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.
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.
