Is $d$ topologically equivalent to the usual metric on $\mathbb{C}$?

Question

Let $d$ be a function defined on $\mathbb{C} \times \mathbb{C}$ by

$$d(z,z') = \begin{cases} 0 &\text{if} \; z= z' \\\\ |z| + |z'| &\text{if}\; z \neq z' \end{cases}$$

Is $d$ topologically equivalent to the usual metric on $\mathbb{C}$?

My attempt: I know that $d$ is a metric on $\mathbb{C}$ but here im confusing that "Is $d$ topologically equivalent to the usual metric on $\mathbb{C}$?

Any hints/solution ?

Answer 1

You should probably review the definition of "topologically equivalent". But one characterization is that two metrics are topologically equivalent iff they have the same convergent sequences.

Consider the sequence $x_n = 1 + \frac{1}{n}$. Does it converge to $1$ with respect to the usual metric on $\mathbb{C}$? Does it converge to $1$ with respect to this new metric $d$?

For that matter, with respect to $d$, is there any sequence that converges to $1$? For any $z_0 \ne 0$, is there any sequence that converges to $z_0$?

Another characterization of topologically equivalent metrics is that they have the same open sets. Is $\{1\}$ an open set with respect to the usual metric? With respect to $d$?