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A scattered space if a space $X$ that contains no nonempty dense-in-itself subset. Equivalently, every nonempty subset $A$ of $X$ contains a point isolated in $A$.

Note that in general the union of two scattered sets is not scattered. For example, if $X=\{a,b\}$ with the indiscrete topology, $\{a\}$ and $\{b\}$ are both scattered, but their union, $X$, is not scattered as it has no isolated point.

It is shown in Kuratowski's book (p. 79) that in a $T_1$ space, the union of two scattered sets is scattered (he assumes spaces are $T_1$ unless mentioned otherwise). I think the following is more generally true:

Theorem: In a $T_0$ space, the union of two scattered sets is scattered.

How can this be proved?

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  • $\begingroup$ What's the idea behind Kuratowski's original proof? Knowing how to solve this for the $T_1$ case might make the $T_0$ case easier (I cannot figure out either as of yet) $\endgroup$ Oct 7, 2020 at 21:34
  • $\begingroup$ @BrandonduPreez Kuratowski deduces it from another result: In a dense-in-itself space, every scattered set is nowhere dense, and the complement of a scattered set is dense-in-itself. I have not checked if that holds in T0 spaces, finding it easier to just prove the union result directly. $\endgroup$
    – PatrickR
    Oct 8, 2020 at 2:48

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This, and a number of similar results, are proven in Scattered spaces, compactifications and an application to image classification problem by M. Al-Hajri et. al.

The proof they give is roughly as follows: Let $A$ and $B$ be scattered subspaces, and $S\subseteq A\cup B$. We show that $S$ has an isolated point. Let $S_A \ = S\cap A$ and $S_B = S\cap B$, and note that $S_A$ and $S_B$ are scattered. Since $S_A$ is scattered, there is an $a\in S_A$ and and an open set $U$ such that $\{a\} = S_A\cap U$. If $U\cap S_B = \emptyset$, we're done. So assume not. Since $U\cap S_B \subseteq S_B$, it has an isolated point. So there exists $b$ in $U\cap S_B$ and an open set $V$ such that $\{v\} = U\cap S_B \cap V$.

Thus the open set $U\cap V$ is either $\{b\}$ (in which case we're done), or $U\cap V = \{a,b\}$, in which case we use the fact that our space is $T_0$ to isolate one of $a$ or $b$, completing the proof.

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    $\begingroup$ Thanks Brandon for your answer and for the reference. $\endgroup$
    – PatrickR
    Oct 8, 2020 at 2:43
  • $\begingroup$ (your answer is basically the same as mine, but I had mine hidden to solicit more answers. thanks again) $\endgroup$
    – PatrickR
    Oct 8, 2020 at 2:50
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Suppose $X$ is $T_0$. Let $A$ and $B$ be two nonempty scattered subsets of $X$ and assume by contradiction that $A\cup B$ is not scattered. So we can find a nonempty $C\subseteq A\cup B$ with $C$ dense-in-itself.

$C$ cannot be entirely contained in $B$, otherwise it would have an isolated point. So $C\setminus B$ is a nonempty subset of $A$ (scattered) and has an isolated point $a$. There is an open nbhd $V$ of $a$ such that $V\cap(C\setminus B)=\{a\}$.

But $a$ cannot be an isolated point of $C$. So $V$ meets $C\cap B$, which must be a nonempty subset of $B$. Since $B$ is scattered, $C\cap B$ has an isolated point $b$. So we can find an open $U\subseteq V$ such that $U\cap(C\cap B)=\{b\}$.

  • Case $a\notin U$. Then $U\cap C=\{b\}$ and $C$ has an isolated point. Contradiction.
  • Case $a\in U$. Then $U\cap C=\{a,b\}$. Since $X$ is $T_0$, there is an open set containing one of the points and not the other. Intersecting further with $U$ shows that one of the two points is isolated in $C$. Again, contradiction.
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