Why are quotient metric spaces defined this way?

Question

From Wikipedia:

If $M$ is a metric space with metric $d$, and $\sim$ is an equivalence relation on $M$, then we can endow the quotient set $M/{\sim}$ with the following (pseudo)metric. Given two equivalence classes $[x]$ and $[y]$, we define $$ d'([x],[y]) = \inf\{d(p_1,q_1)+d(p_2,q_2)+\dotsb+d(p_{n},q_{n})\} $$ where the infimum is taken over all finite sequences $(p_1, p_2, \dots, p_n)$ and $(q_1, q_2, \dots, q_n)$ with $[p_1]=[x], [q_n]=[y], [q_i]=[p_{i+1}], i=1,2,\dots, n-1$. In general this will only define a pseudometric, i.e. $d'([x],[y])=0$ does not necessarily imply that $[x]=[y]$. However for nice equivalence relations (e.g., those given by gluing together polyhedra along faces), it is a metric.

1. I wonder why the quotient metric is defined that way?
2. Instead, how about using the distance between two subsets of the metric space $$ d'([x],[y]) = \inf\{d(p,q)\} $$ the infimum is taken over all $(p,q)$ such that $[p]=[x], [q]=[y]$?

Thanks and regards!

Answer 1

This is to ensure the triangle inequality. In your proposal, it can happen that $[p]$ and $[q]$ have nearby representatives and $[q]$ and $[r]$ have nearby representatives, but the two representatives of $[q]$ involved are different, so this doesn't guarantee that $[p]$ and $[r]$ have nearby representatives, so the triangle inequality may be violated.

With the Wikipedia definition, on the other hand, it's straightforward to verify the triangle inequality, since any chain of points from $[p]$ to $[q]$ and any chain of points from $[q]$ to $[r]$ can be concatenated to form a chain of points from $[p]$ to $[r]$, so the triangle inequality follows from the individual triangle inequalities. In particular, in the above situation, we can "hop" from one representative of $[q]$ to the other without extra cost.

Answer 2

Clearly, using the definition from Wikipedia, you always get a smaller or equal distance than under your proposal. If the definitions disagree and your approach is employed, the triangle inequality is violated. The definition you find in Wikipedia guarantees that the triangle inequality is valid. So I will give an example in which the triangle inequality is violated under your propsal.

Let $A=\{(0,x):x\in[0,1]\}$, $C=\{(1,x):x\in[0,1]\}$, and $B=\{(x,x):x\in(0,1)\}$. Let $X=A\cup B\cup C$ and endow it with the Euclidean metric. Partition $X$ into $A$, $B$, and $C$. One can make a path from $A$ to $C$ have lenght arbitrarily close to $0$, but $d(A,C)=1$ under your proposal.

Answer 3

Hopefully an example will show the problem:

Consider $X = \{ ±1, ±2, 1.1, 2.1 \}$ with $d(x,y)=|x-y|$ and $\sim\, = \{\{1,-1\},\{2,-2\},\{1.1,2.1\}\}$.

Then $$d([1],[2]) \leq d([1],[1.1]) + d([1.1],[2]),$$ but if we use any sort of infimum definition, then $$d([1],[1.1]) \leq |1-1.1| = 0.1$$ and $$d([1.1],[2])=d([2.1],[2]) \leq |2.1-2|=0.1,$$ so $$d([1],[2]) \leq 0.2 < |1-2|= 1$$

Using a supremum would have the odd effect that $d([1],[1])=d([1],[-1]) \geq 2$, so infimum seems natural, even if it is a little broken (only a pseudo-metric).

(Joriki's answer indicates the same example.)

Answer 4

Regarding

If $M$ is a metric space with metric $d$, and $\sim$ is an equivalence relation on $M$, then we can endow the quotient set $M/{\sim}$ with the following (pseudo)metric.

2. Instead, how about using the distance between two subsets of the metric space $$d'([x],[y]) = \inf\{d(p,q)\}$$ the infimum is taken over all $(p,q)$ such that $[p]=[x]$, $[q]=[y]$?

As said in A Course in Functional Analysis by Conway (2nd edition) Chapter 3 Section 4:

Let $X$ be a normed space, let $\mathcal{M}$ be a linear manifold in $X$, and let $Q:X\rightarrow X/\mathcal{M}$ be the natural map $Qx = x + \mathcal{M}$. We want to make $X/\mathcal{M}$ into a normed space, so define $$(4.1)\hspace{20pt}\| x + \mathcal{M} \| = \inf\{ \| x + y \| : y\in\mathcal{M}\}.$$ Note that because $\mathcal{M}$ is a linear space, $\|x + \mathcal{M}\| = \inf\{\| x - y \| : y\in\mathcal{M}\} = \text{dist}(x, \mathcal{M})$, the distance from $x$ to $\mathcal{M}$. It is left to the reader to show that (4.1) defines a seminorm on $X/\mathcal{M}$. But if $\mathcal{M}$ is not closed in $X$, (4.1) cannot define a norm. (Why?) If, however, $\mathcal{M}$ is closed, then (4.1) does define a norm.

I personally am just beginning my undergrad Functional Analysis course so I, as of right now, will not more help past this but I asked my professor regarding your second question (as I was confused since that was actually the definition we were using as we are dealing with linear manifolds, aka linear subspaces which are not necessarily closed) and this is what we discussed and what he pointed to in our textbook. I hope it helps or at least shows possible cases where your second question does hold as the previous answers showed cases where it does not.