# Does every finite topological space map to a family of pairwise disjoint subsets of the reals under the usual topology with closure preserved?

For a simple example, suppose $$X=\{1,2,3\}$$ under the partition topology $$\mathcal{T}=\{\varnothing,\{1\},\{2,3\},X\}.$$ The map $$\mu$$ taking $$1$$ to $$\{1\}$$, $$2$$ to $$[2,3]\cap\mathbb{Q}$$, and $$3$$ to $$[2,3]\setminus\mathbb{Q}$$ clearly satisfies $$cl(\{x\})\mapsto cl(\mu(x))$$ for each $$x\in X,$$ and since topological closure distributes over finite unions, $$cl(A)\mapsto cl(\mu(A))$$ for each $$A\subseteq X.$$

For further examples involving only connected finite spaces, see the alternative version of essentially this same question that I asked here a few years ago. The present version is much simpler and lifts the connectedness restriction on $$X.$$

Are there any theorems in general topology that ensure the existence of the map $$\mu$$ for every finite topological space $$X?$$ The assertion looks neither provable nor disprovable to me.

Yes, you can do this by induction on $$|X|$$. Let me first restrict to $$T_0$$ spaces; we will construct such a map $$\mu:X\to\mathcal{P}(\mathbb{R})$$ with the additional property that $$\mu(x)$$ is discrete for each $$x$$. If $$x,y\in X$$, we write $$x\leq y$$ for $$x\in\overline{\{y\}}$$ (the specialization order).

The base case $$|X|=0$$ is trivial. If $$|X|>0$$, pick a point $$x\in X$$ which is maximal with respect to $$\leq$$ (i.e., $$\{x\}$$ is open; here is where we use the assumption that $$X$$ is $$T_0$$) and let $$Y=X\setminus\{x\}$$. By the induction hypothesis, there exists such a $$\mu:Y\to\mathcal{P}(\mathbb{R})$$, and we just have to define $$\mu(x)$$ to extend it to $$X$$. Specifically, we need to define $$\mu(x)$$ such that its closure is $$\bigcup_{y\leq x}\mu(y)$$. Let $$S$$ be the set of maximal elements of $$\{y\in Y:y\leq x\}$$; then by the induction hypothesis, the closure of $$\bigcup_{y\in S}\mu(y)$$ is $$\bigcup_{y< x}\mu(y)$$. Also, since the elements of $$S$$ are incomparable with respect to $$\leq$$, $$\bigcup_{y\in S} \mu(y)$$ is discrete (it is a finite union of discrete sets, none of which accumulate on each other).

We assume for convenience that $$\bigcup_{y\in S} \mu(y)$$ is infinite; if it is finite the argument is only easier. Enumerate $$\bigcup_{y\in S} \mu(y)$$ as $$\{r_n\}_{n\in\mathbb{N}}$$ and pick a sequence of disjoint open intervals $$U_n$$ such that $$r_n\in U_n$$ for each $$n$$ and the lengths of the $$U_n$$ converge to $$0$$. In each $$U_n$$, pick a sequence of points (disjoint from $$\mu(y)$$ for all $$y\in Y$$) which converge to $$r_n$$, and let $$\mu(x)$$ be the union of all of these sequences. Then clearly $$\mu(x)$$ is discrete and the closure of $$\mu(x)$$ contains $$\bigcup_{y\in S} \mu(y)$$ and thus also contains $$\bigcup_{y\leq x}\mu(y)$$. On the other hand, if a sequence in $$\mu(x)$$ converges to a point other than some $$r_n$$, then there is a subsequence which consists of points in $$U_n$$ for distinct values of $$n$$, and then this sequence must converge to the limit of the corresponding $$r_n$$ since the lengths of the $$U_n$$ go to $$0$$. This limit is in the closure of $$\bigcup_{y\in S} \mu(y)$$ and thus is in $$\bigcup_{y\leq x}\mu(y)$$. Thus $$\mu(x)$$ has all the desired properties.

