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I got this question on an internal today,

Check whether

$e(x,y)$ = $d(f(x),f(y))$ for any function $f:X \rightarrow X$ is a metric on $(X,d)$.

I think that I have messed it up.

My argument was that, because the identity map is always injective therefore it should be a metric.

But apparently some of my classmates thought otherwise, so i have become a little doubtful of my argument. Could someone tell me whether i am right or not ?

johny
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  • HINT: Is it an isometry? Does the function preserve distance? Check for this. WE ARE NOT TELLING YOU THE ANSWER WITHOUT WORK. – Don Larynx Oct 17 '13 at 10:53
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    @DonLarynx: Let's assume $d(x,y)$ is known to be a metric. By definition if $f$ were an isometry, then $e(x,y)=d(f(x),f(y))$ would be the same metric. But this is not necessary to conclude $e(x,y)$ is a metric, so it may not be the best hint for checking the general case. – hardmath Oct 17 '13 at 11:03
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    The general case asks whether "any function $f$" will give a metric $e(x,y)$. Instead of checking the identity function, think about a function such that $f(x)=f(y)$ for $x\neq y$. – hardmath Oct 17 '13 at 11:27
  • @hardmath So do you mean that "f" need not necessarily be injective ? – johny Oct 17 '13 at 11:30
  • @hardmath Alright, after reading your comment again, i have realized that probably my argument is actually not correct. – johny Oct 17 '13 at 11:42
  • @hardmath How do you check it then? – Don Larynx Oct 17 '13 at 11:58

2 Answers2

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Consider $D(x,y) \to D[i(x), i(y)]$ Is this an isometry?

HINT: Where does the identity function map elements to?

Don Larynx
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  • i believe it is an isometry.The metric should be preserved under the identity map. – johny Oct 17 '13 at 11:01
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    @DonLarynx Sorry but this is not pointing the OP in the right direction. – Did Oct 17 '13 at 11:20
  • @DonLarynx My argument was that, because the identity map is always injective therefore it should be a metric.But apparently some of my classmates thought otherwise, so i have become a little doubtful of my argument. Could you tell me whether i am right or not ? – johny Oct 17 '13 at 11:20
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It is not always true that for any function $f$ from metric space $(X,d)$ to itself that:

$$ e(x,y) = d(f(x),f(y)) $$

will also be a metric. The simplest counterexample would be if $f$ is not injective, say $f(x)=f(y)$ for some $x\neq y$. Then $e(x,y) = 0$ but this contradicts the property of a metric that having zero distance implies equality of two points.

The question of which maps $f$ will give us a metric $e(x,y)$ is an interesting one. The above shows $f$ must be injective, a necessary condition. If $f$ were an isometry, then by definition $e(x,y)=d(x,y)$ gives back the same metric, a sufficient condition. In between these extremes there are other possibilities to investigate.

hardmath
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