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Let X be a set and (M, d) be a metric space. Let f : X → M be an injective map. Then we define f∗d := d ◦ (f × f): X × X →(o,infinity]

i.e. f∗d(x, y) = d(f(x), f(y)) for x, y ∈ X .

I want to prove that f∗d is a distance function on X (“induced metric”). here's what ive tried so far ...

taking all of the above as fact ;

f is injective therefore f(x)=m1 and f(y)=m2 where m1 does not equal m2 and m1,m2 are elements of M.

=> fd(x,y)=d(m1,m2) which is defined already as a metric space X C(M,d) therefore fd is also a distance function.

...I feel like thats too easy and i must be missing something obvious ? any tips would be much appreciated but please dont answer the full question as i want to learn this on my own :)

excalibirr
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  • how about the triangle inequality? – Will Jagy Oct 17 '17 at 00:18
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    Also, if you were given a npn-injective $f,$ why would the result not be a metric space? – Will Jagy Oct 17 '17 at 00:24
  • I know that any distance function has to satisfy the triangle inequality but is it enough to say as i have that as fd(x,y)=d(m1,m2) and d(m1,m2) already is a metric space that as d(m1,m2) satisfies this by definition that fd(x,y) must also satisfy it ? – excalibirr Oct 17 '17 at 00:26
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    I'm sorry, but it seems you proved nothing. Try to call your induced metric something like $g$ where $g:X\times X\longrightarrow\mathbb{R}$ and prove the following three properties $g(x,y)=0$ iff $x=y$, $g(x,y)=g(y,x)$ and $g(x,z)\leq g(x,y)+g(y,z)$ for all $x,y,z\in X$. Actually this is a very simple exercises, so I suppose it is intended as a training for proving and formal thinking. Good luck :) P.S. Being $f$ an injection, you can see $X$ somewhat "inside" $M$ (try to think at $X':=im(f)$) so actually it can be seen as a subspace of $M$ (and the metric $d$ naturally restrict to $X'$) – Tancredi Oct 17 '17 at 00:27
  • thank you Tancredi this is what i wanted to know. I just wasnt sure if we had to go through the 3 properties of a distance function, or if there was some other relation between the metric and the induced metric that we were supposed to use to prove this :) – excalibirr Oct 17 '17 at 00:30
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    @jamiemma1995 note that $d(m_1,m_2)$ is not a metric, it is just a real (positive) number. $d$ is a metric. It is worth noting it is a function – Tancredi Oct 17 '17 at 00:30

1 Answers1

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$f^*d(x,y)=d(m_1,m_2)$ which is defined already as a metric space ... therefore $f^*d$ is also a distance function.

This sentence doesn't make any sense. To what is the “which” referring to? The only possible antecedent in the sentence is the number $f^*d(x,y)=d(m_1,m_2)$. A number isn't defined as a metric space.

If you're asserting that the facts that $d$ is a metric and $f^*d$ is defined in terms of $d$, make $f^*d$ a metric on $X$—well, that's exactly what you need to prove.

Go back to the definitions of metric:

  1. You must show $f^*d(x,y) \geq 0$ for all $x,y \in X$, and if $f^*d(x,y) = 0$, then $x=y$.
  2. You must show $f^*d(x,y) = f^*d(y,x)$ for all $x,y \in X$.
  3. You must show the triangle inequality: For all $x$, $y$, $z\in X$, $$ f^*d(x,y) + f^*d(y,z) \geq f^*d(x,z) $$

Each of those statements can be expressed in terms of $f$ and $d$. You'll know you're doing it right when you find that you need to use each of the requirements that must be satisfied for $d$ to be a metric, and the assumption that $f$ is injective.....

excalibirr
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Matthew Leingang
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  • Thank you for your detailed answer. It was very helpful. I still havent quite the hang of metric spaces, well at all really.At least you guys are here to help me if i make any glaringly silly mistakes. hopefully that will happen less and less though :) – excalibirr Oct 17 '17 at 00:36