Exercise 2, § 1

a) Let $X$ be an ordered set. Show that the set of intervals

is a base of a topology on $X$; this topology is called the right (resp. left) topology of $X$. In the right topology, any intersection of open sets is an open set, and the closure of $\{x\}$ is the interval $] \leftarrow, x]$.

b) A topological space is said to be a Kolmogoroff space if it satisfies the following condition: given any two distinct points $x, x'$ of $X$, there is a neighbourhood of one of these points which does not contain the other. Show that an ordered set with the right topology is a Kolmogoroff space.

c) Let $X$ be a Kolmogoroff space in which every intersection of open sets is an open set. Show that $x \in \{x'\}$ is an order relation between $x$ and $x'$ in $X$, and that if this relation is written as $x \leqslant x'$ then the given topology on $X$ is identical with the right topology determined by this ordering.

d) Deduce from c) that if $X$ is a Kolmogoroff space then every finite non-empty subset of $X$ has at least one isolated point. Hence, if $X$ has no isolated point, every non-empty open subset of $X$ is infinite.