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I was hoping that somebody can explain to me the definition of quotient metric spaces

I got the following definition 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 $[p_1]=[x], [q_n]=[y],[q_i]=[p_{i+1}], i=1,2,\dots, n-1$.

From another discussion on this website I understand that we use this definition, instead of simply the infimum over d(p,q) for all possible combinations for p and q, to guarantee the triangle inequality. But it is not entirely clear to me how to (geometrically) interpret this definition and how to actually compute distances with it.

I tried to work with the following example:

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

and compute the distance between -1 and 1 and also the distance between -1 and 1.1.

Could somebody please be so kind to give me a step by step walk-through on how to use the definition and compute the distances for these two examples.

Thanks!

Gijs Dubbelman

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I think it might aid in understanding (especially given the error underlying your examples) if you didn't use the same letter to denote the two metrics -- you could denote the quotient metric by $d_\sim([x],[y])$, like I did in my answer. –  joriki Oct 31 '12 at 21:53
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1 Answer

This is one of my favourite definitions. You can think of the equivalence classes as networks of teleporters. You can enter any teleporter in a given network (equivalence class) and jump to any other teleporter in the same network (equivalence class), and it doesn't take you any time/distance. All you have to pay for is the distance you cover by foot. The infimum is taken over all possible sequences of teleportations. In this way, points in the same equivalence class become a single point, and you can freely choose which of its incarnations to enter and which one to leave.

Your examples are fundamentally flawed in that you're asking for the distance between points of $X$, but these aren't points of $X/\sim$, so you can't compute their distance in the quotient metric $d_\sim$. You can ask what the distance from $[-1]$ to $[1]$ is, and the answer is $0$, since these are the same points (equivalence classes) of $X/\sim$. For the second one, you can ask for the distance from $[-1]$ to $[1.1]$. To find this, enter the teleporter at $-1$, jump to $1$ for free, and walk to $1.1$ by foot, for a total distance $d_\sim([-1],[1.1])=d_\sim([1],[1.1])=d(1,1.1)=0.1$.

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ok thanks joriki that was already very useful. but in this case do we then not have that d([1],[2]) = 1 but d([1],[1.1])+d([2.1],[2]) = 0.2, i.e. are we in violation of the triangle inequality. what am i doing wrong? –  Gijs Dubbelmanm Oct 31 '12 at 21:54
    
@Gijs: You're welcome! No, for $d([1],[2])$, the optimal teleportation sequence is to walk to $1.1$, then teleport to $2.1$, then walk to $2$ for a total walking distance of $0.2$. In fact this is a good example for understanding how the definition guarantees the triangle inequality. –  joriki Oct 31 '12 at 21:57
    
aha I see! Now it becomes clear to me. The definitions allows us to use [1.1] when computing the distance between [1] and [2]. This would not be allowed when using the definition of the infimum over d(p,q) for all possible combinations for p and q. Thanks again! –  Gijs Dubbelmanm Oct 31 '12 at 22:02
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