Exercise 1, § 8
General Topology, I, 134
a) Let be a topological space. Show that the following three statements are equivalent:
(Q) If are any two distinct points of , there is a neighbourhood of which does not contain .
(Q') Every subset of which consists of a single point is closed in .
(Q'') For every the intersection of the neighbourhoods of consists of the point alone.
A space is said to be accessible if it satisfies these conditions.
b) An ordered set with the right topology (with respect to which is a Kolmogoroff space [§ 1, Exercise 2]) is accessible if and only if no two distinct elements of are comparable, and then the right topology on is the discrete topology.
Solution
No solution has been written yet.