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I have some confusion concerning the definition of the following quotient metric (see, for example 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 $[p_1]=[x], [q_n]=[y],[q_i]=[p_{i+1}], i=1,2,\dots, n-1$.

Suppose $M = \{(x,y) \in \mathbb{R}^2 : x,y \ge 0\} \setminus \{(0,0)\}$ denotes the positive orthant of the plane, less the origin, with the usual euclidean metric.

Suppose I consider the equivalence relation $\sim$ on $M$ given by two points lying on the same ray from the origin: $$ x \sim y \iff \exists \lambda > 0 \textrm{ s.t. } \lambda x = y. $$ I think about the quotient $M /\sim$ as looking like the northeast quarter of the unit circle $S^+$, where $q(x) = \frac{x}{\|x\|}$ is the projection. But, on the other hand, the quotient pseudo-metric defined in the above quote says that if $x, y \in S^+$ (corresponding to equivalence classes $[x]$, $[y]$): $$ d_{S^+}(x,y) = 0, $$ by taking a limit of a sequence of degenerate paths where $p_1 \in [x]$ and $q_1 \in [y]$ where we let the norm of the choice of representative $p_1$ equal that of $q_1$ and let these go to zero. In other words $p_1^n = \frac{x}{n}$ and $q_1^n = \frac{y}{n}$. This yields a (very) different topology on $S^+$ than the quotient topology. Are these not supposed to be the same? Where am I going wrong?

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  • $\begingroup$ You have just confirmed what it says in the wikipedia article you mentioned: The resulting quotient "metric" is in general only a pseudometric. $\endgroup$
    – tkf
    Sep 1 '20 at 16:23
  • $\begingroup$ @tkf I'm not hung up about it being a pseudometric, but rather that the topoloy generated by this pseudometric appears to not agree with the intuitive quotient topology. $\endgroup$ Sep 1 '20 at 16:25
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The pseudometric topology on the quotient is in general only weaker than the quotient topology and not necessarily equal. Your example is one where it is strictly weaker since the pseudometric induces the trivial topology.

See also quotient topology versus pseudo metric topology (for the general argument why it is weaker) and Igor Belegradek's comment on this question on MO for a class of examples where it is strictly weaker.

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  • $\begingroup$ That would make sense. I figured it had to be something like that. Thanks a lot. $\endgroup$ Sep 1 '20 at 16:50

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