w). $$

Thus, if $d_{\operatorname{euc}}(z_n,z) \to 0$ we see that we also have $d(i(z_n), i(z)) = d(z_n, z) \leq 2d_{\operatorname{euc}}(z_n,z) \to 0$ and so $i$ is continuous.