# Metric Space (Elementary Analysis)

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.

(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?

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

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.

• 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. – InfimumMaximum Jul 17 '13 at 3:56
• Why do u show d(x,y) = d(x′,y′) ? – InfimumMaximum Jul 17 '13 at 3:56
• @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 $d([x],[y])$. – Pedro Tamaroff Jul 17 '13 at 4:04
• @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 $d([x],[y])$. – Pedro Tamaroff Jul 17 '13 at 4:05
• 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! – InfimumMaximum Jul 17 '13 at 4:28

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 $\sim$