**Proof.**
Properties (3) and (5) imply that a point $x$ corresponds to an finite equivalence class $\{ u_1, \ldots , u_ n\} \subset U$ of the equivalence relation. Suppose that $x' \in X$ is a second point corresponding to the equivalence class $\{ u'_1, \ldots , u'_ m\} \subset U$. Suppose that $u_ i \leadsto u'_ j$ for some $i, j$. Then for any $r' \in R$ with $s(r') = u'_ j$ by (4) we can find $r \leadsto r'$ with $s(r) = u_ i$. Hence $t(r) \leadsto t(r')$. Since $\{ u'_1, \ldots , u'_ m\} = t(s^{-1}(\{ u'_ j\} ))$ we conclude that every element of $\{ u'_1, \ldots , u'_ m\} $ is the specialization of an element of $\{ u_1, \ldots , u_ n\} $. Thus $\overline{\{ u_1\} } \cup \ldots \cup \overline{\{ u_ n\} }$ is a union of equivalence classes, hence of the form $\pi ^{-1}(Z)$ for some subset $Z \subset X$. By (1) we see that $Z$ is closed in $X$ and in fact $Z = \overline{\{ x\} }$ because $\pi (\overline{\{ u_ i\} }) \subset \overline{\{ x\} }$ for each $i$. In other words, $x \leadsto x'$ if and only if some lift of $x$ in $U$ specializes to some lift of $x'$ in $U$, if and only if every lift of $x'$ in $U$ is a specialization of some lift of $x$ in $U$.

Suppose that both $x \leadsto x'$ and $x' \leadsto x$. Say $x$ corresponds to $\{ u_1, \ldots , u_ n\} $ and $x'$ corresponds to $\{ u'_1, \ldots , u'_ m\} $ as above. Then, by the results of the preceding paragraph, we can find a sequence

\[ \ldots \leadsto u'_{j_3} \leadsto u_{i_3} \leadsto u'_{j_2} \leadsto u_{i_2} \leadsto u'_{j_1} \leadsto u_{i_1} \]

which must repeat, hence by (2) we conclude that $\{ u_1, \ldots , u_ n\} = \{ u'_1, \ldots , u'_ m\} $, i.e., $x = x'$. Thus $X$ is Kolmogorov.
$\square$

## Comments (4)

Comment #742 by Wei Xu on

Comment #754 by Johan on

Comment #6771 by Alejandro González Nevado on

Comment #6773 by Johan on