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In any metric space, $(X,D)$. Let $f$ be self-mapping function that is one-to-one. Set $D(x,y)=D(f(x),f(y))$. Prove that $D$ is a metric on $X$

I understand that $D(x,y)=D(f(x),f(y)) \implies |y-x|=|f(x)-f(y)|$ but since we are dealing with a self mapping function, $f:X \rightarrow X$. Doesnt reflexivity, symmetry, and transitivity follow as a trivial consequence? Cleary, $f(x)=f(y) \implies x=y$ and this is true $\forall x \in X$.

Any help and hints(no solutions) would be appreciated.

Cody S
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1 Answers1

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What do the axioms of an equivalence relation have to do with it? Do you mean the axioms here: https://en.wikipedia.org/wiki/Metric_%28mathematics%29

Maybe as a conceptual hint, you can think about the more general situation where $Y$ is a injection into a metric space $(X,D)$. Then you can try to put a metric on $Y$ by thinking of it as a subset of $X$.

Elle Najt
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  • Yes, I thought that when proving D is a metric; You would have to show that that it satisfies the equivalence relations. Is this not the case here? – Cody S Feb 21 '16 at 19:29
  • @John A metric is not an equivalence relation. It is a function $X \times X \to [0, \infty)$. – Elle Najt Feb 21 '16 at 19:30