4
$\begingroup$

Let $d: X \times X \to \Bbb R$ is a function satisfying all properties of a metric space but $d(x,y)=0 \implies x = y$.

If we define $\sim$ on $X$ by $x\sim y \iff d(x,y) = 0$,

prove that $D([x], [y]) = d(x,y)$ where $[x] = \{z \in X \mid z\sim x\}$ is well-defined on equivalence classes and makes the set of equivalence classes into a metric space.

Some helps please!!

(I showed that $\sim$ is equivalence relation)

Thank you!

+++

I followed your advice and just solved this problem. Could you tell me if there is any weakness or error in my proof?

enter image description here

enter image description here

$\endgroup$
  • 4
    $\begingroup$ You may at some point find it useful to know that a function like $d$ is called a pseudometric. $\endgroup$ – Brian M. Scott Jul 14 '13 at 19:17
3
$\begingroup$

In essence you're identifying all the "indistiguishable" points, that is, pairs such that $x\neq y$ yet $d(x,y)=0$ into one point $\bar x=\{y:d(x,y)=0\}$.

You ought to prove two things:

$(1)$ The new metric $d(\bar x,\bar y):=d(x,y)$ where we choose $x\in\bar x,y\in \bar y$ is "well-defined", meaning that the output does not depend on the representative we take in $\bar x,\bar y $ to feed into $d(x,y)$. Thus, prove that $x\sim x'$ and $y\sim y'\implies d(x,y)=d(x',y')$.

$(2)$ This alleged metric is indeed one.

Hint Assume $d(x,x')=d(y,y')=0$. $$\begin{align}d(x,y)&\leq d(x,x')+d(x',y')+d(y',y)\\d(x,y)&\leq \;\;\;\;0\;\;\;\;+\;\;\;\;0\;\;\;\;\;+d(y',y)\\d(x,y)&\leq d(x',y')\end{align}$$

It remains to show under the same assumption that $d(x,y)\geq d(x',y')$.


When working with say the space of all square Lebesgue integrable functions over some interval, $\mathscr L^2(I)$, one usually uses the above. Concretely, one defines $f\simeq g\iff f=g \text{ a.e. on } I$ to work with a metric space instead of a semi-metric space.

$\endgroup$
  • $\begingroup$ I can't understand (2) part. As Cameron said below, I think the only thing left to show is that D([x],[y])=0 implies that [x]=[y], and this follows from the definition of D. $\endgroup$ – InfimumMaximum Jul 17 '13 at 3:56
  • $\begingroup$ Why do u show d(x,y) = d(x′,y′) ? $\endgroup$ – InfimumMaximum Jul 17 '13 at 3:56
  • $\begingroup$ @InfimumMaximum Because we want to show the new distance function is well defined, that is, it does not depend on the choice of $a\in [x], b\in[y]$ we pick to compute $d([x],[y])$. $\endgroup$ – Pedro Tamaroff Jul 17 '13 at 4:04
  • 1
    $\begingroup$ @InfimumMaximum Because we want to show the new distance function is well defined, that is, it does not depend on the choice of $a\in [x], b\in[y]$ we pick to compute $d([x],[y])$. $\endgroup$ – Pedro Tamaroff Jul 17 '13 at 4:05
  • $\begingroup$ Thank you! I followed your advice and just solved this problem. I uploaded my solution images. Could you tell me if there is any weakness or error in my proof please? Thank you! $\endgroup$ – InfimumMaximum Jul 17 '13 at 4:28
2
$\begingroup$

You're well on your way! You've shown that $\sim$ is an equivalence relation. Now, you still need to prove that $D$ is well-defined (that our choice of representatives from the equivalence classes doesn't matter).

By definition of $D$, it is straightforward to prove that $D$ has all the metric properties that $d$ does, so the only thing left to show is that $D([x],[y])=0$ implies that $[x]=[y],$ but again this follows fairly directly from the definition of $D$ and $\sim$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.