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How to prove the following claim:

A space $X$ can be represented as the union of two disjoint subset $A$ and $B$, where $A$ is a scattered and $B$ satisfies $B=B'=:\{x: x \text{ is the accumulation point of B}\}$.

A scattered space is a space for which every not empty subset has an isolated point.

My idea: Let $A$ is the set of all isolated points, $B=X\setminus A$. However, I'm not sure $B=B'$.

Thanks for your help.

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Paul
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  • You may want to require that $B$ is either empty or not scattered at all (and allow $A$ to be empty as well). – Asaf Karagila Apr 18 '13 at 13:21
  • @Paul I've tried to edit your title to describe your question better. It is better to use more descriptive title than just How to prove the following claim? – Martin Sleziak Apr 18 '13 at 14:05

1 Answers1

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Problem 1.7.10 in Engelking's General Topology gives the following hint how to solve prove this result:

Show that if each member of a family $\mathcal A$ of subset of a space $X$ is dense in itself, then the union $\bigcup \mathcal A$ is dense in itself. Note that if $A\subset X$ is dense in itself, then the closure $\overline A$ is dense in itself. Deduce from the above that every topological space can be represented as the union of two disjoint sets, one of which is perfect and the second one is scattered.

A set is called dense in itself if $A\subseteq A'$.
perfect set = dense in itself and closed
scattered set = contains no non-empty dense in itself subset

If you simply take $A$ to be the set of all isolated points than for $X=\{0\}\cup\{1/n; n=1,2,\dots\}$ (with the usual Euclidean topology) you would bet $A=\{1/n; n=1,2,\dots\}$ and $B=\{0\}$, which is not the desired decomposition.

Martin Sleziak
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