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

Are these the almost discrete topological spaces?

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    $\begingroup$ 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 $\endgroup$ – Alessandro Apr 10 at 17:38
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    $\begingroup$ 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 $\endgroup$ – Alessandro Apr 10 at 17:40
  • $\begingroup$ There's an error: A quotient map $\pi:X \rightarrow Y$ is not necessarily open, so the second half is wrong $\endgroup$ – Alessandro Apr 10 at 20:16
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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.

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

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

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