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.


1 Answer 1


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.


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