Quotient (Pseudo)Metric Topology vs Quotient Topology

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?

• You have just confirmed what it says in the wikipedia article you mentioned: The resulting quotient "metric" is in general only a pseudometric.
– tkf
Sep 1 '20 at 16:23
• @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. Sep 1 '20 at 16:25