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.


2 Answers 2


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. $\square$

  • 1
    $\begingroup$ 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 $A_2$. $\endgroup$ Apr 28, 2020 at 21:49
  • 1
    $\begingroup$ 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.) $\endgroup$ Apr 28, 2020 at 21:55
  • $\begingroup$ @EricWofsey You are right - it seems I only preserve the "is closed"-property, not the closure operator ... $\endgroup$ Apr 29, 2020 at 6:14

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