Now here is how you can modify the construction to handle non-$$T_0$$ spaces. First, use the construction above on the $$T_0$$ quotient $$X'$$ of $$X$$, except that you replace each point by a Cantor set. So, each $$\mu(x)$$ will be homeomorphic to a disjoint union of Cantor sets (rather than a disjoint union of points). The points $$r_n$$ will be the countably many endpoints of each of the Cantor sets making up $$\mu(y)$$ for each $$y\in S$$; note that these $$r_n$$ will no longer form a discrete set, but we can still pick disjoint open intervals $$U_n$$ such that each $$U_n$$ has $$r_n$$ as an endpoint (take an interval in the "hole" of the Cantor set at $$r_n$$). Instead of picking just a sequence in $$U_n$$ approaching each $$r_n$$ to put in $$\mu(x)$$, you pick a sequence of disjoint shrinking Cantor sets in $$U_n$$ that approach $$r_n$$.

Finally, to get a $$\mu$$ that works for $$X$$ itself rather than its $$T_0$$ quotient $$X'$$, just take each of the Cantor sets making up $$\mu(x)$$ for $$x\in X'$$ and split it as a union of finitely many dense subsets, one for each preimage of $$x$$ in $$X$$.

Lemma. Let $$A\subseteq \Bbb R$$ have no isolated points. Then there there are $$A_1,A_2$$ with

• $$A_1\cap A_2=\emptyset$$
• $$A_1\cup A_2=A$$
• $$\overline{A_1}=\overline{A_2}=\overline A$$

Proof. Let $$\mathscr Z=\{\,(U,V)\mid U\cap V=\emptyset, U\cup V\subseteq A,\overline U=\overline V\,\}.$$ Then $$\mathscr Z$$ is partially ordered by inclusion, i.e., we say $$(U,V)\preceq (U',V')$$ if $$U\subseteq U'$$ and $$V\subseteq V'$$. Let $$\mathscr C\subseteq \mathscr Z$$ be a totally ordered subset. Let $$\hat U=\bigcup_{(U,V)\in \mathscr C}U$$ and $$\hat V=\bigcup_{(U,V)\in \mathscr C}U$$. If $$x\in \hat U\cap \hat V$$, then $$x\in U$$ and $$x\in V'$$ for soem $$(U,V)(U',V')\in\mathscr C$$. But either $$U\subseteq U'$$ or $$V'\subseteq V$$ so that $$x\in U\cap V$$ or $$x\in U'\cap V'$$ - but both are impossible. Hence $$\hat U\cap \hat V=\emptyset$$. Also clearly $$\hat U\cup \hat V\subseteq A$$. If an open set $$O$$ is disjoint from $$\hat U$$, then it is disjoint from $$U$$ for every $$(U,V)\in\mathscr C$$, hence also disjoint from $$V$$ for every $$(U,V)\in\mathscr C$$, i.e., disjoint from $$\hat V$$. Together with the symmetric conclusion, we find that $$\overline{\hat U}=\overline{\hat V}$$. In summary, $$(\hat U,\hat V)\in \mathscr Z$$. By Zorn's lemma, we conclude that $$\mathscr Z$$ has a maximal element $$(A_1,A_2)$$.

Assume $$a\in A\setminus(A_1\cup A_2)$$. If $$a\in \overline{A_1}$$ then $$(A_1\cup\{a\},A_2)$$ contradicts maximality of $$(A_1,A_2)$$. Hence there is an open neighbourhood $$(a-r,a+r)$$ of $$a$$ that is disjoint from the $$\overline{A_i}$$. We construct a sequence of sets of pairwise disjoint open intervals, where each interval is in $$(a-r,a+r)$$ and centered around a point of $$A$$. We start with $$S_0=\{(a-r,a+r)$$. Given $$S_n$$ and $$(\xi-\rho,\xi+\rho)\in S_n$$, we know that $$\xi$$ is not isolated, hence can pick a sequence $$\xi_i\to \xi$$ in $$A$$ where wlog the $$\xi_i$$ are distinct and in $$(\xi-\rho,\xi+\rho)$$, and by discreteness of this sequence can pick $$\rho_i$$ such that the $$(\xi_i-\rho_i,\xi_i+\rho_i)$$ are pairwise disjoint and are in $$(\xi-\rho,\xi+\rho)$$. We pick such a sequence for each $$(\xi-\rho,\xi+\rho)\in S_n$$ and let $$S_{n+1}$$ be the set of all these intervals. By construction, the intervals in $$S_{n+1}$$ are pairwise disjoint. Now let $$B_1$$ be the set of interval midpoints of intervals in some $$S_n$$ with odd $$n$$, and similarly $$B_2$$ for even $$n$$. By construction, $$B_1\subseteq \overline{B_2}$$ and $$B_2\subseteq\overline{B_1}$$, so that $$(A_1\cup B_1, A_2\cup B_2)$$ contradicts the maximality of $$(A_1,A_2)$$.

