Let $X$ be a compact Hausdorff space (I don't mind assuming it's metrizable).
Let $A\_i$ $i\in \mathbb{N}$ be a collection of disjoint closed subsets of $X$.


My question: Does there exist a Hausdorff quotient $Y$ such that the quotient $\phi: X \to Y$ satisfies:


a)$\phi(A\_i)$ is a single point for each $i$.


b)For distinct $i$ and $j$, $\phi(A\_i)\neq \phi(A\_j)$.


Remark: The "obvious" equivalence relation obtained by collapsing each of the $A\_i$ does not necessarily give a Hausdorff quotient, since a sequence of points $a\_i \in A\_i$ can accumulate to a point outside the union of the $A\_i$. Also the closure of this relation does not appear to be transitive. 



Hausdorff quotient collapsing and separating a prescribed collection of disjoint closed subsets
No. Consider $Y=\{0,n^{-1}\mid n\in \mathbf N\setminus \{0\}\}$, $X=Y\times \{-1,1\}$ and $A\_n=\{(n-1)^{-1}\}\times \{-1,1\}$ for $n>1$, $A\_0=\{(0,-1)\}$ and $A\_1=\{(0,1)\}$.


Since $X=\bigcup\_n A\_n$, there is only one quotient such that images of $A\_n$ are disjoint singletons, namely a convergent sequence with two limits. This quotient is not Hausdorff.


