# In a $T_0$ space the union of two scattered sets is scattered

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?

• 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) Oct 7, 2020 at 21:34
• @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. Oct 8, 2020 at 2:48

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.

• Thanks Brandon for your answer and for the reference. Oct 8, 2020 at 2:43
• (your answer is basically the same as mine, but I had mine hidden to solicit more answers. thanks again) Oct 8, 2020 at 2:50

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.