We conclude that $$A_1\cup A_2=A$$ and then also $$\overline A=\overline{A_1}=\overline{A_2}$$. $$\square$$

Corollary. For every $$n\ge1$$, a set $$A$$ as in the lemma, there are sets $$A_1,\ldots, A_n$$ with

• $$A_i\cap A_j)=\emptyset$$ for $$i\ne j$$.
• $$\bigcup_{i=1}^nA_i=A$$
• $$\overline{A_i}=\overline A$$ for all $$i$$

Proof. Induction, where we use the lemma to split $$A_n$$ into two subsets. $$\square$$

Remark. Of course, the $$A_i$$ in the lemma as well as in the corollary are also without isolated points.

Proposition. For every finite topological space $$X$$, there exists a compact set $$C\subset \Bbb R$$ and a map $$\mu\colon X\to\mathcal P(C)$$ such that

• $$\mu(x)\cap \mu(y)=\emptyset$$ if $$x\ne y$$
• $$\bigcup_{x\in A}\mu(x)$$ is closed iff $$A$$ is closed
• $$\mu(x)$$ has no isolated points

Proof. Induction on $$|X|$$, the case $$|X|=0$$ being trivial.

Let $$A$$ be a maximal closed subset $$\ne X$$, i.e. $$A$$ is closed and the only closed subset properly containing $$A$$ is $$X$$. By induction hypothesis, there exists a compact $$C_A$$ and a map $$\mu_A\colon A\to \mathcal P(C_A)$$ as in the proposition.

For each $$x\in X\setminus A$$, we have $$\overline{\{x\}}\cup A=X$$. Let $$B=\bigcap_{x\in X\setminus A}\overline{\{x\}}$$. Then clearly $$B=\overline{\{x\}}$$ for all $$x\in X\setminus A$$.

Now pick a closed interval $$I$$ disjoint from $$C_A$$. Let $$C=C_A\cup I$$. Enumerate $$X\setminus A=\{x_1,\ldots, x_m\}$$. Use the corollary to split $$I$$ into $$m$$ sets $$I_1,\ldots, I_m$$. Likewise, for each $$x\in A$$, split $$\mu_A(x)$$ into $$m+1$$ sets $$\mu_A(x)_0,\ldots, \mu_A(x)_m$$. Now define $$\mu\colon X\to\mathcal P(C)$$ as $$\mu(x)=\begin{cases}\mu_A(x)&\text{if }x\in A\setminus B\\ \mu_A(x)_0&\text{if }x\in A\cap B\\ I_i\cup\bigcup_{x\in B\cap A}\mu_A(x)_i&\text{if } x=x_i\in X\setminus A\end{cases}$$ One verifies (straightforward, but with a couple of case distinctions) that $$\mu$$ has the desired properties. $$\square$$

• For a simpler proof of the Lemma, just enumerate the open intervals with rational endpoints that intersect $A$ and put a point of each of them into $A_1$ and $A_2$. Apr 28, 2020 at 21:49
• Your construction at the end doesn't work, though. The problem is that if $x\in A\cap B$, then the closure of $\mu_A(x)_0$ contains $\mu_A(x)_i$ for all $i>0$, but the closure of $\{x\}$ in $X$ does not intersect $X\setminus A$. (For a really simple example, think about what your inductive procedure gives if $X$ is the Sierpinski 2-point space.) Apr 28, 2020 at 21:55
• @EricWofsey You are right - it seems I only preserve the "is closed"-property, not the closure operator ... Apr 29, 2020 at 6:14