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Let $(X,d)$ be a metric space and $\sim$ be an equivalence relation on $X$, then we can form the quotient space $X/\sim$. We can also define a pseudo metric on the set of equivalence classes as follows: 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 $p_1\sim x, q_n\sim y, q_i\sim p_{i+1}, i=1,2,\dots, n-1$. I guess this definition is the same as considering the complete graph whose vertex set is $X/\sim$, put weight $\inf\{d(z,w)\mid z\sim x, w\sim y\}$ on the edge connecting $[x]$ and $[y]$, and declare the distance between $[x]$ and $[y]$ to be the infimum of length of paths from $[x]$ to $[y]$.

It can be shown that $d'$ is a pseudo metric and the topology it induces is coarser than $X/\sim$, i.e., it contains fewer open sets. My questions are:

  1. When is $d'$ compatible with $X/\sim$? I can see they are compatible when $X/\sim$ is compact and the topology induced by $d'$ is Hausdorff, namely $d'$ is a metric, in which case the identity map from $X/\sim$ to topology induced by $d'$ is a homeomorphism. The answer to this post seems to prove that $X/\sim$ is metrizable if $X$ is compact and $X/\sim$ is Hausdorff, but I do not see whether the metric can be taken to be $d'$.

  2. Consider a collection of circles $S_n$, each of circumference $1$ with usual metric (the length of great arc connecting two points). The disjoint union $\bigsqcup_n S_n$ is metrizable by declaring points from different circles have distance $2$. Take the wedge sum $\bigvee_n S_n$. It seems to me that the pseudo metric $d'$ on $\bigvee_n S_n$ is a metric, so it cannot be compatible with the quotient topology which is not first countable. It does not seem like the Hawaiian earring either. What is it?

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  • $\begingroup$ nitpick: less -> fewer $\endgroup$ – Henno Brandsma Feb 24 '20 at 23:36

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