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Recall that for a sequence $\{x_{n}\}$ in a metric space, if $\{x_{n}\} \rightarrow x$ while $x_{n} \neq x \ \ \forall n$, then $\{x_{2k}\} \rightarrow x$ and $\{x_{2k+1}\} \rightarrow x$. As a corollary, for arbitrary limit point $a$ of a countable set $A$, there exists an n-partition $\{A_{1}, ... ,A_{n} \} $of $A$ s.t. $t \in \overline{A_{i}},\forall i \in {1,...,n}$.

I wish to generalize this to arbitrary compact Hausdorff spaces and arbitrary sets (without assuming countability). More precisely, I wish to prove:

Given a compact Hausdorff space $X$, a subset $A$ of $X$, and a point $x \in \overline{A}-A$. There exists a partition $A = A_{1} \sqcup A_{2}$ such that $x \in \overline{A_{1}} \cap \overline{A_{2}}$.

Note that direct application of sequential methods fails since $X$ might not be first-countable.

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This is not true. For instance, let $X=\beta\mathbb{N}$, the Stone-Cech compactification of $\mathbb{N}$ with the discrete topology, let $A=\mathbb{N}$, and let $x$ be any point of $\beta\mathbb{N}\setminus\mathbb{N}$. Suppose $A=A_1\sqcup A_2$ is any partition. Then there is a continuous function $f:\mathbb{N}\to[0,1]$ which is $0$ on $A_1$ and $1$ on $A_2$, and this function extends continuously to $X$. If $x$ were in the closure of both $A_1$ and $A_2$, then the continuous extension would have to take both the value $0$ and the value $1$ at $x$.

In general, if $x\in\overline{A}$, the set of neighborhoods of $x$ intersected with $A$ forms a filter $F$ on $A$. If $B\subseteq A$, then $x\in\overline{B}$ iff the complement of $B$ is not in $F$. So, your question is whether there exists a set $A_1\subseteq A$ such that neither $A_1$ nor its complement is in $F$. Such an $A_1$ exists iff $F$ is not an ultrafilter.

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