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Let $ d_1 $ $ d_2 $ be equivalent metrics in an non empty set X

  1. If $U$ is $ d_1 $ open then $U$ is $d_2$ open.
  2. If $U$ is $ d_1 $ closed then $U$ is $d_2$ closed.
  3. If $U$ is $ d_1 $ bounded then $U$ is $d_2$ bounded.
  4. Constant function is $ d_1$ - $d_2 $ continuous.
  5. identity function is $ d_1$ - $d_2 $ continuous.
  6. If $U$ is $ d_1 $ open ball then $U$ is $d_2$ open ball.
  7. $ v(x,y) $ = $ | d_1(x,y) -d_2(x,y)|\,$ is a metric on $X$.
Javi
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Infinity78
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    Just write down the definition of equivalent metrics, and say what's the problem with 4). I'm pretty sure, you can do this! – SBF Jun 07 '13 at 17:05
  • Do you know any equivalent metrics? Here's an example: the euclidean metric and the taxicab metric on $\mathbb{R}^2$ are equivalent. Play around with them - it might help you to get a feel for things. (Actually, there are two notions of equivalence, and I don't know which you're talking about, but these two metrics are equivalent in both senses.) – Billy Jun 07 '13 at 17:11
  • @Ilya: Why is question 4. a problem? – copper.hat Jun 07 '13 at 17:34
  • Since the constant function is contious with respect to any mertic 4) is correct right? Even if the metrics weren't equivalent – Infinity78 Jun 07 '13 at 18:01
  • @copper.hat in fact, I meant 3) which was less trivial but still easy nuff to start thinking of how to deal with equivalent metrics. Typo – SBF Jun 08 '13 at 09:00

1 Answers1

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I assume you say

DEF Two metrics $d_1,d_2$ are equivalent on some set $X$ if they generate the same topology.

You should see that $1.$ and $2.$ are true using the definition of equivalence of metrics.

For $3.$, take any unbounded metric $d(x,y)$ and set $d'(x,y)=\min\{d(x,y),1\}$.

For $6.$ consider the Euclidean metric versus the $\max$ metric on $\Bbb R^n$ to come up with a counterexample.

What do you mean by the identity being $d_1-d_2$ continuous? Do you mean $\operatorname{id}:(X,d_1)\to(X,d_2)$ is continuous? If so, again look at the definition of equivalence of metrics. Take an open set $G$ in $(X,d_2)$. It's preimage is the same set. Is it open in $(X,d_1)$?

For $4.$, recall that constant functions are always continuous.

For $7.$, can you find two points such that $x\neq y$ but $d_1(x,y)=d_2(x,y)$? In such a case, you will find this violates that $v(x,y)=0\iff x=y$.

Pedro
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  • Yes, that’s what’s meant in (4) and (5). And of course in (4) it doesn’t really matter what’s meant, since constant functions are always continuous. You might add a hint for (7): What if $d_1=d_2$? – Brian M. Scott Jun 07 '13 at 21:26
  • @BrianM.Scott Aha. ${}{}{}$ The $\TeX$-ed $-$ was confusing! – Pedro Jun 07 '13 at 21:28