# What is the name for a topological space whose Kolmogorov quotient is discrete?

What is the name for a topological space whose Kolmogorov quotient is discrete?

Are these the almost discrete topological spaces?

• I don't know the name, but let's take a space $X$ whose Kolmogorov quotient $Y$ is discrete. It can be proved that the class of equivalence of a point $x\in X$ in $Y$ is the closure of $x$ in $X$ (let $\mbox{cl}(x)$ be the closure). Since the quotient map $\pi:X\rightarrow Y$ is continuous and $Y$ discrete $\mbox{cl}(x)=\pi^{-1}(\pi(x))$ is open, so every closed subset $C$ is open (since it contains $\mbox{cl}(x)$, which is open, of every $x\in C$). So a space having discrete Kolmogorov quotient is an almost discrete space – Alessandro Apr 10 at 17:38
• On the other hand, if $X$ is almost discrete, then $\mbox{cl}(x)$ is closed, so it's open. Since the quotient map $\pi:X\rightarrow Y$ is open, then $\pi(\mbox{cl}(x))=\mbox{class of equivalence of }x$ is open and so the Kolmogorov quotient is discrete – Alessandro Apr 10 at 17:40
• There's an error: A quotient map $\pi:X \rightarrow Y$ is not necessarily open, so the second half is wrong – Alessandro Apr 10 at 20:16

## 2 Answers

We prove that a space $$X$$ is almost discrete (every closed subspace is open) iff its Kolmogorov quotient $$K(X)$$ (the quotient space of $$X$$ by the relation $$x\equiv y\leftrightarrow\overline{x}=\overline{y}$$, where $$\overline{x}$$ is the closure of $$x$$) is discrete.

We start with a space $$X$$ with discrete Kolmogorov quotient. The equivalence class of a point $$x\in X$$ is the closure of $$x$$, that is, $$[x]=\{y\in X:\overline x=\overline y\}=\overline x$$ so, since $$K(X)$$ is discrete, $$\overline x=\pi^{-1}([x])$$ (where $$\pi$$ is the projection onto the quotient $$X\rightarrow K(X)$$) is an open set, so every closed $$C$$ contains the open neighborhood $$\overline x$$ of the generic point $$x\in C$$, that is, every closed subspace is open.

On the other hand, suppose $$X$$ is almost discrete. Since $$\overline x$$ is closed, it is open and $$\overline x=\pi^{-1}([x])$$, so $$\{[x]\}\subseteq K(X)$$ is open: The quotient topology on $$K(X)$$ is made of those subsets $$U$$ such that $$\pi^{-1}(U)$$ is open in $$X$$, so $$[x]$$, having inverse image the open set $$\overline x$$, is open and $$K(X)$$ is discrete.

• Thank you, clearly explained, I appreciate it. – fundagain Apr 10 at 20:45
• I think this is often called a partition space, as it has a base that forms a partion of $X$. – Henno Brandsma Apr 10 at 21:09

(I’m adding this primarily for my own benefit, so as to have it easily available should I want it at some point.)

Theorem: Let $$\langle X,\tau\rangle$$ be a topological space. Then the following are equivalent:

1. $$X$$ is almost discrete.
2. $$\tau$$ is a partition topology on $$X$$.
3. The Kolmogorov quotient of $$X$$ is discrete.

Proof: Suppose that $$\langle X,\tau\rangle$$ is almost discrete. For each $$x\in X$$ let $$C(x)=\operatorname{cl}\{x\}$$. Every open nbhd of $$x$$ is closed and therefore contains $$C(x)$$, and $$C(x)$$ is an open nbhd of $$x$$, so $$C(x)=\bigcap\{U\in\tau:x\in U\}$$: it is both the smallest closed set containing $$x$$ and the smallest open set containing $$x$$. If $$y\in C(x)$$, then clearly $$C(y)\subseteq C(x)$$. Moreover, $$x\in C(y)$$, as otherwise $$C(x)\setminus C(y)$$ would be an open nbhd of $$x$$ contradicting the minimality of $$C(x)$$, so $$C(x)\subseteq C(y)$$, and hence $$C(y)=C(x)$$. Thus, if $$\mathscr{C}=\{C(x):x\in X\}$$, then $$\mathscr{C}$$ is a partition of $$X$$ and clearly also a base for $$\tau$$.

Now suppose that $$\mathscr{C}$$ is a partition of $$X$$ that is a base for $$\tau$$, and for $$x\in X$$ let $$C(x)$$ be the unique member of $$\mathscr{C}$$ containing $$x$$. Clearly $$x,y\in X$$ are topologically indistinguishable iff $$C(x)=C(y)$$, and

$$\tau=\left\{\bigcup\mathscr{U}:\mathscr{U}\subseteq\mathscr{C}\right\}\,,$$

so the Kolmogorov quotient of $$X$$ is simply $$\langle\mathscr{C},\wp(\mathscr{C})\rangle$$, i.e., $$\mathscr{C}$$ with the discrete topology.

Now suppose that the Kolmogorov quotient of $$\langle X,\tau\rangle$$ is discrete, and for $$x\in X$$ let $$C(x)$$ be the equivalence class of $$x$$, i.e., the set of points of $$X$$ that are topologically indistinguishable from $$x$$, and let $$\mathscr{C}=\{C(x):x\in X\}$$. Since the Kolmogorov quotient of $$X$$ is discrete, it’s clear that

$$\tau\supseteq\left\{\bigcup\mathscr{U}:\mathscr{U}\subseteq\mathscr{C}\right\}\,.$$

If $$U\in\tau\setminus\left\{\bigcup\mathscr{U}:\mathscr{U}\subseteq\mathscr{C}\right\}$$, then there is an $$x\in X$$ such that

$$U\cap C(x)\ne\varnothing\ne C(x)\setminus U\,.$$

Let $$y\in U\cap C(x)$$ and $$z\in C(x)\setminus U$$; then $$y$$ and $$z$$ are topologically distinguished by $$U$$, which is impossible, since both are in the equivalence class $$C(x)$$. Thus,

$$\tau=\left\{\bigcup\mathscr{U}:\mathscr{U}\subseteq\mathscr{C}\right\}\,,$$

and $$X$$ is almost discrete.

• Thank you. You really put in the effort. And thanks for the partition topology – fundagain Apr 11 at 9:44
• @fundagain: You’re welcome. – Brian M. Scott Apr 11 at 22:05