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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 ?

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jasmine
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    Two metrics are "topologically equivalent" if they give the same open sets. And you can show that two metrics are "topologically equivalent" if and only if for two open sets, U and V, defined by one metric, such that V is a subset of U, there exist an open set, W, defined by the other metric, such that W is a subset of U and V is subset of W: $V\subset W\subset U$. – user247327 Mar 31 '19 at 01:26
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    Hint: how many points are distance $1$ away from $2i$? – Theo Bendit Mar 31 '19 at 01:28
  • @user247327 i thinks u r talking about comparison – jasmine Mar 31 '19 at 01:29
  • @TheoBendit we will get infinite point – jasmine Mar 31 '19 at 01:30
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    @jasmine Find me one such point then. (Don't say $i$, because $d(i, 2i) = |i| + |2i| = 3$). – Theo Bendit Mar 31 '19 at 02:06
  • @TheoBendit i was eating breakfast,,,,sorry for late reply,,, that mean there is no points so d is not topologically equivalent to the usual metric on $\mathbb{C}$??? – jasmine Mar 31 '19 at 03:16

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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 $d$?

Nate Eldredge
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  • i thinks both ${x_n}$ converges to $1 $ in new metric and usual metric but$ {1} $ is not open in usual @Nate eldredge – jasmine Mar 31 '19 at 04:43
  • @jasmine: Look again. $x_n$ does not converge to $1$ in the metric $d$. If it did, you would have $d(x_n, 1) \to 0$, but what is $d(x_n, 1)$? And you're right that ${1}$ is not open in the usual metric, but in the $d$ metric, it is open. – Nate Eldredge Mar 31 '19 at 17:05