quotient topology versus pseudo metric topology I'm a little confused about something. Let's say M is a compact metric space, and we endow it with an equivalence relation ~. the, the quotient space M/~ is compact in the quotient topology. What about the quotient pseudo metric and the topology M/~ inherits from that? would M/~ still be compact in that topology? Also, is the open ball in the quotient pseudo metric topology also open in the quotient topology?
Thank you for any help.
 A: The topology inherited from the quotient pseudometric is contained in the quotient topology. To see this,  let $\epsilon > 0$ and $q(x)$ be a point in the quotient set. The $\epsilon$-ball about $q(x)$ will be all the $q(y)$ so that there is a chain of points $$x = x_1, y_1 \sim x_2, y_2 \sim x_3, \ldots, y_{n-1} \sim x_n, y_n = y $$ such that $$\sum_i d(x_i,y_i) < \epsilon.$$ Now, the set of $y$ that have such a chain is an open set in $M$, since if $\epsilon_0$ is the value of that sum and $d(y,y') < \epsilon - \epsilon_0$, then we may extend our chain by setting $x_{n+1} = y, y_{n+1}  = y'$ to get a chain from $x$ to $y'$ such that $$\sum_{i=1}^{n+1} d(x_i,y_i) = \epsilon_0 + d(y,y') < \epsilon.$$ So $B_\epsilon(q(x))$ has open inverse under $q$, and so is open in the quotient topology.
Now, this implies that the identity map $$(M/\sim, \mbox{quotient}) \rightarrow (M/\sim, \mbox{pseudometric})$$ is a continuous function. The continuous image of a compact topological space is compact, and the quotient topology is compact, so the pseudometric topology is compact.
