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?
